Vectors Part Trois
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Vectors Part Trois
Vector Geometry
OP a
OR b
RQ
Consider this parallelogramQ
O
P
Ra
b
PQ
Opposite sides are Parallel
OQ OP PQ
OQ OR RQ
OQ is known as the resultant of a and b
a+b
b+a
a+b b+a
Resultant of Two Vectors
Is the same no matter which route is followed
Use this to find vectors in geometrical figures
Example
Q
O
P
Ra
b
SS is the Midpoint of PQ
Work out the vector OS
frac12OS OP PQ
= a + frac12b
Alternatively
Q
O
P
Ra
b
SS is the Midpoint of PQ
Work out the vector OS
OS OR RQ QS
= a + frac12b
= b + a - frac12b
= frac12b + a
Example
AB
C
p
q
M M is the Midpoint of BC
Find BC
AC= p AB = q
BC BA AC= += -q + p
= p - q
Example
AB
C
p
q
M M is the Midpoint of BC
Find BM
AC= p AB = q
BM frac12BC=
= frac12(p ndash q)
Example
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= q + frac12(p ndash q)
AM + frac12BC= AB
= q +frac12p - frac12q
= frac12q +frac12p = frac12(q + p) = frac12(p + q)
Alternatively
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= p + frac12(q ndash p)
AM + frac12CB= AC
= p +frac12q - frac12p
= frac12p +frac12q = frac12(p + q)
Why We represent objects using
mainly linear primitives points lines segments planes polygons
Need to know how to compute distances transformationshellip
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Vector Geometry
OP a
OR b
RQ
Consider this parallelogramQ
O
P
Ra
b
PQ
Opposite sides are Parallel
OQ OP PQ
OQ OR RQ
OQ is known as the resultant of a and b
a+b
b+a
a+b b+a
Resultant of Two Vectors
Is the same no matter which route is followed
Use this to find vectors in geometrical figures
Example
Q
O
P
Ra
b
SS is the Midpoint of PQ
Work out the vector OS
frac12OS OP PQ
= a + frac12b
Alternatively
Q
O
P
Ra
b
SS is the Midpoint of PQ
Work out the vector OS
OS OR RQ QS
= a + frac12b
= b + a - frac12b
= frac12b + a
Example
AB
C
p
q
M M is the Midpoint of BC
Find BC
AC= p AB = q
BC BA AC= += -q + p
= p - q
Example
AB
C
p
q
M M is the Midpoint of BC
Find BM
AC= p AB = q
BM frac12BC=
= frac12(p ndash q)
Example
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= q + frac12(p ndash q)
AM + frac12BC= AB
= q +frac12p - frac12q
= frac12q +frac12p = frac12(q + p) = frac12(p + q)
Alternatively
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= p + frac12(q ndash p)
AM + frac12CB= AC
= p +frac12q - frac12p
= frac12p +frac12q = frac12(p + q)
Why We represent objects using
mainly linear primitives points lines segments planes polygons
Need to know how to compute distances transformationshellip
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Resultant of Two Vectors
Is the same no matter which route is followed
Use this to find vectors in geometrical figures
Example
Q
O
P
Ra
b
SS is the Midpoint of PQ
Work out the vector OS
frac12OS OP PQ
= a + frac12b
Alternatively
Q
O
P
Ra
b
SS is the Midpoint of PQ
Work out the vector OS
OS OR RQ QS
= a + frac12b
= b + a - frac12b
= frac12b + a
Example
AB
C
p
q
M M is the Midpoint of BC
Find BC
AC= p AB = q
BC BA AC= += -q + p
= p - q
Example
AB
C
p
q
M M is the Midpoint of BC
Find BM
AC= p AB = q
BM frac12BC=
= frac12(p ndash q)
Example
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= q + frac12(p ndash q)
AM + frac12BC= AB
= q +frac12p - frac12q
= frac12q +frac12p = frac12(q + p) = frac12(p + q)
Alternatively
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= p + frac12(q ndash p)
AM + frac12CB= AC
= p +frac12q - frac12p
= frac12p +frac12q = frac12(p + q)
Why We represent objects using
mainly linear primitives points lines segments planes polygons
Need to know how to compute distances transformationshellip
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Example
Q
O
P
Ra
b
SS is the Midpoint of PQ
Work out the vector OS
frac12OS OP PQ
