Analysis Question: Projection.

Question

Let $S$ be a sphere $x^2 + y^2 + z^2 = 1$. $\forall$ point $\vec{p}=(x,y,z) \in S$, consider the line $L_{\vec{p}}$ that passes through the pole $(0,0,1) \in S$ and by the point $\vec{p}$. Let $\vec{q}$ be the interception of $L_{\vec{p}}$ and the plane $XY$. The function $f$ that takes $\vec{p}$ to $\vec{q}$ is called stereographic projection. If $(u,v) = f(x,y,z)$, then what is the value of $(u,v)$.

I really don't get it, it has being hard even to interprete the question, to be honest.

The solution is: $(u,v) = (\frac{x}{1-z}, \frac{y}{1-z})$.

Answer 1

The line that passes through $(0,0,1)$ and $\mathbf p=(x,y,z)$ in parametric form is $$L_{\mathbf p}=\left\{\begin{align} &x'=xt\\ &y'=yt\\ &z'=1+\left(z-1\right)t \end{align}\right.,\ t\in\mathbf R$$ Imposing $z'=0$ we'll find $t$ when the line passes through the $XY$ plane, i.e., $$z'=1+\left(z-1\right)t=0\iff t=\frac{1}{1-z}$$ Evaluating $L_{\mathbf p}$ at that $t$ we get $$L_{\mathbf p}\left(t=\frac{1}{1-z}\right)=\left\{\begin{align} &x'=\frac{x}{1-z}\\ &y'=\frac{y}{1-z}\\ &z'=0 \end{align}\right\}= \begin{pmatrix} u\\ v\\ 0 \end{pmatrix}$$

Answer 2

Before studying stereographic projection in space, it is better to study what is also known as stereographic projection in the plane. Then it's easy to go back to the 3D-space and furthermore, provided you with the answer of @Joan S.Guillamet.

We call $I=(0,1)$ the pole, we have a point $P=(x,y)\in S=\{(x,y)\in \mathbb R^2:x^2+y^2=1\}$

1. First, we're going to look at the $l=IP$ line : $M=(\xi,\eta)\in l\iff \vec {IM}\parallel \vec{IP}$

$\vec {IM}\begin{pmatrix}\xi-0 \\\eta-1\end{pmatrix},\overrightarrow{IP}\begin{pmatrix}x \\y-1\end{pmatrix}$

Then $M\in l\iff \begin{vmatrix}\xi & x \\\eta-1 & y-1\end{vmatrix}=0\iff\xi(y-1)-x(\eta-1)=0$

2. Now let's apply the assumption that for $Q, \eta=0$ We have then $$\begin{cases}\xi(y-1)-x(\eta-1)=0 \\ \eta=0 \end{cases}$$

So, $$\begin{cases}\xi(1-y)=x \\ \eta=0 \end{cases}$$and finally $$\begin{cases}\xi=\frac{x}{1-y} \\ \eta=0 \end{cases}$$

3. Otherwise written with notations $Q=(u,v)$,$$\begin{cases}u=\frac{x}{1-y} \\ v=0 \end{cases}$$