Today I started studying Rotation matrices and derived a rotation matrix like this:
\begin{equation*}
\begin{pmatrix}
x & y
\end{pmatrix}
\begin{pmatrix}
\cos(\beta) & \sin(\beta) \\
-\sin(\beta) & \cos(\beta)
\end{pmatrix}
\end{equation*}
When I was googling for the solution to check if this was ok, I found almost everybody expressed it like this:
\begin{equation*}
\begin{pmatrix}
\cos(\beta) & -\sin(\beta) \\
\sin(\beta) & \cos(\beta)
\end{pmatrix}
\begin{pmatrix}
x \\ y
\end{pmatrix}
\end{equation*}
If I'm not mistaken, this leads to the same equations, but is there anything else special about expressing it in these 2 different forms? Has this something to do with Handedness?
Best Answer
The difference is that you used a row vector and the other expression uses a column vector. And they are the same if you transpose one of them. \begin{align} \left( \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} \cos(\beta) & \sin(\beta) \\ -\sin(\beta) & \cos(\beta) \end{pmatrix} \right)^T &= \begin{pmatrix} \cos(\beta) & \sin(\beta) \\ -\sin(\beta) & \cos(\beta) \end{pmatrix} ^T \begin{pmatrix} x & y \end{pmatrix} ^T \\ &= \begin{pmatrix} \cos(\beta) & -\sin(\beta) \\ \sin(\beta) & \cos(\beta) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \end{align}