Math 251-003
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Lecture Notes #2
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Spring 2008
David Maslanka
Direction Cosines
The direction angles α, β, and γ associated with a nonzero vector v determine three circles on the unit sphere as described
below. Each of these inscribed circles is perpendicular to one of the coordinate axes and their common point of intersection
determines the direction of the vector uniquely.
> restart:
> with(plots):with(plottools):
> s:=sphereplot(1,theta=0..2*Pi,phi=0..Pi,style=wireframe,color=green):
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Circle1:=spacecurve([1/2,sqrt(3)*cos(t)/2,sqrt(3)*sin(t)/2],t=0..6*Pi,color=navy):
for n from 0 to 63 do
y||n:=sqrt(3)*cos(n*Pi/32)/2;
z||n:=sqrt(3)*sin(n*Pi/32)/2;
c||n:=line([1/2,y||n,z||n],[0,0,0],color=navy);
end do:
C1:=seq(c||i,i=0..63):
> Circle2:=spacecurve([sqrt(13)*cos(t)/6,sqrt(23)/6,sqrt(13)*sin(t)/6],t=0..6*Pi,color=
red):
> for n from 0 to 63 do
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x||n:=sqrt(13)*cos(n*Pi/32)/6;
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z||n:=sqrt(13)*sin(n*Pi/32)/6;
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c||n:=line([x||n,sqrt(23)/6,z||n],[0,0,0],color=red);
> end do:
> C2:=seq(c||i,i=0..63):
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Circle3:=spacecurve([sqrt(8)*cos(t)/3,sqrt(8)*sin(t)/3,1/3],t=0..6*Pi,color=magenta):
for n from 0 to 63 do
x||n:=sqrt(8)*cos(n*Pi/32)/3;
y||n:=sqrt(8)*sin(n*Pi/32)/3;
c||n:=line([x||n,y||n,1/3],[0,0,0],color=magenta);
end do:
C3:=seq(c||i,i=0..63):
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l1 := line([0,0,0], [1.8,0,0],color=black):
l2 := line([0,0,0], [0,1.8,0],color=black):
l3 := line([0,0,0], [0,0,1.5],color=black):
t1:=textplot3d([1.7,0,-0.15,`x`],font=[TIMES,BOLDITALIC,36],color=black):
t2:=textplot3d([0,1.6,-0.15,`y`],font=[TIMES,BOLDITALIC,36],color=black):
t3:=textplot3d([0,0.15,1.4,`z`],font=[TIMES,BOLDITALIC,36],color=black):
l4 := line([0,0,0], [1/2,sqrt(23)/6,1/3],color=black,thickness=3,linestyle=2):
l5 := line([1,1*sqrt(23)/3,2/3], [1/2,sqrt(23)/6,1/3],color=black,thickness=3):
c:= sphere([1/2,sqrt(23)/6,1/3],0.015,color=black):
t:=0.025:
P:=[0.9+t*1/3,3*sqrt(23)/10+t*2/3,6/10-t*(sqrt(23)/3+1/2)]:
t:=-0.025:
Q:=[0.9+t*1/3,3*sqrt(23)/10+t*2/3,6/10-t*(sqrt(23)/3+1/2)]:
R:=[1,1*sqrt(23)/3,2/3]:
ar:=polygon([P,Q,R],color=black):
t4:=textplot3d([1,1*sqrt(23)/3,3/4,`v`],font=[TIMES,BOLDITALIC,38],color=black):
arx:=polygon([[1.8,0,0],[1.65,0.1,0],[1.65,-0.1,0]],color=black):
ary:=polygon([[0,1.8,0],[0.1,1.65,0],[-0.1,1.65,0]],color=black):
arz:=polygon([[0,0,1.5],[0.1,0,1.35],[-0.1,0,1.35]],color=black):
> display(C1,Circle1,C2,Circle2,C3,Circle3,l1,l2,l3,l4,l5,t1,t2,t3,t4,c,s,ar,arx,ary,arz,
orientation=[44,74],scaling=constrained);
z
v
y
x
Note that every position vector which terminates on the boundary of Circle1 makes an angle of
1
= arccos(
) = 60o with the positive x - axis.
2
Each such vector terminating on the boudary of Circle2 makes an angle of
23
) = 36.9o with the positive y - axis,
6
and every such vector terminating on the boundary of Circle3 makes an angle of
= arccos( 1 ) = 70.5o with the positive z - axis.
3
Observe that there is a unique point common to all three circles.
= arccos(
Thus, the direction angles:
, , and
determine a unique direction for the position vector.