= a + frac12b
Alternatively
Q
O
P
Ra
b
SS is the Midpoint of PQ
Work out the vector OS
OS OR RQ QS
= a + frac12b
= b + a - frac12b
= frac12b + a
Example
AB
C
p
q
M M is the Midpoint of BC
Find BC
AC= p AB = q
BC BA AC= += -q + p
= p - q
Example
AB
C
p
q
M M is the Midpoint of BC
Find BM
AC= p AB = q
BM frac12BC=
= frac12(p ndash q)
Example
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= q + frac12(p ndash q)
AM + frac12BC= AB
= q +frac12p - frac12q
= frac12q +frac12p = frac12(q + p) = frac12(p + q)
Alternatively
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= p + frac12(q ndash p)
AM + frac12CB= AC
= p +frac12q - frac12p
= frac12p +frac12q = frac12(p + q)
Why We represent objects using
mainly linear primitives points lines segments planes polygons
Need to know how to compute distances transformationshellip
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Alternatively
Q
O
P
Ra
b
SS is the Midpoint of PQ
Work out the vector OS
OS OR RQ QS
= a + frac12b
= b + a - frac12b
= frac12b + a
Example
AB
C
p
q
M M is the Midpoint of BC
Find BC
AC= p AB = q
BC BA AC= += -q + p
= p - q
Example
AB
C
p
q
M M is the Midpoint of BC
Find BM
AC= p AB = q
BM frac12BC=
= frac12(p ndash q)
Example
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= q + frac12(p ndash q)
AM + frac12BC= AB
= q +frac12p - frac12q
= frac12q +frac12p = frac12(q + p) = frac12(p + q)
Alternatively
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= p + frac12(q ndash p)
AM + frac12CB= AC
= p +frac12q - frac12p
= frac12p +frac12q = frac12(p + q)
Why We represent objects using
mainly linear primitives points lines segments planes polygons
Need to know how to compute distances transformationshellip
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Example
AB
C
p
q
M M is the Midpoint of BC
Find BC
AC= p AB = q
BC BA AC= += -q + p
= p - q
Example
AB
C
p
q
M M is the Midpoint of BC
Find BM
AC= p AB = q
BM frac12BC=
= frac12(p ndash q)
Example
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= q + frac12(p ndash q)
AM + frac12BC= AB
= q +frac12p - frac12q
= frac12q +frac12p = frac12(q + p) = frac12(p + q)
Alternatively
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= p + frac12(q ndash p)
AM + frac12CB= AC
= p +frac12q - frac12p
= frac12p +frac12q = frac12(p + q)
Why We represent objects using
mainly linear primitives points lines segments planes polygons
Need to know how to compute distances transformationshellip
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Example
AB
C
p
q
M M is the Midpoint of BC
Find BM
AC= p AB = q
BM frac12BC=
= frac12(p ndash q)
Example
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= q + frac12(p ndash q)
AM + frac12BC= AB
= q +frac12p - frac12q
= frac12q +frac12p = frac12(q + p) = frac12(p + q)
Alternatively
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= p + frac12(q ndash p)
AM + frac12CB= AC
= p +frac12q - frac12p
= frac12p +frac12q = frac12(p + q)
Why We represent objects using
mainly linear primitives points lines segments planes polygons
Need to know how to compute distances transformationshellip
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Example
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= q + frac12(p ndash q)
AM + frac12BC= AB
= q +frac12p - frac12q
= frac12q +frac12p = frac12(q + p) = frac12(p + q)
Alternatively
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= p + frac12(q ndash p)
AM + frac12CB= AC
= p +frac12q - frac12p
= frac12p +frac12q = frac12(p + q)
Why We represent objects using
mainly linear primitives points lines segments planes polygons
Need to know how to compute distances transformationshellip
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Alternatively
AB
C
p
q
M M is the Midpoint of BC
Find AM
AC= p AB = q
= p + frac12(q ndash p)
