BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 1 3.1 Vector
Concepts Magnitude and Direction of a Vector Measurements in
sciences are of two types, i.e. scalar and vector quantities. A
scalar quantity has magnitude, but not direction. For example, a
pen may have length "10 cm". The length 10 cm is a scalar quantity
- it has magnitude, but no direction is involved. Other examples of
scalar quantities are volume, density, temperature, mass, speed,
time, length, distance, work and energy. Each of these quantities
has magnitude only, and do not involve direction. A vector is a
quantity that has both magnitude and direction. (Magnitude just
means 'size'.). Examples of Vector Quantities: - I travel 30 km in
a Northerly direction (magnitude is 30 km, direction is North -
this is a displacement vector) - The train is going 80 km/h towards
Sydney (magnitude is 80 km/h, direction is 'towards Sydney' - it is
a velocity vector) - The force on the bridge is 50 N acting
downwards (the magnitude is 50 Newton and the direction is down -
it is a force vector) Other examples of vectors include
acceleration, momentum, angular momentum, magnetic and electric
fields. Each of these examples involves magnitude and direction.
3.1.1 The Fundamentals of Vectors a) Vector Notation To distinguish
between vector and scalar quantities, various ways are used. These
include: - bold capital letter. For example, a force vector could
be written as F , AB. - An arrow above the vector name, F, AB. - A
line over the top or under the vector name ; F, F, F, F, AB, AB, AB
- Cartesian component = 2 i +3 j . - You will also see vectors
written using matrix-like notation. For example, the vector acting
from (0, 0) in the direction of the point (2, 3): Column vector =
23| | |\ . or 23 ( ( ; Row vector = (2,3) or 2,3 ; b) Vector
Representation Graphically, a vector is represented by an arrow,
defining the direction, and the length of the arrow defines the
vector's magnitude. This is shown in Figure 1. If we denote one end
of the arrow by the origin O and the tip of the arrow by Q. Then
the vector may be represented algebraically by q or OQ. BA501
ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 2 c) Equality of Vector
Two vectors are equal if they have the same magnitude and
direction. In Figure 2 vectors AB and CD are equal even though
their positions differ. We can write: AB = CD. d) Negative Vectors
A vector having the same magnitude as AB but in the opposite
direction to AB is denoted by - AB . We can write: AB = - CD as
shown in Figure 3. e) Unit Vectors A unit vector has length 1 unit
and can take any direction. A one-dimensional unit vector is
usually written i. If u is a vector with magnitude u , then uuis a
unit vector having the same direction as u . If u has length 5, for
example, then a unit vector in the direction of u is clearly 1u5 .
Figure 1 OQ= q CD Figure 2 A B AB C D AB Figure 3 A B AB A B BA501
ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 3 f) Magnitude of a
Vector We indicate the magnitude of a vector using vertical lines
on either side of the vector name. The magnitude of RSis written RS
. So for example, vector a above has magnitude 5 units. We would
write the magnitude of vector a as: | a | = 5. If vector xAB xi
yjy| |= + = |\ .then the magnitude of AB is 2 2AB x y = + g)
Position Vector If O is the origin, OA is known as the position
vector of A (relative to O). 3.2 The operation of Vectors 3.2.1
Addition and Subtraction of Vectors a) Adding 1-dimensional Vectors
Subtraction of Vectors The 2 vectors are now acting in opposite
directions. v u u + v B O A From the diagram; OB - OA = AB OA + AB
= OB v u v u u - v BA501 ENGINEERING MATHEMATICS IV TOPIC 3 :
VECTOR 4 b) Adding Vectors Using a Triangle Law The resultant
vector of the vectors a and b can be obtained by connecting the
starting point of b with the endpoint of a as shown in Figure 4. c)
Adding Vectors Using a Parallelogram Law The resultant vector of
the vectors a and b can be obtained by connecting both vectors of a
