Dr. Neal, WKU MATH 117 Vectors and Polar Formpeople.wku.edu/david.neal/117/Unit5/Vectors.pdf · Dr. Neal, WKU MATH 117 Vectors and Polar Form Any point (x, y) in the x y plane forms

Dr. Neal, WKU MATH 117 Vectors and Polar Form Any point ( x , y ) in the x y plane forms a directed line segment from the origin (0, 0) to the point ( x , y ). Such a segment is called a vector. When we want to consider the vector and not just the point, then we generally label it as v = ( x , y ).

v = (7, 4)

!

Length and Direction A vector v = ( x , y ) has a length (or norm) denoted by v which is simply the distance to the origin given by the hypotenuse. The direction ! is the standard angle determined by ( x , y ) as measured from the positive x axis. Thus we have

v = x

2+ y

2 and tan ! =y

x

To find the direction ! , compute tan !1(y / x) and then adjust the angle to the proper quadrant. A vector written in terms of its length and direction is then in polar form. Example 1. Let u = (–6, 8) and v = (–2, –10). Find the length and direction of each vector and write the vectors in polar form. Solution. Vector u has length u = 62+ 8

2 = 10. Here, tan !1( y / x) is tan

!1(8 / !6) ≈ –53.13º. So in Quad. II, ! =

!53.13º +180º ≈ 126.87! .

So u = (10, 126.87! ) in polar form.

u

v Vector v has length v = 2

2+ 10

2 = 104 . Its angle in Quadrant III is given by ! = tan !1(10 / 2) +180º ≈ 258.69! . Then v = ( 104 , 258.69! ) in polar form.

Dr. Neal, WKU

Converting Polar Form Back to Rectangular Form If a vector is given in polar form ( v , ! ), then we recover the x and y coordinates by

x = v cos! and y = v sin!

Here, v is taking the place of the radius r , where x = r cos! and y = r sin! . Example 2. Find the rectangular form of the vectors u = (30, 120º) and v = (20, 330º).

Solution. For u = (30, 120º), we have x = 30 cos120º= 30 ! "1

2

#

$ %

&

' ( = "15 and y

=30 sin120º= 30 ! 3

2

"

# $ $

%

& ' ' = 15 3 ; so u = (–15, 15 3 ).

For v , x = 20 cos330º= 20 ! 3

2

"

# $ $

%

& ' ' = 10 3 and y = 20 sin330º= 20 ! "

1

2

#

$ %

&

' ( = "10 ; so

then v = (10 3 , –10).

Adding Vectors Given two vectors u = (x1, y1) and v = (x2, y2) , both in rectangular form, we obtain the sum of vectors by adding component wise: u + v = (x1 + x2, y1 + y2 ) . If we make a parallelogram out of the vectors u and v , then vector u + v is the diagonal that starts at the origin. Example 3. Let u = (3, 6) and v = (–8, –2). Graph u , v , and u + v . What is the length and direction of u + v ? Solution. First, u + v = (3 + (–8), 6 + (–2)) = (–5, 4), which we see to be the diagonal of the parallelogram determined by u and v . (See graphs on next page.) Then u + v = 5

2+ 4

2= 41 ! 6. 4 , and the direction of u + v is given by

! = tan"1("4 / 5) +180º # 141.34º .

Dr. Neal, WKU

u = (3, 6)

v = (–8, –2)

u + v

u

u

v

v

u + v

Adding Forces Often, a force is given in terms of its magnitude and direction. In order to add two forces, we convert each to rectangular form, add the x and y components to get the sum, then convert the result back to polar form. The sum of two forces is called the resultant force. Example 4. Let F1 be a force of 50 Newtons in the direction 30º East of South, and let F2 be a force of 80 Newtons in the direction 10º South of East. Find the magnitude and direction of the resultant F1 + F2 . Solution. First, the angle for F1 is 30º + 270º = 300º, and the angle for F2 is 360º – 10º = 350º. Next, the x and y components for each force and the resultant are:

F1 = (50 cos300º , 50sin 300º )

F2 = (80cos350º , 80sin350º )

S

E

F1

10º

30ºF2

So the resultant force is

F1 + F2 = (50 cos300º + 80cos350º , 50sin 300º + 80sin350º ) = (103.78462, –57.19312) So F1 + F2 has magnitude F1 + F2 = 103.78462

2+ 57.19312

2! 118.5 Newtons . Its

direction is in the 4th Quadrant is tan !1(!57.19312 / 103.78462) + 360º ≈ 331.142º, or about 28.858º South of East.

