angle of rotation of a $3\times 3$ matrix about axis of $3\times 1$ vector.

Question

Matrix $$A=\begin{bmatrix} 0 & 0 & 1\\ 1 & 0 &0\\ 0& 1& 0\end{bmatrix}$$ rotates a vector about axis $B=\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}$ by an angle of rotation? Where I can find relevant information. and elementary methods to solve this kind of problem.

Answer 1

Define $B'$ as the image of $B$:

$B' = A\cdot B$

Now calculate the angle $\beta$ between vectors $B$ and $B'$ using

$\cos\beta=\frac{B\cdot B'}{||B||\cdot ||B'||}$

Answer 2

Notice that the vector B is fixed by the transformation; it is the the axis about which the rotation occurs. Now note that \begin{eqnarray*} \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \longrightarrow \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} \longrightarrow \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} \longrightarrow \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \end{eqnarray*} Does that help ?

EDIT: The following three vectors are all in the plane whose normal is $B$ \begin{eqnarray*} \begin{bmatrix} 1 \\ -1 \\ 0 \\ \end{bmatrix} \longrightarrow \begin{bmatrix} 0 \\ 1 \\ -1 \\ \end{bmatrix} \longrightarrow \begin{bmatrix} -1 \\ 0 \\ 1 \\ \end{bmatrix} \longrightarrow \begin{bmatrix} 1 \\ -1 \\ 0 \\ \end{bmatrix} \end{eqnarray*} Now use $ \cos \theta = \frac{a \cdot b}{ \mid a \mid \mid b \mid}$ to calculate the angle between them.

Answer 3

Given the information you provided, we can find the image of B on A: $AB = \begin{pmatrix} 0&0&1\\ 1&0&0 \\0&1 &0\end{pmatrix} \begin{pmatrix} 1\\ 1 \\1 \end{pmatrix} = \begin{pmatrix} 1\\ 1 \\1 \end{pmatrix}$

Now, $|AB||B|\cos(\theta) = (AB).B \Rightarrow \theta = \arccos(1) \therefore \theta = 0 \text{ rad}, 2\pi \text{ rad}$