and b from the common starting point. Complete the parallelogram
OABC as shown in Figure 5. d) Adding Vectors Using a Polygon method
The resultant vector of three or more vectors can be using the
polygon law. The resultant vector of a, b,c and d can be determined
by constructing a polygon as shown in Figure 6: a A O a b + b b C O
a A A Figure 5 C B C A O A A O C a b + a b b a Figure 4 BA501
ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 5 Resultant vector = a
b c d + + + = ( )AB BC CD DE + + + = (AC CD) DE + + = AD DE AE + =
d) Adding Vectors Using Method of Components We can write any
2-dimensional vector in terms of the unit vectors i and j. Example
In an earlier example, we had the following vector: 3.2.2 Scalar
Multiplication We can increase or decrease the magnitude of a
vector by multiplying the vector by a scalar. When a vector a is
multiplied by a scalar k, the product is ka. Scalar Multiplication
- Example 1 C B A D E a b c Figure 6 d We can write these vector
components using subscripts as follows: Vx = 6 units; Vy = 3 units
We have: Magnitude of V = | V | = (62 + 32)= 45= 6.71 units We
could write the components of the vector V as follows. Vx = 6 i ;
Vy = 3 j ,So we can write the vector V using unit vectors as
follows: V = 6 i + 3 j BA501 ENGINEERING MATHEMATICS IV TOPIC 3 :
VECTOR 6 In the examples we saw earlier, vector B (2 units) is half
the size of vector A (which is 4 units) . We can write: B = 0.5 A
This is an example of a scalar multiple. We have multiplied the
vector A by the scalar 0.5. Properties of addition and scalar
multiples of vector For any vectors p, q ,r and scalar and . 1. p +
q = q + p Commutative Law for Addition 2. p + ( q + r) = ( q + p )
+ r Associative Law for Addition 3. ( p) = ( ) p Associative Law
for muliplication 4. (p + q) = p + q Distributive Law for
muliplication 5. ( + ) p = p+ p Distributive Law for muliplication
3.3 Dot Product (Scalar Product) of Two Vectors 3.3.1 Definition of
scalar product If we have any 2 vectors P and Q, the dot product of
P and Q is given by: Where | P | and | Q| are the magnitudes of P
and Q respectively, and is the angle between the two vectors. The
dot product of the vectors P and Q is also known as the scalar
product since it always returns a scalar value. 3.3.2 The
properties for scalar products are given in the following list. If
P and Q are parallel vectors, P Q P Qcos0 P Q - = = The scalar
product is commutative The distributive rule Where m is a scalar u
P Q P Q P Qcos - = u BA501 ENGINEERING MATHEMATICS IV TOPIC 3 :
VECTOR 7 If P and Q are perpendicular (orthogonal) vectors, 0 P Q P
Qcos90 - = = 3.3.3 Calculation Of The Scalar Product Dot Products
of Unit Vectors For the unit vectors i (acting in the x-direction)
and j (acting in the y-direction), we have the following dot (i.e.
scalar) products (since they are perpendicular to each other): i j
i j 0 - = - = and i i j j 1 - = - = Recall that vectors can be
written using scalar products of unit vectors. If we have 2 vectors
P and Q defined as: P = a i + b j ; Q = c i + d j , Where a, b, c,
d are constants; i is the unit vector in the x-direction; and j is
the unit vector in the y-direction, then it can be shown that the
dot product (scalar product) of P and Q is given by: P Q ac bd - =
+ P Q - = (a i + b j ) (c i + d j )= ( )ac i i - + ( ) ( ) ( )ad i
j bc j i bd j j - + - + - = ac + bd This result can be generalized
for P and Q in any orientation. Example Find P Q - if P = 6 i + 5 j
and Q = 2 i 8 j Answer P Q - = (6 i + 5 j ) (2 i 8 j ) = (6 2) + (5
-8) = 12 40 = 28 Now we see another use for the dot product finding
the angle between vectors. Angle Between Two Vectors P Q P Qcos - =
u P QcosP Q-u = P Qar cosP Q-u = Example Find the angle between the
vectors P = 3 i - 5 j and Q = 4 i +6 j BA501 ENGINEERING
MATHEMATICS IV TOPIC 3 : VECTOR 8 We use the formula we just
derived: 2 2 2 2(3x4) ( 5x6) 18115.334 523 ( 5) 4 6P Qarccos arccos
arccosP Q + = = = | | | |+ + | |\ . \ .-u = Therefore the angle
