What is the union of all the tangent plane at every point of a sphere?

Question

Let $S = \{x \in \mathbb{R}^3: ||x||_2 = 1\}$

Then pick a point on $S$.

The tangent space to the point is the plane that is perpendicular to the vector from origin to that point.

What is the union of all these planes for all the points on $S$?

Answer 1

It’s $\left\{x\in\Bbb R^3:\|x\|_2\ge 1\right\}$. It’s clear that the set in question includes $S$ itself. If $\|x\|_2>1$, let $P$ be any plane that contains $x$ and the origin. $S\cap P$ is a great circle on $S$, and it should be clear that there are two lines through $x$ that are tangent to that circle. Let $y$ be one of the points of tangency; then the tangent plane to $S$ at $y$ contains $x$.

Answer 2

If $S$ is any convex smooth solid, the union of all the tangent planes to $S$ is all of space except for the interior of $S$.

For any point P outside of $S$, the set of lines through $P$ tangent to $S$ form a generalized cone. To find one of these lines (which is all you need), find the two points on $S$ closest and farthest from $P$. By continuity, the tangent planes at these points are perpendicular to the lines from $P$ to the points. Now, rotate one of the tangent planes until it coincides with the other plane, keeping tangent to $S$. At the start, $P$ is on one side of the plane, and, at the end, $P$ is on the other side of the plane. Therefore, somewhere in between, the plane passed through $P$.

Anyway, this satisfies me, but please do not ask me to make it rigorous. My differential geometry days are long gone and forgotten.