Equation for 3D vector rotation

350 Views Asked by At

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.

1

There are 1 best solutions below

4
On BEST ANSWER

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.