between the vectors P and Q is, = 115.3 The 3-dimensional
Co-ordinate System We can expand our 2-dimensional (x-y) coordinate
system into a 3-dimensional coordinate system, using x-, y-, and
z-axes. We normally use the 'right-hand orientation' for the 3 axes
as shown below. 2 2 2distance OP OP a b c = = + + Vectors in 3-D
Space The vector OP has initial point at the origin O (0, 0, 0) and
terminal point at P (2, 3, 5). We can draw the vector OP as
follows: 2OP 2i 3j 4k 34| | |= + + = | |\ . BA501 ENGINEERING
MATHEMATICS IV TOPIC 3 : VECTOR 9 Magnitude of a 3-Dimensional
Vector For the vector OP above, the magnitude of the vector is
given by: 2 2 2OP 2 3 5 38 6.164 units = + + = = Direction Cosines
Direction of OPdetermined by cos , cos and cos are known as
direction cosines and , and are direction angles. Adding
3-dimensional Vectors Example Two anchors are holding a ship in
place and their forces acting on the ship are represented by
vectors A and B as follows: A = 2i + 5j 4k and B = 2i 3j 5k Answer
The problem just requires us to add the vectors to get the single
resultant vector. A +B = (2 + 2) i + (5 3)j + (4 5)k = 0 i + 2 j 9
k = 2 j 9 k Dot Product of 3-dimensional Vectors Example - Dot
Product Using Magnitude and Angle Find the dot product of the
vectors P and Q given that the angle between the two vectors is 35
and | P | = 25 units and | Q | = 4 units Answer P Q = |P| |Q| cos ;
we have: is the angle between OP and the x-axis, is the angle
between OP and the y-axis is the angle between OP and the z-axis.
OP = cos i + cos j + cos k y x zcos , cos , cosOP OP OPo = | = = P
BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 10 P Q = |P| |Q|
cos = 25 4 cos 35 = 81.92 Example - Dot Product if Vectors are
Multiples of Unit Vectors Find the dot product of the vectors A and
B. A = 2i + 5j 4k and B = 2i 3j 5k Answer A B = (2i + 5j 4k) (2i 3j
5k) = (2 -2) + (5 -3) + (-4 -5)= -4 + -15 + 20 = 1 Example a) Find
the angle between the vectors P= 4i + 7k and Q = -2i + j + 3k.
Answer Using the formula ; P QarccosP Q-u = 2 2 2 2 2(4x( 2)) (0x1)
(7x3) 1364.4765 144 7 ( 2) 1 3P Qar cos ar cos ar cosP Q + += = = |
| | |+ + + | |\ . \ .-u = Therefore the angle between the vectors P
and Q is = 64.47 b) Find the angle between the vectors P = 3i + 4j
7k and Q = -2i + j + 3k. Answer 3.4 Cross Product (Vector Product)
of 2 Vectors 3.4.1 Definition of Vector Product Suppose we have 2
vectors A and B. These 2 vectors lie on a plane and the unit vector
n is normal (at right angles) to that plane. The cross product
(also known as the vector product) of A and B is given by: A B A B
sin = u n The right hand side represents a vector at right angles
to the plane containing vectors A and B. Note: Some textbooks use
the following notation for the cross product: AB. BA501 ENGINEERING
MATHEMATICS IV TOPIC 3 : VECTOR 11 3.4.2 Properties Of Vector
Product - The length of the cross product of two vectors : -
Anticommutativity: - Multiplication by scalars: - Distributivity:
3.4.4 The Area of a parallelogram 3.5 Triple Product In
mathematics, the triple product is a product of three vectors. The
name "triple product" is used for two different products, the
scalar-valued scalar triple product and, less often, the
vector-valued vector triple product. 3.5.1 Scalar Triple Product
Area of parallelogram = OC x OA OC x OA sin = u Area of triangle
OCA = 1OC x OA2 Unit vector that is perpendicular to both vector OC
x OAOC and OAOC x OA= = H=OC sinu u O C B A Three vectors defining
a parallelepiped The scalar triple product (also called the mixed
or box product) is defined as the dot product of one of the vectors
with the cross product of the other two. Properties Switching the
two vectors in the cross product negates the triple product, i.e.:
. BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 12 b~ h | a~
Geometric interpretation Geometrically, the scalar triple product
a~- ( b~ x c~) = ( a~ x b~)- c~=b~- ( c~x a~) is the (signed)
volume of the parallelepiped defined by the three vectors given.
