# Find point $X$ such that line through plane $E$ and sphere $S$ meet at $(0,0,1)$ (stereographic projection)

Find the point $X$ such that the line going through the plane $E$ and sphere $S$ meet at the point $(0,0,1)$ (stereographic projection).

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 $X(u,v)$.

1) 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 $S$.]

For this bit, I said using the fact that

$$\pmatrix{x\\y\\x} = \lambda_0 \cdot \pmatrix{u \\ v \\ 0} + (1 - \lambda_0)\cdot \pmatrix{0 \\ 0 \\ 1},$$

we get

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

Putting it in $x^2 + y^2 + z^2 = 1$ and solving gives us $\lambda_0 = 0, \frac{2}{1 + u^2 + v^2}$. When $\lambda_0 = 0$, we get the point $(0,0,1)$, and according to the question the point that meets $S$ but isn't $(0,0,1)$ and so we pick $\lambda_0 = \frac{2}{1 + u^2 + v^2}$. So we end up getting

$$X(u,v) = \left( \frac{2u}{1 + u^2 + v^2}, \frac{2v}{1 + u^2 + v^2}, 1 - \frac{2}{1 + u^2 + v^2}\right).$$

2) Show that the map $X: \mathbb{r}^2 \rightarrow \mathbb{R}^3$ determines a refular surface patch. [HINT: Prove that $X_u \circ X_v = 0$, then show that $a \circ b =$ implies that the vectors $a$ and $b$ are linearly independent]

Here $a \circ b$ is the dot product between $a$ and $b$ and $X_u$ is the partial derivative of $X$ with respect to $u$. So the first thing to do was the partial derivatives and I got them to be

$$X_u = \left(\frac{2}{1 + u^2 + v^2} - \frac{4u^2}{(1 + u^2 + v^2)^2}, -\frac{4uv}{(1 + u^2 + v^2)^2}, \frac{4u}{(1 + u^2 + v^2)^2} \right)$$ $$X_v = \left( -\frac{4uv}{(1 + u^2 + v^2)^2}, \frac{2}{(1 + u^2 + v^2)} - \frac{4v^2}{(1 + u^2 + v^2)^2}, \frac{4v}{(1 + u^2 + v^2)^2} \right)$$

but when I then do the dot product, I don't get them to be $0$. I get it to be

$$\left(\frac{2}{1 + u^2 + v^2} - \frac{4u^2}{(1 + u^2 + v^2)^2} \right) \cdot \left( -\frac{4uv}{(1 + u^2 + v^2)^2} \right) + \left(-\frac{4uv}{(1 + u^2 + v^2)^2} \right) \cdot \left(\frac{2}{(1 + u^2 + v^2)} - \frac{4v^2}{(1 + u^2 + v^2)^2} \right) + \left(\frac{4u}{(1 + u^2 + v^2)^2} \right) \cdot \left(\frac{4v}{(1 + u^2 + v^2)^2} \right),$$

which gives me

$$\frac{16u^3v}{(1 + u^2 + v^2)^4} - \frac{8uv}{(1 + u^2 + v^2)^3} + \frac{16v^3u}{(1 + u^2 + v^2)^4} - \frac{8uv}{(1 + u^2 + v^2)^3} + \frac{16uv}{(1 + u^2 + v^2)^4}$$ $$= \frac{16u^3v}{(1 + u^2 + v^2)^4}+ \frac{16v^3u}{(1 + u^2 + v^2)^4} - \frac{16uv}{(1 + u^2 + v^2)^3} + \frac{16uv}{(1 + u^2 + v^2)^4} \neq 0$$

Where am I making my mistake?

3) How much of the sphere is covered by the parametrization $X$?

I also haven't got a clue how to do this bit. Maybe once I sort the first two parts out, I might get an idea.

EDIT: Also, I'm thinking that my $X$ is wrong as when I do $x^2 + y^2 + z^2$, I don't get it to equal $1$. I'm not sure if it should or not. If it did, then it would lie on the surface of the sphere and not inside it. The question doesn't necasarrily say that it needs to lie on the surface, but I thought if I used the constraint $x^2 + y^2 + z^2 = 1$ then I've moved to the point on the line that does lie on the surface and so I should get $x^2 + y^2 + z^2 = 1$ for my $x,y,z$ coordinates for $X$, right?

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## 1 Answer

I do calculate zero: \begin{align} &\frac{16u^3v}{(1 + u^2 + v^2)^4}+ \frac{16v^3u}{(1 + u^2 + v^2)^4} - \frac{16uv}{(1 + u^2 + v^2)^3} + \frac{16uv}{(1 + u^2 + v^2)^4} \\ =&\frac{16u^3v +16uv^3 + 16uv}{(1 + u^2 + v^2)^4} - \frac{16uv(1 + u^2 + v^2)}{(1 + u^2 + v^2)^4} \\ =&\frac{16u^3v +16uv^3 +16uv - 16uv - 16u^3v - 16uv^3}{(1 + u^2 + v^2)^4}\\ =&\frac{0}{(1 + u^2 + v^2)^4}\\ =& 0 \end{align}

Calculating the norm squared of $X(u,v)$: \begin{align} \Vert X(u,v) \Vert^2 &= \frac{4u^2}{(1 + u^2 + v^2)^2} + \frac{4v^2}{(1 + u^2 + v^2)^2}+ \left(\frac{1 + u^2 + v^2}{1 + u^2 + v^2} - \frac{2}{1 + u^2 + v^2}\right)^2 \\ &= \frac{4u^2}{(1 + u^2 + v^2)^2} + \frac{4v^2}{(1 + u^2 + v^2)^2} + \frac{(-1 + u^2 + v^2)^2}{(1 + u^2 + v^2)^2} \\ &= \frac{4n_2 +(- 1 +n_2)^2}{(1 + n_2)^2}\quad\text{(Let $n_2 = u^2 + v^2$)} \\ &= \frac{4n_2 + 1 - 2n_2 + n_2^2}{(1 + n_2)^2} \\ &= \frac{2n_2 + 1 + n_2^2}{1+2n_2 + n_2^2} \\ &= 1 \\ \end{align}

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Oh dang! I just assumed it would've all cancelled out really easily and didn't think about it properly, rushing too much :( Thanks for the help :) Do you know anything about the third question? – Kaish Feb 23 at 19:37
$x^2 + y^2 + z^2 = 1$ should indeed be true as you say. It should be as it is on the sphere. I imagine it is nice and complicated thus easy to make a mistake, but I will give it a shot in an edit. As far as the parameterization goes, it covers the entire sphere, except for $(0,0,1)$ unless you allow for either $u$ or $v$ to go to infinity. Proving it though, I don't know off-hand, though it should be easy enough. – adam W Feb 23 at 21:19
Edit finished, I do calculate $1$ as well... the key that avoided many possible mistakes there was to substitute $n_2= u^2 + v^2$ so that there were not too many terms to manipulate. – adam W Feb 23 at 21:33
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