Let us call the curves c1, c2 and c3
It is evident that:
c1 = (cos t, sin t, 0)
c2 = (cos t, 0, sin t)
c3 = (0, sin t, cos t)
[I am considering the circles to be centered at the origin and of radius 1. This is just to make the algebra simple]
You wanted me to prove that c1 intersects c2 making an angle of pi/2 and c2 intersects c3 making an angle of pi/2 and c3 intersects c1 making an angle of pi/2
Consider the intersection of c1 and c2. At this point, c1(t) = c2(t)
Or, (cos t, sin t, 0) = (cos t, 0, sin t)
Solving, you get t = 0
Now, c1(0) = c2(0) = (1, 0, 0)
Differentiating the curves,
c1' = (-sin t, cos t, 0)
c2' = (-sin t, 0, cos t)
At t=0, c1'(0) = (0, 1, 0) and c2'(0) = (0, 0, 1)
Let x1 be the tangent of c1 at the intersection point
We have, x1= c1(0) + t(c1'(0)) = (1, 0, 0) + t(0, 1, 0) = (1, t, 0)
Let x2 be the tangent of c2 at the intersection point
We have, x2= c2(0) + t(c2'(0)) = (1, 0, 0) + t(0, 0, 1) = (1, 0, t)
By the angle between c1 and c2, we obviously mean the angle between x1 and x2.
Let that angle be θ
x1 . x2 = (1, t, 0) . (1, 0, t) = 1*1 + t*0 + t*0 = 1
|x1| = Sqrt[1+t^2]
|x2| = Sqrt[1+t^2]
At the intersection point, t=0. So,
Similarly, you can go on proving the same thing for the two other intersection points (but that has been left to the reader as an exercise )
QED
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show that all the angles of the internal vertices of at least one of the spherical triangles formed by the intersection of x^2+y^2=r^2 and x^2+z^2=r^2 and y^2+z^2=r^2 on x^2+y^2+z^2=r^2 is pi/2
]]>How do I show that xy+x+y=0 and xz+x+z=0 are perpendicular?
]]>y = 1 - x , put into parametric form.
Say x = t now y = 1 - t
Do you see an advantage of this?
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