(Exer 3.27) Consider the plane $H$ determined by the equation $x + y -z = 0$.
What is a unit normal vector to $H$?
Compute the image of $X:=H\cap \mathbb S^{2}$ under the stereographic projection $\Phi$.
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- For 2, I computed $X$ to be a 3D circle, parametrised here, and its image to be $Y:= \Phi(X) = \{|z-(1+i)|^2 = 3\}$, but now I ask:
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For 1, What's the relevance of asking about the unit normal vector?
I computed the unit normal vectors to be $[1,1,-1]\frac{\pm 1}{\sqrt{3}}$. I observe their terminal points to be on the unit sphere.
Here are the parametrisations:
$Y:= \Phi(X) = \{|z-(1+i)|^2 = 3\}$ is parametrised:
$$\begin{bmatrix} y_1(t)\\ y_2(t)\\ y_3(t) \end{bmatrix} = \begin{bmatrix} \sqrt{3}\cos(t) + 1\\ \sqrt{3}\sin(t) + 1\\ 0 \end{bmatrix} = \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} + \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\sqrt{3}\cos(t)+ \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}\sqrt{3}\sin(t)$$
$X$ is parametrised:
$$\begin{bmatrix} x_1(t)\\ x_2(t)\\ x_3(t) \end{bmatrix} = \begin{bmatrix} \sqrt{\frac 2 3} \cos[t]\\ -\sqrt{\frac 2 4} \sin[t] - \sqrt{\frac 2 {12}} \cos[t]\\ -\sqrt{\frac 2 4} \sin[t] + \sqrt{\frac 2 {12}} \cos[t] \end{bmatrix} = \begin{bmatrix} \sqrt{\frac 1 3}\\ -\sqrt{\frac 1 {12}}\\ \sqrt{\frac 1 {12}} \end{bmatrix}\sqrt{2}\cos(t)+ \begin{bmatrix} 0\\ -\sqrt{\frac 1 {4}}\\ -\sqrt{\frac 1 {4}} \end{bmatrix}\sqrt{2}\sin(t)$$
$$H = \{x + y -z = 0\} = \{[1,1,-1] \cdot [x,y,z]=0\} = \{[1,1,-1]\frac{1}{\sqrt{3}} \cdot [x,y,z]=0\}$$ is parametrised:
$$\begin{bmatrix} h_1(r,s)\\ h_2(r,s)\\ h_3(r,s) \end{bmatrix}=\begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}r + \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}s$$
Here is one possible explanation. We are to find the image of $H\cap{\Bbb S}^2$ under $\Phi$.
A solution may go like this: $$ Y\in\Phi(H\cap{\Bbb S}^2)\iff X=\Phi^{-1}(Y)\in H\cap{\Bbb S}^2 \subseteq H. $$ The latter can be described by $ax_1+bx_2+cx_3+d=0$, where $(a,b,c)$ is a normal vector, giving an equation for $Y$ once the mapping $\Phi^{-1}$ is explicitly calculated.
EDIT: What I mean here is that $\Phi$ is the mapping $$ \Phi\colon \underbrace{(x_1,x_2,x_3)}_{X}\in{\Bbb S}^2\to \underbrace{(y_1,y_2,y_3)}_{Y}=\left(\frac{x_1}{1-x_3},\frac{x_2}{1-x_3},0\right)\in{\Bbb R}^2. $$ It is a bijection and $$ \Phi^{-1}\colon(y_1,y_2,0)\to\Big(\underbrace{\frac{2y_1}{y_1^2+y_2^2+1}}_{x_1},\underbrace{\frac{2y_2}{y_1^2+y_2^2+1}}_{x_2},\underbrace{\frac{y_1^2+y_2^2-1}{y_1^2+y_2^2+1}}_{x_3}\Big). $$ The necessary and sufficient condition for $Y$ to belong to the image is that $X$ (on the sphere) belongs to the plane $H$ (if a point is on the sphere, but not on $H$, then it does not belong to $H\cap{\Bbb S}^2$, hence, not mapped to the image due to one-to-one mapping), i.e. orthogonal to the normal vector $$ x_1+x_2-x_3=0. $$ When we substitute (and get rid of the denominator) we get the circle $$ 2y_1+2y_2-(y_1^2+y_2^2-1)=0\iff (y_1-1)^2+(y_2-1)^2=3. $$