Dr. Neal, WKU

Subtraction and Distance Between Vectors Given two vectors u = (x1, y1) and v = (x2, y2) , both in rectangular form, we obtain the vector from u to v by the difference v ! u which is obtained by subtracting component wise: v ! u = (x2 ! x1, y2 ! y1) . If we make a parallelogram out of the vectors u and v , then vector v ! u is equivalent to the diagonal from the end of u to the end of v after it is picked up and moved to the origin. The length of v ! u (or u ! v ) gives the distance between the endpoints of u and v , which we call the distance between the vectors. This length of v ! u is equivalent to the common “distance formula” in the x y plane:

u = (x1, y1) v = (x2, y2) v ! u = (x2 ! x1, y2 ! y1)

distance = (x2 ! x1)2+ (y2 ! y1)

2= v ! u

Example 5. Let u = (3, 6) and v = (–8, –2). Find the vector from u to v and the distance between u and v . Graph all the vectors. Solution. By subtracting component wise, we get v ! u = (–11, –8). Now we can pick up the vector (–11, –8) and place it between the original u and v to see that v ! u is the diagonal from u to v .

u = (3, 6)

v = (–8, –2)

v ! u

u = (3, 6)

!u = (–3, –6)

v

v ! u

(!11, !8)

v = (!8,!2)

The distance between u and v is v ! u = 112+ 8

2= 185 " 13.6 .

Dr. Neal, WKU

Dot Product and Angle Between Vectors Given vectors u = (x1, y1) and v = (x2, y2) , the dot product is given by

u ! v = x1 x2 + y1 y2

The dot product can be used to find the angle ! between the vectors u and v . Because v ! u is the length of the third side of a triangle and is opposite angle ! , the Law of Cosines gives

v ! u2 = u 2

+ v2! 2 u v cos"

We now can solve for cos! and simplify the expression:

u

v

v ! u

"

cos! =u 2 + v 2 " v " u 2

2 u v

=x12 + y1

2( ) + x22 + y2

2( ) " (x2 " x1)2 + ( y2 " y1)

2( )2 u v

=" "2 x1 x2 " 2 y1 y2( )

2 u v

=x1 x2 + y1 y2

u v=

u # v

u v.

Thus, cos! =u " v

u v and ! = cos

"1 u # v

u v

$

% &

'

( ) .

Example 6. Let u = (3, 6) and v = (–8, –2). Use the dot product to find the angle between u and v . Use the angles of each vector to verify the result.

Dr. Neal, WKU Solution. For u = (3, 6) and v = (–8, –2), the dot product is

u ! v = x1 x2 + y1 y2 = (–24) + (–12 )= –36. The lengths of the two vectors are u = 3

2+ 6

2 = 45 and v = 82+ 2

2 = 68 . So the angle between the vectors is

! = cos"1 u # v

u v

$

% &

'

( ) = cos"1

"36

45 68

$

% &

'

( ) * 130. 6º

We also can find the angles for v and u separately. The angle for v in Quadrant III is tan !1(2 / 8) +180º ≈ 194.036º. The angle for u in Quadrant I is tan

!1(6 / 3) ≈ 63.435º. So the angle in

between is 194.036º – 63.435º ≈ 130.6º. u

v

194.036º

63.435º

Exercise

Let u = (10, –4) and v = (–2, 8). (a) Find the lengths and directions of the vectors u , v and the length and direction of the resultant u + v . (b) Find the vector from u to v and the distance between u and v . (c) Use the dot product to find the angle between u and v . Verify the result using the angles found in Part (a).

Dr. Neal, WKU

Answers (a) u = ( 116 , 338.1986º) and v = ( 68 , 104.0362º) u + v = (8, 4) = ( 80 , 26.565º) (b) v ! u = (–12, 12); distance = v ! u = 288 (c) u ! v = –52; Angle in between is ! = cos!1(!52 / ( 116 " 68)) = 125.8376º

Using the angles found in (a), we have

338.1986º

104.0362º

!

! = 360º – (338.1986º – 104.0362º ) = 125.8376º