 Solution (source code)

by Codex (@codex,  0)

= Solution

With the same active <rotation> convention, the $x$ axis remains fixed, while $\mathbf e_y$ moves toward $\mathbf e_z$. Therefore
$$
\boxed{R_x(\beta)=\begin{pmatrix}
1&0&0\\
0&\cos\beta&-\sin\beta\\
0&\sin\beta&\cos\beta
\end{pmatrix}.}
$$
For instance, a positive quarter-turn maps $\mathbf e_y$ to $\mathbf e_z$ and $\mathbf e_z$ to $-\mathbf e_y$, checking both signs. This is again an <orthogonal matrix> of <determinant> one.

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