# Rotation around the axis

I wish to rotate a body that is located on $$\hat z$$ axis.

If I rotate the body at angle $$\alpha$$ around the $$\hat x$$ axis and then at angle $$\beta$$ around the $$\hat y$$ axis then I think I should get:

$$R_y(\beta)R_x(\alpha)=\begin{bmatrix}\cos(\beta) && 0 && \sin(\beta)\\ 0 && 1 && 0 \\ -\sin(\beta) && 0 && \cos(\beta)\end{bmatrix} \begin{bmatrix}1 && 0 && 0\\ 0 && \cos(\alpha) &&-\sin(\alpha) \\ 0 && \sin(\alpha) && \cos(\alpha) \end{bmatrix}$$

According to my TA this should give:

$$\begin{bmatrix} \cos(\beta) && \cos(\alpha)\sin(\beta) && \sin(\alpha)\sin(\beta) \\ 0 && -\sin(\alpha) && \cos(\alpha) \\ -\sin(\beta) && \cos(\alpha)\cos(\beta) && \cos(\beta)\sin(\alpha) \end{bmatrix}$$

but when I multiply it myself I get something else.

Either I don't know how to multiply matrices, I did not use the the right rotation matrices, or the TA is wrong.

• The TA's answer cannot possibly be correct because its determinant is not $1$, which must be the case for two rotations. – Chrystomath Feb 17 at 9:11
• If the body is located on $z$ (?), there is no need to compute the whole matrix. – Yves Daoust Feb 17 at 9:22

The correct answer is$$\begin{bmatrix} \cos (\beta) & \sin (\alpha) \sin (\beta) & \cos (\alpha) \sin (\beta) \\ 0 & \cos (\alpha) & -\sin (\alpha) \\ -\sin (\beta) & \cos (\beta) \sin (\alpha) & \cos (\alpha) \cos (\beta) \end{bmatrix}.$$If this is what you got, then your TA is wrong.