The intersection of a plane and a sphere proof

[ky2345]

Homework Statement

When a plane intersects a sphere at more than two points, it is a circle (given). Let x^2+y^2+z^2=1 be a sphere S, and P be a plane that intersects S to make a circle (called C). Let q:[a,b] -> R^3 be a unit speed parameterization whose trace is C. Prove that the second derivative of q, q''(t), is orthogonal to S for all t iff P passes through the origin.

Homework Equations

- q''(t) is orthogonal to S at a specific point q(t_0) if q''(t) is orthogonal to the tangent plane of S at q(t_0)
- two vectors are orthogonal iff their dot product is zero (I'm letting * denote dot product)
- A curve is parameterized by arclength exactly when it has unit speed

The Attempt at a Solution

let P be the plane that intersects the origin, and let n be a vector that is orthogonal to P emitting from the origin. Then, an equation for the plane is n*x=0. The vector equation for the sphere is |x|^2=1, or x*x=1. Then, I can say that x*x+n*x=1 is an equation for the circle, or q(t)*q(t)+n*q(t)=1 if we plug in the parameterization. Now, differentiating with respect to t two times, I get that 2[q(t)*q''(t)+q'(t)*q'(t)]+n*q''(t)=0. I know that q'(t)*q'(t)=|q'(t)|^2=1 because q is unit speed. I'm stuck here because I don't know how to do the following two things:
(1) prove that q(t)*q''(t)=-1
(2) Then prove that since n and q''(t) are orthogonal for all t, q"(t) is orthogonal to the tangent plane at any given t_0 in [a,b]

 

[ystael]

You can simplify your computations somewhat by applying a rotation so that [tex]P[/tex] is parallel to the [tex]xy[/tex]-plane. Now you should be able to write down the parametric equation for [tex]C[/tex] explicitly.

It is not correct to say that [tex]\vec{x} \cdot \vec{x} + \vec{n} \cdot \vec{x} = 1[/tex] is an equation for [tex]C[/tex]; this is an equation for a quadric surface which contains [tex]C[/tex]. For example, if [tex]P[/tex] is the [tex]xy[/tex]-plane and [tex]\vec{n} = (0, 0, 1)[/tex], the surface is the sphere [tex]x^2 + y^2 + z^2 + z = 1[/tex], or [tex]x^2 + y^2 + (z + \textstyle\frac12)^2 = \textstyle\frac54[/tex].

 

[ky2345]

If I apply a rotation, and I know that (cos(t), sin(t)) is the arclength parameterization for a circle, then can I just use those formulas to prove that the acceleration is tangent to the sphere? How do I apply a rotation, should I just say I'm applying a rotation and viewing P as the xy plane, or do I have to somehow apply formulas to apply the rotation? If I use the (cos(t), sin(t) equations, I know that the acceleration is (-cos(t), -sin(t)). Now, how would I come up with the formula for the tangent plane to try to show that the acceleration is orthogonal?

Also, I have to do it in the other direction, that is, prove that if the acceleration vector is tangent to the sphere, then the plane P passes through the origin. However, since I can't assume my premise, I can't apply a rotation that time.