volume of the parallelepiped The area of base of =l b~ x c~l, and
the height , h = ac x bc x b~~~~~- Volume of the parallelepiped =
|b~ x c~ |x ac x bc x b~~~~~- = |( b~ x c~) - a~| unit3 Example
Given a~ = 5k j, b~ = 3i + j and c~= i + j + k. Find the volume of
the parallelepiped Solution a~x b~ = 0 1 35 1 0 k j i = (0 5)i (0
15)j + (0 + 3)k = -5i + 15j + 3k |a~x b~|- c~ = 3131113155unit
=|||.|
\|-|||.|
\| z y x c~ u A n c a b BA501 ENGINEERING MATHEMATICS IV TOPIC 3
: VECTOR 13 Example Find the volume of the parallelepiped formed by
the points A(5,4,5), B(4, 10, 6), C(1,8,7), and D(2,6,9) Solution
The position vectorin unit vector; OA = 5 i + 4 j + 5k , OB = 4 i +
10 j +6k , OC = i + 8 j + 7k , OD = 2 i + 6 j + 9k AB = AO + OB = -
OA + OB = - i + 6 j + k AC = AO + OC = - OA + OC = -4 i + 4 j + 2k
AD = AO + OD = - OA + OD = -3 i + 2 j + 4k Volume = The area of
base x height = |( a~ x b~) - c~ | = |( AB x AC) - AD| ABx AC = i j
k1 6 14 4 2 = 8 i 2 j + 20k ( AB x AC) - AD = 524232028=((((
-((((
unit3 EXERCISE 1. If A~ = i~+ 2 j~+ 3k~ , B~ = 2 i~- 3 j~+k~, C~
= 3 i~+ j~- 2Find A~ - ( B~ x C~) , B~ - ( A~x C~) and A~ - ( B~ x
A~) Answer A~ - ( B~ x C~) = 2 1 31 3 23 2 1 = 1(6 - 1) 2(- 4 - 3)
+ 3(2 + 9) = 52 B~ - ( A~x C~) = 2 1 33 2 11 3 2 =2(- 4 - 3) + 3(-2
- 9) + 1(1- 6) = -52 A~ - ( B~ x A~) = 3 2 11 3 23 2 1 =1(-9 - 2) -
2(6 - 1) + 3(4 + 3) = 0 BA501 ENGINEERING MATHEMATICS IV TOPIC 3 :
VECTOR 14 2. If A~ = 2 i~-3 j~+ k~ , B~ = i~+ 2 j~- k~, C~ = 3 i~+
j~+ 3k~ . Find A~x (B~ x C~) . Answer B~ x C~ = 3 1 31 2 1k~j~i~ =
1 32 1k~3 31 1j~3 11 2i~ + = i~k~(6 + 1) - j~(3+3) + k~(1 6) = 7
i~- 6 j~ - 5 k~ Then A~x (B~ x C~) = = 5 6 71 3 2k~j~i~ 6 73 2k~5
71 2j~5 61 3i~+ = i~k~(15 + 6) - j~(-10 - 7) + k~(-12 + 21) = 21
i~+17 j~ + 9k~ 3. If a 2,1, 1 = and b 3,4,1 = compute each of the
following. (a) a x b (b) b x a Answer (a) Here is the computation
for this one. (b) And here is the computation for this one. 4.
Given that P (-3, 6), Q (2, 8) and R (2, 7), find 3PQ QR in terms
of unit vectors iand j . 5. Givena 2i j 3k and b=3i-2j+k = + + ,
find the angle between a and b . 6. The points A, B and C have
position vectors 2i j 3k , 3i-j+k and i-2j+k + + respectively,
Find: a. The length of AB and AC b. The angle BAC, correct to the
nearest degree. BA501 ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR
15 c. The area of triangle ABC. 7. If a~= k~j~5 i~ , b~ = k~j~5 i~2
+ + and k~2 j~i~3 c~ + = . Find : i. a~ + c~ + b~ ii. |a~ + c~ -
b~| 8. Given k j i OA + = 3 2 and k j i OB 3 2 + = ; find the
direction angle of AB. 9. For what values or value of h will the
vectors u 2i 4 j 5k = + and v 4 i 8 j hk = + be parallel? 10. The
triangle ABC has vertices at (2,3,4) , (-2,1,0) and ( 4,0,2). Find;
i. BAC Z ii. The area of the triangle 11. If a 2 = , b 3 = , and
the angle between a and b is 3t, find a 2b . 12. Simplify; ii. a~ -
( a~ Xb~) ii. ( a~ - b~) X a~ 13. The angle between P and Q is 3t.
