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 xy plane to the point. !"x#(x, y, z)r(x, y, 0)z 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). The z coordinate: From the relationship sin!=z", we have z = !sin". The radius r on the xy plane: From the relationship cos!=r", we have r =!cos" 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 120º30º12uyxz For point u, ! = 12, ! = 120º, and ! = 30º. z = 12 sin 30º = 12 !12= 6 x = 12 cos30 º cos 120º = 12 !32!12= 3 3 y = 12 cos 30º sin120º = 12 !32!32= 9 Thus, u = (3 3, 9, 6). vy270º–45ºxy20 For v, ! = 20, ! = 270º, and ! = –45º. z = 20 sin(!45º) = 20 "! 22= !10 2 x = 20 cos(!45º) cos270º = 20 "22" 0 = 0 y = 20 cos(!45º ) sin 270º = 20 "22" !1 = !10 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 = (!1cos"1cos#1, !1cos"1sin#1, !1sin"1) and v = (!2cos"2cos#2, !2cos"2sin#2, !2sin"2) The dot product is then u ! v ="1cos#1cos$1%"2cos#2cos$2+"1cos#1sin$1%"2cos#2sin$2+"1sin#1%"2sin#2="1"2cos#1cos$1cos#2cos$2+ cos#1sin$1cos#2sin$2+ sin#1sin#2( )="1"2cos#1cos#2cos$1cos$2+ sin$1sin$2( )+ sin#1sin#2( )="1"2cos#1cos#2cos($1&$2) + sin#1sin#2( ) Because u and v are on spheres with radii !1 and !2, we have u = !1 and v = !2. Thus, u! vu v="1"2cos#1cos#2cos($1%$2) + sin#1sin#2( )"1"2= cos#1cos#2cos($1%$2) + sin#1sin#2 So the angle in between u and v is != cos"1cos#1cos#2cos($1"$2) + sin#1sin#2( ) Example 2. Find the angle between u = (12, 120º, 30º) and v = (20, 270º, –45º). Solution. The lengths 12 and 20 of the two vectors are not relevant to the angle in between. All we need are the directional angles: For u: !1 = 120º in the x y plane and !1 = +30º (upward). For v: !2 = 270º in the x y plane and !2 = –45º (downward).Dr. Neal, WKU 120º30º12uvy270º–45ºxy20 The angle in between u and v is != cos"1cos30º cos("45º ) cos(120º "270º ) + sin 30º sin("45º )( )= cos"1cos30º cos("45º ) cos("150º ) + sin 30º sin("45º )( )= cos"132#22#" 32+12#" 22$ % & ' ( ) = cos"1"3 28"24$ % & ' ( ) = cos"1"5 28$ % & ' ( ) * 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"1cos#1cos#2cos($1"$2) + sin#1sin#2( ) with !1 = 60º !2 = –30º !1 = 100º !2 = –20º != cos"1cos60º cos("30º ) cos(100º "("20º )) + sin 60º sin("30 º )( )= cos"1cos60º cos("30º ) cos(120º ) + sin 60º sin("30º )( )= cos"112#32#"12+32#"12$ % & ' ( ) = cos"1" 38"34$ % & ' ( ) = cos"1"3 38$ % & ' ( ) *