Dr. Neal, WKU
MATH 117
Spherical to Rectangular Coordinates
A point in three-dimensional space is uniquely determined by its spherical coordinates
( ! , ! , ! ), where ! is the length (or radius of the sphere), ! is the sideways angle
measured from the positive x -axis, and ! is the vertical angle measured upward or
downward from the x y plane to the point.
(x, y, z)
!
z
#
"
r
x
(x, y, 0)
We generally take –90º ≤ ! ≤ 90º and 0 ≤ ! ≤ 360º, although sometimes we take –
180º ≤ ! ≤ 180º. Of course, ! ≥ 0. Now given the spherical coordinates ( ! , ! , ! ), we
wish to find the rectangular coordinates ( x , y , z ).
z
, we have z = ! sin " .
"
r
The radius r on the xy plane: From the relationship cos ! = , we have r = ! cos "
"
The z coordinate: From the relationship sin ! =
The x and y coordinates: As usual, x = r cos ! and y = r sin ! . By using r = ! cos " ,
we have x = ! cos " cos # and y = ! cos " sin # .
The rectangular coordinates of a point ( ! , ! , ! ) are given by
x = ! cos " cos #
y = ! cos " sin #
z = ! sin "
Example 1. Compute the rectangular coordinates of the two points u = (12, 120º, 30º)
and v = (20, 270º, –45º).
Dr. Neal, WKU
For point u , ! = 12, ! = 120º, and ! = 30º.
z
z = 12 sin30º = 12 !
u
12
x = 12 cos30º cos120º = 12 !
30º
y
120º
y = 12 cos30º sin120º = 12 !
1
=6
2
3 1
! =3 3
2
2
3
3
!
=9
2
2
Thus, u = ( 3 3 , 9, 6).
x
270º
y
y
–45º
20
v
For vx, ! = 20, ! = 270º, and ! = –45º.
z = 20 sin(!45º) = 20 "
! 2
= !10 2
2
x = 20 cos(!45º) cos270º = 20 "
y = 20 cos(!45º ) sin 270º = 20 "
2
"0= 0
2
2
" !1 = !10 2
2
Thus, v = (0, !10 2 , !10 2 ).
Dr. Neal, WKU
The Angle Between Two Points in Spherical Coordinates
Let u = ( !1 , !1 , !1 ) and v = ( ! 2 , ! 2 , ! 2 ) be two given points. We will now see how to
find the angle between u and v based upon just the given angles. To do so, we will use
the formula for the rectangular coordinates and then simplify the dot product. The
rectangular coordinates of u and v are
u = ( !1 cos "1 cos #1 , !1 cos "1 sin #1 , !1 sin "1 )
and
v = ( ! 2 cos " 2 cos# 2 , ! 2 cos " 2 sin #2 , ! 2 sin " 2 )
The dot product is then
u ! v = "1 cos #1 cos $1 % " 2 cos # 2 cos $2 + "1 cos #1 sin $1 % "2 cos #2 sin $ 2 + "1 sin #1 % " 2 sin # 2
= "1 "2 ( cos #1 cos $1 cos # 2 cos $ 2 + cos #1 sin $1 cos # 2 sin $ 2 + sin #1 sin # 2 )
= "1 "2 ( cos #1 cos #2 (cos $1 cos $ 2 + sin $1 sin $2 ) + sin #1 sin #2 )
= "1 "2 ( cos #1 cos #2 cos($1 & $2 ) + sin #1 sin # 2 )
Because u and v are on spheres with radii !1 and ! 2 , we have u = !1 and v =
! 2 . Thus,
" " (cos #1 cos # 2 cos($1 % $2 ) + sin #1 sin # 2 )
u! v
= 1 2
= cos #1 cos # 2 cos( $1 % $ 2 ) + sin #1 sin # 2
u v
"1 "2
So the angle in between u and v is
! = cos "1( cos #1 cos #2 cos($1 " $2 ) + sin #1 sin # 2 )
Example 2. Find the angle between u = (12, 120º, 30º) and v = (20, 270º, –45º).
Solution. The lengths 12 and 20 of the two vectors are not relevant to the angle in
between. All we need are the directional angles: For u : !1 = 120º in the x y plane and
!1 = +30º (upward). For v : ! 2 = 270º in the x y plane and ! 2 = –45º (downward).
Dr. Neal, WKU
u
12
270º
30º
y
–45º
y
120º
20
v
x angle in between u and v is
The
! = cos "1 ( cos30º cos("45º ) cos(120º "270º ) + sin 30º sin("45º ) )
= cos "1 ( cos30º cos("45º ) cos("150º ) + sin 30º sin("45º ) )
$ 3
2 " 3 1 " 2'
)
= cos "1 &
#
#
+ #
% 2
2
2
2
2 (
$ "3 2
$ "5 2 '
2'
= cos "1 &
"
) = cos"1 &
)
% 8
4 (
% 8 (
* 152.1144º
Exercise. Find the angle between u = (10, 100º, 60º) and v = (6, –20º, –30º).
(Answer on next page)
Dr. Neal, WKU
Use ! = cos "1( cos #1 cos #2 cos($1 " $2 ) + sin #1 sin # 2 )
with
!1 = 60º
! 2 = –30º
!1 = 100º
! 2 = –20º
! = cos "1 ( cos60º cos("30º ) cos(100º "("20º )) + sin 60º sin("30 º ) )
= cos "1 ( cos60º cos("30º ) cos(120º ) + sin 60º sin("30º ))
$1
3 "1
3 "1'
= cos "1 & #
#
+
# )
%2
2
2
2
2(
$" 3
$ "3
3'
= cos "1 &
"
) = cos"1 &
% 8
4 (
% 8
* 130.5º
3'
)
(