AM + frac12CB= AC
= p +frac12q - frac12p
= frac12p +frac12q = frac12(p + q)
Why We represent objects using
mainly linear primitives points lines segments planes polygons
Need to know how to compute distances transformationshellip
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Why We represent objects using
mainly linear primitives points lines segments planes polygons
Need to know how to compute distances transformationshellip
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Basic definitions
Points specify location in space (or in the plane)
Vectors have magnitude and direction (like velocity)
Points Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
vector + vector = vector
Parallelogram rule
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
point - point = vector
A
BB ndash A
A
BA ndash B
Final - Initial
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
point + point not defined
Makes no sense
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Map points to vectors
If we have a coordinate system with
origin at point OWe can define correspondence between
points and vectors
vOv
OPPP
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Position VectorsA position vector for a point X (with respect to the origin 0) is the fixed vector OX
O
XOX
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Components of a Vector Between Two Points
Given and then 1 1 1P x y 2 2 2P x y 1 2 2 1 2 1PP x x y y
Given and then 1 1 1 1 P x y z 2 2 2 2 P x y z 1 2 2 1 2 1 2 1 PP x x y y z z
Proof
1 2 1 2
2 1
2 2 2 1 1 1
2 1 2 1 2 1
PP PO OP
OP OP
x y z x y z
x x y y z z
1P
2PO
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding 1 2 1 2 2 1If 2 34 and 5 26 then find and P P PP P P
1 2 5 26 2 34
712
PP
2 1 2 34 5 26
7 1 2
P P
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
If and find 2 3 3 5 5 4 3u i j k v i j k u v
2 2 2
4 3 4 2 3 3 3 5 5
4 23 1 3 3 55
812 4 9 1515
1727 19
17 27 19
1379
u v i j k i j k
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
1 2 3If a nd prove that u u u u k R ku k u
1 2 3 ku k u u u
2 2 2 21 2 3k u u u
2 2 2 21 2 3k u u u
k u
1 2 3 ku ku ku
2 2 2
1 2 3ku ku ku
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
Determine the coordinates of the point that divides AB in the ratio 53 where A is (2 -1 4) and B is (3 1 7)
n mOP OA OB
m n m n
If APPB = mn then
3 52 14 317
5 3 5 33 5
2 14 3178 8
21 2 47
8 8 8
OP
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
P Q R
0
3 5
3
8PQ PR
OQ OP PQ
3
8OQ OP PR
3
8OQ OP PO OR
3
8OQ OP OP OR
3 3
8 8OQ OP OP OR
5 3
8 8OQ OP OR
Since
Then
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
UnderstandingConsider collinear points P Q and R consider also a reference point O Write OQ as a linear combination of OP and OR if
a) Q divides PR (internally) in the ratio 35b) Q divides PR (externally) in the ratio -27
Q P R
0
-2 5
2
5PQ PR
OQ OP PQ
2
5OQ OP PR
2
5OQ OP PO OR
2
5OQ OP OP OR
2 2
5 5OQ OP OP OR
7 2
5 5OQ OP OR
Since
Then
7
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
Determine the point Q that divides line segment PR in the ratio 58 if P is (152) and R is (-412)
152 412
8 20 40 5 16 10
13 13 13 13 13 13
12 45 26
13 13
1
1
8
13 3
3
5OQ
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Inner (dot) product
Defined for vectors | || | cosθv w v w
Lv
w
| |v w L v
wcos(θ) is the scalar projection of w on vector v which we will call L
the dot product can be understood geometrically as the product of the length of this projection and the length of v
The dot product or scalar product allows us to determine a product the magnitude of one vector by the magnitude of the component of the other vector thatrsquos parallel to the first (The dot product was invented specifically for this purpose)
