Write the transformation as a map $ \ T:\mathbb{R}^3 \to \mathbb{R}^3 \ $

Question

Let $ \ T: \mathbb{R}^3 \to \mathbb{R}^3 \ $ be the linear transformation obtained by $ \ 120^{\circ} \ $ rotation about the line $ l: \ x=\lambda , \ y=\lambda , \ z=-\lambda \ ; \ \ where \ \ \lambda \in \mathbb{R} $

(a) Write the transformation as a map $ \ T:\mathbb{R}^3 \to \mathbb{R}^3 \ $ given by $ \ T(\vec u)=A \cdot \vec u \ $

(b) Find the image of the plane $ \ P: \ x+2y-2z=6 \ $ under the transformation $ \ T \ $. What Geometrical object do you obtain?

Answer:

I need the linear transformation $ \ T \ $ obtained by $ \ 120^{\circ} \ $ rotation about the line $ \ l: x=\lambda , \ y=\lambda , \ z=-\lambda \ $

I know the rotation matrix about a point , about an axis but not about a line

I need help right here.

Answer 1

We first set up a new orthonormal basis $(f_1,f_2,f_3)$ adapted to the given $T$. To this end choose $$f_3:={1\over\sqrt{3}}(1,1,-1), \quad f_1:={1\over\sqrt{2}}(-1,1,0), \qquad f_2:=f_3\times f_1={1\over\sqrt{6}}(1,1,2)\ .$$ The transform matrix from the standard basis $(e_1,e_2,e_3)$ to $(f_1,f_2,f_3)$ then is $$S:=\left[\matrix{-{1\over\sqrt{2}}&{1\over\sqrt{6}}&{1\over\sqrt{3}}\cr {1\over\sqrt{2}}&{1\over\sqrt{6}}&{1\over\sqrt{3}}\cr 0&{2\over\sqrt{6}}&-{1\over\sqrt{3}}\cr}\right]\ .$$ With respect to the basis $f:=(f_1,f_2,f_3)$ the given transformation $T$ has the matrix $$[T]_f=\left[\matrix{-{1\over2}&-{\sqrt{3}\over2}&0\cr {\sqrt{3}\over2}&-{1\over2}&0\cr 0&0&1\cr}\right]$$ (we have chosen the sense of rotation here). According to the rules of linear algebra the matrix of $T$ with respect to the standard basis then is given by $$[T]_e=S\> [T]_f\>S^\top\ .$$ As for the second problem: Coordinatewise the map ${\bf x}\mapsto {\bf x}':=T{\bf x}$ is given by $${\bf x'}=[T]_e\>{\bf x}\ .$$ This implies $${\bf x}=[T]_e^{-1}{\bf x}'=[T]_e^\top{\bf x}'\ .\tag{1}$$ This allows to express the coordinates $(x,y,z)$ of a preimage point ${\bf x}$ by the coordinates $(x',y',z')$ of the image point ${\bf x}'$. The point ${\bf x}$ lies in the plane $P$ iff its coordinates satisfy $x+2y-2z=6$. Plug the expressions in terms of $x'$, $y'$, $z'$ resulting from $(1)$ into this equation, and you obtain the equation of the image plane $T(P)$.

Answer 2

$R=\begin{pmatrix}\cos\frac{2\pi}3&\sin\frac{2\pi}3&0\\-\sin\frac{2\pi}3&\cos\frac{2\pi}3&0\\0&0&1\end{pmatrix}$ is rotation about the $z$-axis. Similarly one can do the $x$ and $y$ axes...

We will form an orthogonal matrix,$$R_1=\begin{pmatrix}\frac1{\sqrt6}&\frac1{\sqrt6} &\frac2{\sqrt6}\\\frac1{\sqrt2}&-\frac1{\sqrt2}&0\\\frac1{\sqrt3}&\frac1{\sqrt3}&-\frac1{\sqrt3}\end{pmatrix}$$. The top row will be the cross-product of the other two. This matrix makes the last row the new $z$-axis. (Note: there's a question of orientation here. I chose $(1,1,-1)$. The opposite vector could have been chosen.)

Finally, the rotation will be given by $R_1^{-1}RR_1$...

Of course, $(R_1)^{-1}=R_1^t$.

Check, for instance, that $(2,2,-2)$ is fixed...

For part $2$, the image should be another plane (the rotation of a plane is a plane).