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

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?

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.

• Your attempt at a solution is a restatement of the first part of the problem. Do you have any thoughts on how you might solve it?
– amd
Jun 29 '18 at 1:52
• The problem statement is a bit ambiguous as to the direction of rotation. Are we meant to assume some orientation of the line from its description?
– amd
Jun 29 '18 at 1:53
• In what direction are you rotating clock wise of counter clockwise? Jun 29 '18 at 2:04
• counterclockwise ?
– user484305
Jun 29 '18 at 2:09
• So $A$ is the matrix representation of $T$ wrt the standard basis? Use the theorem that tells you how to do this, starting with the image under $T$ of each basis vector. Jun 29 '18 at 2:13

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 $T(P)$.

$$R=\begin{pmatrix}\cos\frac{2\pi}3&\sin\frac{2\pi}3&0\\-\sin\frac{2\pi}3&\cos\frac{2\pi}3&0\\0&0&1\end{pmatrix}$$ 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).

• two more possibilities are also there no?
– user464147
Jun 29 '18 at 2:35
• I guess you're right.
– user403337
Jun 29 '18 at 2:36
• okay. I think he has to specify the direction. Else, question is incomplete.
– user464147
Jun 29 '18 at 2:52