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Dot product in coordinates( )
( )
v v
w w
v w v w
v x y
w x y
v w x x y y
v
w
xv
yv
xw
yw
x
y
O
22
2 2
2 2
2
2 cos
w c v
c v w
c v w
v v w w
v w v w
An interesting proof
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Proof of Geometric and Coordinate Definitions Equality
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
u u i u j
u u
u u
22 2 2 21 2
2 2 2 2 2 21 2
2 21 2
ˆ ˆ
1 0 0 1
v v i v j
v v
v v
2 2 2
1 1 2 2
2 2 2 21 1 1 1 2 2 2 2
2 2 2 21 2 1 2 1 1 2 2
2 2
1 1 2 2
2 2
2
2
c v u
c v u v u
v v u u v v u u
v v u u v u v u
u v v u v u
22 2
2 cosc u v u v
From the Cosine Law
therefore
1 1 2 2 cosu v v u u v
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Vector Properties
u v v u
2 u u u
ku v k u v u kv
u v w u v u w
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
Given and and 8 find1 47 6 12 u vu v
cos 126u v u v
81 47 cos 126u v
224u v
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
Prove that for any vector u 2
u u u
Proof
cos 0u u u u
21u
2u
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understandinga) If and what values can take4 2 u v u v
b) Prove the Cauchy-Schwarz Inequality u v u v
cosu v u v
4 2 cosu v
8cosu v
Cosine has the range of -1 to 18 8u v
cosu v u v
cosu v u v
cosu v u v
u v u v
1
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
Given that work is defined as (the dot product of force f and distance travelled s A crate on a ramp is hauled 8 m up the ramp under a constant force of 20 N applied at an angle of 300 to the ramp Determine the work done
W f s
30
f
s
W f s
cosW f s
20 8 cos 30W N m
140W J
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
The angle between two vectors u and v is 1100 if the magnitude of vector u is 12 then determine
vproj u
The projection of vector u onto vector v
12cos 110
41vproj u
110
u
v
vproj u
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
The Cross Product
Some physical concepts (torque angular momentum magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that is perpendicular to the first (recall in the dot product we used the parallel component)
The cross product was invented for this specific purposeB
Bsin(θ)
θ
A
A B
sinA B AB
The magnitude (length) of the cross product
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
The Cross ProductUnlike the dot product the cross product is a vector quantity
For vectors and we define the cross product of and asx y x y
ˆsinx y x y n
Where θ is the angle between and and is a unit vector perpendicular to both and such that and form a right handed triangle
x
x
xy
y
yn n
Take your right hand and point all 4 fingers (not the thumb) in the direction of the first vector x Next rotate your arm or wrist so that you can curl your fingers in the direction of y Then extend your thumb which is perpendicular to both x and y The thumbrsquos direction is the direction the cross product will point
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
UnderstandingIf and the angle between the vectors is 30 determine
)
)
8 5
a u v
v ub
u v
ˆsin
ˆ8 5 sin 30
ˆ20
u v u v n
n
n
ˆsin
ˆ5 8 sin 30
ˆ20
v u v u n
n
n
30
u
v
30
u
v
30
u
vInto board
Out off board
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
Prove that for non-zero vectors and
if and only 0 if and are colline ar
u v
u v u v
First suppose u and v are collinear then the angle θ between u and v is 00 Since sin(00)=0 then ˆsin 0
0
u v u v n
Conversely suppose that then Since u and v are non-zero vectors then sin(θ)=0 and hence θ=0 or θ=π
0u v ˆsin 0 0u v n
Therefore u and v are collinear
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Properties
1
2
3
u v w u v u w
u v w u w v w
ku v k u v u kv
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
UnderstandingDetermine the area of the parallelogram below using the cross product
