# Shortest distance to a surface

Let $S$ be a surface in $\mathbb{R}^3$ which is locally defined by a level set of some smooth function. Let $M$ be a point which is not on the surface. First of all, I would like to show that there always exists a unique point $N$ which is on the surface and that minimizes the distance $MN$. Secondly I would like to show that $\overrightarrow{NM}=d\cdot\vec\nu(N)$. I think this is a standard theorem of surface theory but I am not able to prove the following points. Could somebody provide some ideas or references?

Many thanks!

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I don't quite see how you can expect a unique point $N$. I can flex a surface so that there are two such points - or given a point $M$ I can take a plane $P$ not containing it, and find a circle $C$ of points in $P$ all equidistant from $M$. Then I can construct a surface for which the points on $C$ are the ones at minimal distance from $M$ –  Mark Bennet Apr 4 '12 at 20:03
Take the surface $x^2+y^2=1$, and the point $(0,0)$. All points on the surface minimize the distance. –  copper.hat Apr 4 '12 at 21:12
Both of the things you want to show are true as long as $M$ is not in the cutlocus of $S$. As long as $S$ is a regular surface, there is a neighborhood $U$ of $S$ in $\mathbb R^3$ that is diffeomorphic to $S\times[0,\epsilon)$. The diffeomorphism $\phi:S\times[0,\epsilon) \to U$ is given by $$(p,t) \to t\nu(p)$$ where $\nu(p)$ is the normal of $S$ at $p$ (thus either $S$ is oriented or you work locally). The Gauss Lemma implies that dist$((p,t), S) = t$ and so your your second claim is verified as well. This diffeomorphism will however not exist for all $\epsilon$, as an example let $S$ be the unit sphere and choose $M$ to be the origin. Then the uniqueness part of your statement gets messed up.