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Suppose I have a vector $P_0$ in a direct orthonormed system and a rotation of this vector defined by a rotation matrix $RM$.

Which of the two following equations is correct to find the new coordinates of the vector $P_0$ rotated onto a new vector $P_1$ ?

$$ \vec{\mathbf P}_1 \begin{bmatrix} \vec x_1 \\ \vec y_1 \\ \vec z_1 \end{bmatrix} = RM \times \vec{\mathbf P}_0 \begin{bmatrix} \vec x_0 \\ \vec y_0 \\ \vec z_0 \end{bmatrix} \tag1 $$ or $$ \vec{\mathbf P}_0 \begin{bmatrix} \vec x_1 \\ \vec y_1 \\ \vec z_1 \end{bmatrix} = RM \times \vec{\mathbf P}_1 \begin{bmatrix} \vec x_0 \\ \vec y_0 \\ \vec z_0 \end{bmatrix} \quad \rightarrow \quad \vec{\mathbf P}_1 \begin{bmatrix} \vec x_1 \\ \vec y_1 \\ \vec z_1 \end{bmatrix} = RM^{-1} \times \vec{\mathbf P}_0 \begin{bmatrix} \vec x_0 \\ \vec y_0 \\ \vec z_0 \end{bmatrix} \tag2 $$ (Original image of equations)

I have heard both of them from different sources and I am now confused.

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It seems that your confusion is between an active and a passive transformation.

To say that $R_M$ is the matrix that rotates a vector $P_0$ to a vector $P_1$ means that if we represent $P_0$ in the standard basis, than $P_1=R_M P_0$ is the rotated vector in the same standard basis. And this is an active rotation.

The inverse matrix $R_M^{-1}$ represents the inverse rotation in the standard basis, i.e. the rotation such that $P_0=R_M^{-1}P_1$.

This inverse matrix can be used to find the components of the starting vector $P_0$ in a new basis basis $B$ that is rotated by $R_M$ with respect to the standard basis, in the sense that the components of $P_0$ in this basis $B$ are given by $[P_0]_B=R_M^{-1}P_0$. And this is a passive rotation.

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  • $\begingroup$ I don't quite get it. On this page, a change of basis is described as "new coordinates" = RM x "old coordinates" that is to say in the same way as an active transformation or alibi. Shouldn't it be a passive transformation formulated as "new coordinates" = RM-1 x "old coordinates" ? What are the consequences on the results ? $\endgroup$ – Fjaero_ Oct 26 '16 at 14:25
  • $\begingroup$ It is a common confusion derived from notation. You can see that the matrix $ \begin{bmatrix}\cos \theta&\sin \theta\\-\sin \theta & \cos \theta\end{bmatrix}$ is the inverse of the matrix that represents a rotation of a vector of angle $\theta$ ( in standard basis) so it is a passive rotation, that is it represents a vector in a rotated reference system. $\endgroup$ – Emilio Novati Oct 26 '16 at 14:36
  • $\begingroup$ So this is more a transposed matrix as a inverted matrix ? And the impact should only concern the direction of the rotation, not the values of the vector's components ? $\endgroup$ – Fjaero_ Oct 26 '16 at 15:18
  • $\begingroup$ Yes. the point is that $\cos(-\theta)=\cos(\theta)$ and $\sin(-\theta)=-\sin(\theta)$ and $\cos^2 \theta +\sin^2\theta=1$. So inverse and transpose are the same for such matrices. $\endgroup$ – Emilio Novati Oct 26 '16 at 15:23

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