8
13
300
sin
8 13 sin 30
52
area u v
u v
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
)a u v w
)b u v w
)c u v w
)d u v w
)e u v u w
)f u v u w
)g u v u w
State whether each expression has meaning If not explain why if so state whether it is a vector or a scalar quantity
scalar
Non sense
Non sense
Non sense
vector
vector
scalar
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
UnderstandingA right-threaded bolt is tightened by applying a 50 N force to a 020 m wrench as shown in the diagram Find the moment of the force about the centre of the bolt
M r F
ˆsin
ˆ50 020 sin 110
ˆ94
M r F
r F n
n
n
The moment of the bolt is 94 Nm and is directed down
70
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Vector Product in Component Form
Recall ˆsinu v u v n
But how do we obtain the answer in component form
The easiest technique is called the Sarrusrsquo Scheme x y z x y zA a a a B b b b
ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
x y z x y z
i j i jk kA B a a a a a a
b b b b b b
ˆy z z ya b a b i ˆ
z x x za b a b j ˆx y y xa b a b k
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
6 13 254Find
ˆ ˆ ˆ ˆ ˆ ˆ
6 1 3 6 1 3
2 5 4 2 5 4
i j i jk kA B
ˆ ˆ ˆ ˆ ˆ
6 3
ˆ
1
4 2 5 45
6 1 3
2
i j i jk kA B
1 4 3 5 i
ˆ ˆ ˆ ˆ ˆ
6
ˆ
3 6
4 5 42
1 1 3
2 5
i i jk kA B
j
3 2 6 4 j
ˆ ˆ ˆ ˆ ˆ
6 1 3
ˆ
6 3
2 5 4 4
1
2 5
i j k i j kA B
6 5 1 ˆ2 k
19 3028
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
AD
B C
a
3 1 5 321
Area AB AD
Area AB AD
20 4 111 3 1 5AB
432 111 321AD
1 1 5 2 5 3 3 1 3 2 9 12 31 3
ˆ ˆ ˆ ˆ ˆ ˆ
3 1 5 3 1 5
3 2 1 3 2 1
i j i jk kA B
2 22
9 12 3
9 12 3
234
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding The points 111 20 4 12 3 and 432 are vertices of a parallelogram
Find the area of each of the following
a)
b) triangle
A B C D
ABCD
ABC
b Area of triangle is one-half the area of the parallelogram
1234
2Area
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding If 1 21 21 1 and 1 13 then evaluate u v w u v w
ˆ ˆ ˆ ˆ ˆ
1 2 1 1
ˆ
2 1
2 1 1 2 1 1
j i jk kiu v
2 1 1 1 1 2 1 1 1 1 2 2
135
u v
ˆ ˆ ˆ ˆ ˆ
1 3 5 1 3
1 1 3 1 1 3
ˆ
5j i jk k
A wi
3 3 5 1 5 1 1 3 1 1 3 1
14 82
A w
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Understanding
Find a unit vector that is perpendicular to both 123
23 5
u
v
We need to apply the cross product to determine a vector perpendicular to both u and v Then we divide by the length of this vector to make it a unit vector
ˆa b
na b
2 2 2
1 23 23 5ˆ
1 23 23 5
19 1
1
7
19 1
9 1 7
411 411 41
7
1
n
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
Useful Properties
0i i
0j j
0k k
i j k
i k j
j k i
j i k
k i j
k j i
u v v u
u v w u v w
- Vectors Part Trois
- Vector Geometry
- Resultant of Two Vectors
- Example
- Alternatively
- Example (2)
- Example (3)
- Example (4)
- Alternatively (2)
- Why
- Basic definitions
- Point + vector = point
- vector + vector = vector
- point - point = vector
- point + point not defined
- Map points to vectors
- Position Vectors
- Components of a Vector Between Two Points
- Understanding
- Understanding (2)
- Understanding (3)
- Understanding (4)
- Understanding (5)
- Understanding (6)
- Understanding (7)
- Inner (dot) product
- Dot product in coordinates
- Proof of Geometric and Coordinate Definitions Equality
- Vector Properties
- Understanding (8)
- Understanding (9)
- Understanding (10)
- Understanding (11)
- Understanding (12)
- The Cross Product
- The Cross Product (2)
- Understanding (13)
- Understanding (14)
- Properties
- Understanding (15)
- Understanding (16)
- Understanding (17)
- Vector Product in Component Form
- Understanding (18)
- Understanding (19)
- Understanding (20)
- Understanding (21)
- Slide 48
- Useful Properties
-
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