Matrix of rotation in $\mathbb R^{3}$

Question

I'm attempting to teach myself basic multivariable calculus and so far finding some of the 3d visualizations a bit difficult. Here's an example:

Let $f: \mathbb R^{3} \longrightarrow \mathbb R^{3}$ be the rotation by $\pi$ about the line $x_1 = x_2$ and $x_3 = 0$ in the $x_1x_2$ plane. What is the matrix of $f$ with respect to the standard bases?

I suppose the line is the intersection of the plane $x_1 = x_2$ with plane $x_3 = 0$. However, it's unclear to me just what thing (for lack of a better term) is being rotated when we say "rotated about such-and-such in the so-and-so (plane, etc)".

Silly question, but what would be the difference from just rotating the axes $x_1$ and $x_2$ themselves by $\pi$, vs. what is specifically asked to be rotated in this problem?

The more images and visual intuitions one can offer, the better. Thank you.

Answer 1

For any vector $v\in\Bbb R^3$, write it as $v_1+v_2$, where $v_1$ is a multiple of $(1,1,0)$ (that is, $v_1$ belongs to the given line) and $v_2$ is orthogonal to $v_1$. Then $f(v)=f(v_1+v_2)=v_1-v_2$.

So, since $(1,0,0)=\left(\frac12,\frac12,0\right)+\left(\frac12,-\frac12,0\right)$, you have$$f(1,0,0)=\left(\frac12,\frac12,0\right)-\left(\frac12,-\frac12,0\right)=(0,1,0).$$By the same reason, $f(0,1,0)=(1,0,0)$. Finally, $f(0,0,1)=-(0,0,1)=(0,0,-1)$. Therefore, the matrix that you are after is$$\begin{bmatrix}0&1&0\\1&0&0\\0&0&-1\end{bmatrix}.$$

Answer 2

The rotation matrix to rotate polar angle by $\alpha$:

$$ R(\alpha)= \left[ {\begin{array}{ccc} cos(\alpha) & -sin(\alpha) & 0 \\ sin(\alpha) & cos(\alpha) & 0 \\ 0 & 0 & 1 \end{array} } \right]$$

$$ T= \frac{1}{\sqrt{2}} \left[ {\begin{array}{ccc} 0 & 0 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & 0 \end{array} } \right]$$

1. Change the line/axis $x_1 = x_2$, $x_3 = 0$ to $x_1 = x_2 = 0$ line/axis. This is done by multiplying by $T$.
2. Rotate by desired angle $\phi_r$ by multiplying $R(\phi_r)$.
3. change $x_1 = x_2 = 0$ axis to $x_1 = x_2$, $x_3 = 0$ by multiplying by $T^{-1}$

Overall Transform for rotating a vector $v$ angle $\phi_r$ about the $x_1 = x_2$, $x_3 =0$ line is: $$v_{rotated} = T^{-1}R(\phi_r)Tv$$

Answer 3

Construct an orthonormal basis for $\mathbb{R}^3$ as follows

$ v_1 = ( 1, 1, 0 ) / \sqrt{2} $

$ v_2 = (-1, 1, 0 ) / \sqrt{2} $

$ v_3 = (0, 0, 1 ) $

Let $ A = [v_1, v_2, v_3] = \dfrac{1}{\sqrt{2}} \begin{bmatrix} 1 && -1 && 0 \\ 1 && 1 && 0 \\ 0 && 0 && \sqrt{2} \end{bmatrix} $

Then given a point $P$, its representation in the above basis is given by

$ Q = A^T P = [q_1, q_2, q_3] $

Now, rotating $Q$ about the given line, is rotating $Q$ about the first basis vector, and since the rotation angle is $\pi$, the rotated image is

$ Q' = [ q_1, -q_2, -q_3 ] = B Q $ where $ B = \begin{bmatrix} 1 && 0 && 0 \\ 0 && -1 && 0 \\ 0 && 0 && -1 \end{bmatrix} $

The corresponding coordinates in the standard basis are

$ P' = A Q' = A B Q = A B A^T P $

Therefore, the rotation matrix is

$ R = A B A^T = \dfrac{1}{2} \begin{bmatrix} 1 && -1 && 0 \\ 1 && 1 && 0 \\ 0 && 0 && \sqrt{2} \end{bmatrix} \begin{bmatrix} 1 && 0 && 0 \\ 0 && -1 && 0 \\ 0 && 0 && -1 \end{bmatrix} \begin{bmatrix} 1 && 1 && 0 \\ -1 && 1 && 0 \\ 0 && 0 && \sqrt{2} \end{bmatrix} $

Calculating the product of the last two matrices,

$ R = \dfrac{1}{2} \begin{bmatrix} 1 && -1 && 0 \\ 1 && 1 && 0 \\ 0 && 0 && \sqrt{2} \end{bmatrix} \begin{bmatrix} 1 && 1 && 0 \\ 1 && -1 && 0 \\ 0 && 0 && - \sqrt{2} \end{bmatrix}$

Multiplying these two matrices yields

$ R = \begin{bmatrix} 0 && 1 && 0 \\ 1 && 0 && 0 \\ 0 && 0 && -1 \end{bmatrix} $