Can anyone help with this question?
Compute the vector $\vec{r} = 20_{x0}-10_{y0}+30_{z0}$ after it is rotated first around ${x0}$ by 30 degrees, and then around ${y0}$ by 60 degrees.
Attempt at a Solution which is wrong:
$\begin{pmatrix}1 & 0 & 0 \\ 0 & cos \varphi & -\sin \varphi \\ 0 & \sin \varphi & \cos \varphi \end{pmatrix}\begin{pmatrix}\cos\vartheta & 0 & -\sin\vartheta \\ 0 & 1 & 0 \\ \sin\vartheta & 0 & \cos \vartheta \end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & cos \varphi & -\sin \varphi \\ 0 & \sin \varphi & \cos \varphi \end{pmatrix} ^T \begin{pmatrix}1 & 0 & 0 \\ 0 & cos \varphi & -\sin \varphi \\ 0 & \sin \varphi & \cos \varphi \end{pmatrix} \begin{pmatrix} 20\\-10\\ 30\end{pmatrix}=\begin{pmatrix} 35.98\\-7.5\\ -7.009\end{pmatrix}$
$\varphi=30 deg, \vartheta = 60 deg$
$$ $$
Answer is \begin{pmatrix} 14.085\\-11.83\\ -6.83\end{pmatrix}