# Rotation by exchange of components

Consider the rotation matrices around e.g. the $$z$$-axis and the $$y$$-axis

$$R_z(\phi) = \left[ \begin{matrix} \cos \phi & - \sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 &1 \end{matrix} \right] \quad , \qquad R_y(\phi) = \left[ \begin{matrix} \cos \phi & 0 & \sin \phi \\ 0 & 1 & 0 \\ -\sin \phi & 0 & \cos \phi \end{matrix} \right]$$

Deriving those matrices is no problem. However, for the sake of comprehension, I tried to obtain them differently. Let $$v$$ be the vector that I want to rotate. My thinking is, that if I

1. switch $$z$$ and $$y$$ components,
2. then rotate about the $$z$$-axis
3. and switch components back again,

I should get the same result as if I had rotated around the $$y$$-axis. But if I calculate this, I get something different.

$$R_y(\phi) \overset{!}{=} \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right] \cdot R_z(\phi) \cdot \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right]$$

$$\Longrightarrow \quad R_y(\phi) \overset{?}{=} \left[ \begin{matrix} \cos \phi & 0 & \color{red}{-}\sin \phi \\ 0 & 1 & 0 \\ \color{red}{+} \sin \phi & 0 & \cos \phi \end{matrix} \right]$$

The issue is that you did an unusual coordinate transform when you did switch the $$z$$ and $$y$$ components. I assume that you want to preserve rotations, so the determinant should be $$1$$. To achieve this, here are several options (choose only one):
• switch $$x$$ to $$-x$$. You will then get a right hand coordinate system.
• swap $$z$$ and $$-y$$.