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 =
x2 + y2
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
6 2 + 82 = 10.
Here, tan !1 ( y / x) is
tan !1 (8 / !6) ≈ –53.13º. So in Quad. II, ! =
!
!53.13º +180º ≈ 126.87 .
u
!
So u = (10, 126.87 ) in polar form.
v
Vector v has length v
=
2 2 + 102 = 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).
# 1&
(30, 120º), we have x = 30 cos120º = 30 ! % " ( = "15 and y
$ 2'
" 3%
= 30 sin120º = 30 ! $$ '' = 15 3 ; so u = (–15, 15 3 ).
# 2 &
" 3%
# 1&
For v , x = 20 cos330º = 20 ! $$ '' = 10 3 and y = 20 sin330º = 20 ! % " ( = "10 ; so
$ 2'
# 2 &
then v = (10 3 , –10).
Solution.
For u =
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 = 52 + 42 = 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
u+v
u
u+v
u
v
v = (–8, –2)
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:
10º
F2
30º
F1 = ( 50 cos300º , 50sin 300º )
F2 = ( 80cos350º , 80sin350º )
E
F1
S
So the resultant force is
F1 + F2 = ( 50 cos300º + 80cos350º , 50sin 300º + 80sin350º ) = (103.78462, –57.19312)
2
2
So F1 + F2 has magnitude F1 + F2 = 103.78462 + 57.19312 ! 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 = (x 2 ! 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 = (x 2 ! x1, y2 ! y1)
2
2
distance = (x 2 ! x1 ) + (y2 ! y1 ) = 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)
u = (3, 6)
v!u
v = (!8, !2)
v = (–8, –2)
!u = (–3, –6)
v!u
(!11, !8)
v
The distance between u and v is v ! u = 112 + 82 = 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 x 2 + y1 y 2
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!u
2
= u 2 + v 2 !2 u
v cos "
u
v!u
We now can solve for cos ! and simplify
the expression:
"
v
cos ! =
=
u 2 + v 2 " v"u 2
2 u v
( x12 + y12 ) + ( x 22 + y22 ) " ( (x 2 " x1 )2 + ( y2 " y1 )2 )
2 u
v
" ( "2 x1 x 2 " 2 y1 y2 )
2 u v
x x + y1 y2
u# v
= 1 2
=
.
u v
u v
=
Thus, cos ! =
u "v
u v
and
$ u#v '
! = cos "1 &
).
% 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 x 2 + y1 y 2 = (–24) + (–12 )= –36.
The lengths of the two vectors are u
So the angle between the vectors is
=
32 + 62 =
$ u#v '
$
! = cos "1 &
) = cos "1 &
% u v (
%
45 and v
=
82 + 22 =
68 .
"36 '
) * 130. 6º
45 68 (
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
!1
tan (6 / 3) ≈ 63.435º. So the angle in
between is 194.036º – 63.435º ≈ 130.6º.
63.435º
u
v
194.036º
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)
(b) v ! u = (–12, 12); distance = v ! u
u + v = (8, 4) = ( 80 , 26.565º)
= 288
(c) u ! v = –52; Angle in between is ! = cos !1(!52 / ( 116 " 68)) = 125.8376º
Using the angles found in (a), we have
104.0362º
!
338.1986º
! = 360º – (338.1986º – 104.0362º ) = 125.8376º