# Stereographic Projection: Find the point $X$ where the plane $E$ meets the sphere $S$

Let $S$ denote the unit sphere

$$S = \{(x,y,z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 = 1\}$$

and $E$ denote the plane in $\mathbb{R}^3$ given by $z = 0$

$$E = \{(x,y,z) \in \mathbb{R}^3 \mid z = 0\}.$$

If $(u,v,0)$ is a point of $E$ then the line joining $(u,v,0)$ to $(0,0,1)$ meets $S$ in a point other than $(0,0,1)$. Denote this point by $X(u,v)$.

Compute the formula for $X$. [HINT: Any point on the line joining $(u,v,0)$ and $(0,0,1)$ is of the form $\lambda \cdot (u,v,0) + (1 - \lambda) \cdot (0,0,1)$ for some $\lambda \in \mathbb{R}$. We need to determine such $\lambda_0 \in \mathbb{R}$ that $X(u,v) = \lambda_0 \cdot (u,v,0) + (1 - \lambda_0) \cdot (0,0,1)$ lies on $S$.]

So clearly I need to work out that $\lambda_0$ but I'm not exactly sure how to do this. I'm thinking this, in order for it to lie on $S$, the sum of the $x,y,z$ components, when squared, should equal $1$. So if I expand and write it out in terms of $x,y,z$, I get:

$$x = \lambda_0 \cdot u$$ $$y = \lambda_0 \cdot v$$ $$z = 1 - \lambda_ 0 = 0$$

From here, I get $\lambda_0 = \frac{x}{u} = \frac{v}{y}$, but this then doesn't make sense as I want to write it as componenets of $x,y$.

Can someone give me some hints on what to do please?

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Let $r = |(u,v,0)|$ and $t = \tan \alpha = \frac{1}{r}$, then \begin{align} \sin 2\alpha &= \frac{2t}{1+t^2}, \\ \cos 2\alpha &= \frac{1-t^2}{1+t^2}. \end{align}