PROBLEM:
Let $\vec x(t)$ be a path with $\vec x'$x $\vec x'' \ne 0$ and suppose that there is a point $\vec x_0$ that lies on every normal plane to $\vec x$. Show that the image of $\vec x$ lies on a sphere (the normal plane at $t=t_0$ is the plane containing $\vec x(t_0)$ and determined by the normal and binormal vectors $\vec N(t_0)$ and $\vec B(t_0)$).
What's the geometric interpretation of this problem? Can someone solve it? Any help would be greatly appreciated.
Since $$ (\vec x-\vec x_0)^T\vec x'=0 $$ we have: $$ \frac{d}{dt}((\vec x-\vec x_0)^T(\vec x-\vec x_0))=2(\vec x-\vec x_0)^T\vec x'=0 $$ therefore: $$ (\vec x-\vec x_0)^T(\vec x-\vec x_0)=\text{constant} $$