# plane of symmetrie leads to geodesic

Any plane of symmetry intersects a surface in a geodesic.

I am having a little difficulty with the proof of this one. The proof says:

The normal to the surface in such a point must be invariant under reflection in the plane of symmetry and hence lie in that plane. It is orthogonal to the tangent vector of the curve of intersection and so $\vec T'$ points normally.

What I don't understand is the conclusion that $T'$ points normally. I know that we can write $k\vec N=\vec T'=k_n \vec n+k_g \vec n\times\vec T$. We already know that $\vec n$ lies in the plane of symmetry and, as I understand it, $\vec T$ has to lie in the plane, too. So $\vec n\times\vec T$ is perpendicular to the plane of symmetry. So why is $\vec T'$ orthogonal to the surface if it's a combination of a vector in the plane and a vector perpendicular to the plane?

• It's been awhile since I've used this notation, so I'm not sure if this is correct. But...using your equation, note that rearranging, you get $k\vec{N} - k_n\vec{n} = k_g \vec{n}\times\vec{T}$. Since $\vec{N}$ and $\vec{n}$ both lie in the plane, $k_g \vec{n}\times\vec{T}$ must as well. Since you've already argued $\vec{n}\times \vec{T}$ is perpendicular to the plane, this implies $k_g = 0$. Nov 13, 2013 at 21:43
• Oh, yes, I guess I just looked at this the wrong way. Thanks! Nov 13, 2013 at 21:48
• You may want to write it up as a full answer this way this question is officially answered :-) Nov 13, 2013 at 21:52

## 1 Answer

The normal of the surface $\vec n$ has to be invariant under reflection in the plane, hence it has to lie in the plane. For the same reasons the tangent vector has to lie in the plane. Since we can write $k\vec N = \vec T' = k_n\vec n + k_g \vec n\times\vec T \Leftrightarrow k\vec N-k_n\vec n = k_g\vec n\times \vec T$ we get that $\vec n\times\vec T$ has to lie in the plane, too, as $\vec n$ and $\vec N$ do. This cannot be, so $k_g=0$, which tells us that the intersection curve must be a geodesic.