Find x if P= k~j~i~x + + and Q = k~j~x i~ + + . b) Given k~2 j~i~3
A + = , k~2 j~3 i~B + = , k~2 j~2 i~9 C + + = , Find; i. Magnitude
and direction cosine of B A ii. The angle between B A and C . 14.
If OA = k~3 j~4 i~2 + and OB = k~2 j~3 i~ + + . Find :- i. OA - OB
ii. OA x OB iii. The angle between OA and OB 15. If a~ = i~ - j~ +
2 k~, b~ = 3 i~ + 4 j~ - 2k~ and c~ = -5 i~ + j~ - 4 k~, find:- i.
Direction cosine of c~ iv. A unit vector parallel to a~ - b~ + 2 c~
ii. The angle between ( a~ + b~) and ( a~ - b~) v. ( a~ x b~) - (
b~ x c~) iii. a - ( b x c~) 16. If a~= k j i~ ~5~ , b~ = k j i~
~5~2 + + and k j i c~2~ ~3~ + = . Find b~ ( a~ - c~). 17. Find the
value(values) of o, i. a~ = k~j~i~2 + , b~ = k~j~i~2 + + o and b~a~
= ii. a~ = j~i~2 , b~ = j~3 i~ + o and 5 b~a~ = + iii. a~ = k~j~i~
+ + o , b~ = k~j~i~ + o + and the angle between a~ and b~is 3t 18.
Given A~ = 2 i~- j~+ 3k~ , B~ = 3 i~+ 2 j~+ k~, C~ = i~+ p j~+ 4k~,
if A~ - ( B~ x C~) = 0, find the value of p. 19. Given ( ) w~v~u~ -
= 3. Find; i. ( ) v~w~u~ - iii. ( ) w~u~v~ - v. ( ) u~w~v~ - BA501
ENGINEERING MATHEMATICS IV TOPIC 3 : VECTOR 16 ii. ( ) v~w~u~ - iv.
( ) v~u~w~ - vi. ( ) w~w~v~ - 20. Given a~= kji + 2 , b~ = ki4 3 ,
c~ = ji+ and kjid~3 2 + = . Find; i. ( ) c~b~a~ ii. ( ) c~b~a~ -
iii. ( ) ( ) d~b~c~a~ - ANSWER a. (i) 75.96 ii) 12 unit2 b.
|||.|
\|bp aqar cpcq br-|||.|
\|cba = |||.|
\|bp aqar cpcq br- |||.|
\|rqp=0 c. i. 0 ii.-( b~ x a~) = a~ x b~ d. i. 2 i~- j~+ 2k~ ii.
157 e. I. 0 = - AC AB ii. 33.73 unit2 f. x = 0 @ 4 g. 180 ;
1801018081804= = |= o kos ; kos ; kos ii. 90 h. i. 8 ii. k~j~i~2 7
17 + iii. 66.61 i. i. 424421425 = = |= o kos ; kos ; kos ii. 133.3
iii. 10 iv. |||.|
\|4312131 v. 421 j. 650 k. -19 l. 118.48 m. = = | = o 9 51 5 140
9 98 . ; . ; . n. i. 1 = o ii. o= 21 2 iii. o=0,4 o. |||.|
\|555755 p. k~j~i~23 7 6 + + q. k~j~i~6 2 8 + ; 42.83 r. -3 s.
i) ( ) c~b~+21 ; ( ) a~c~b~ + +21 ii. 21 t. ( ) k~j~i~7 2541 + ; 0
14 4 18249712= =|||.|
\| -|||.|
\| u. i. -3 ii. 3 iii. 3 iv. -3 v. -3 vi. 0 BA501 ENGINEERING
MATHEMATICS IV TOPIC 3 : VECTOR 17 v. i. |||.|
\| 766 ii. 9 iii. 4