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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.

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

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  • $\begingroup$ Thank you! wasted a whole hour on that! $\endgroup$ – segevp Feb 17 at 9:15
  • $\begingroup$ But is this what you got or not? $\endgroup$ – José Carlos Santos Feb 17 at 9:18
  • $\begingroup$ yes, this is what I got :) $\endgroup$ – segevp Feb 17 at 9:19

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