5
$\begingroup$

Here is the problem:

Let $S$ $\subset$ $\mathbb{R^{3}}$ be a regular surface with the property that all normal lines meet the z-axis. Prove that S is contained in a revolution surface around z.

I first tried to prove that every curve contained in this surface with a fixed z-coordinate is a circle (by showing that all normal lines to this curve meet at a point, it's a previous exercise), but the normal vector to the surface and to the curve are not the same, and I didn't go further... can you help me?

$\endgroup$

1 Answer 1

3
$\begingroup$

Sounds like you have a good strategy. You are looking at the cross section in some plane $z=z_0$, and you want to prove that the cross section here is a circle.

You are given that any normal line will intersect the $z$ axis. Let's write this line as $\vec{p}+\vec{n}t$ where $\vec{n}$ is the normal vector at the point $\vec{p}$ we are discussing. You can then show that a new line $\vec{p} + (\vec{n}-\hbox{proj}_{\hat{z}}\vec{n})t$ will also intersect the $z$ axis (taking away the z component doesn't change the fact that the line is in a radial direction).

Moreover, this new line will be normal to the cross section at $\vec{p}$: the tangent plane projects to a tangent line of the cross section, so must a normal vector of the surface project to a normal vector of the cross section.

This gives you the criteria you wanted -- all horizontal lines orthogonal to cross sections will intersect at the origin. This tells you roughly that the cross sections are circular, and that the overall shape will be a surface of revolution.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .