TITLE: Continuously extend this function to the unit sphere.
QUESTION [5 upvotes]: Given a nonzero vector $u$, I can define the map
$$
f_{u} : S^2 \to \mathbb{R}^3 \\
f_{u}(x) = \frac{u-(u\cdot x)x}{({(u \cdot x)^2}/(u\cdot u))-1}
$$
which is well-defined everywhere except when $u$ and $x$ are parallel. So, I have two questions:

Does $f_{u}$ have limits at those two points where it is not well-defined, and what are those limits?
Is there a simple geometric interpretation of what $f_{u}$ does to the unit sphere?

REPLY [4 votes]: Note that $f_{cu}(x)=cf_u(x)$ so you may as well let $u\in S^2$.
To see what the function is doing, take the great circle $C$ containing $\frac{u}{|u|},x\in S^2$
Set the coordinates $u=(|u|,0),, x=(\cos\theta,\sin\theta)$ so that $u\cdot x=|u|\cos\theta$.
Then
\begin{eqnarray}
f_u(x)&=\frac{(1,0)-\cos\theta(\cos\theta,\sin\theta)}{\cos^2\theta-1}\cdot |u|\\
      &=\left(-1,\cot\theta\right)\cdot|u|
\end{eqnarray}
Which is a 'vertical' line. So the function maps $S^2$ onto a plane.
Here is a geometrical illustration of how the mapping works.

Addendum: Note that $f$ is discontinuous at $(\pm1,0)$.