Transformation of $(x,y,z)$ to $(x',y',z')$ after spherical coordinate rotation

Question

If I have a a vector $\vec{v}=(x,y,z)$ in cartesian coordinates and I want to rotate my coordinate system by spherical angles $\theta$ and $\phi$ how would I find the resulting vector $\vec{v'}=(x',y',z')$ in cartesian coordinates?

This website seems to contain the answer under the "sum-of-matrices" header, but fails to give any sort of derivation as to how they arrived there.

If anyone can give any insight into how this was derived, that would be greatly appreciated.

Answer 1

In the link you gave it gives the generic form of a rotation matrix of angle $\theta$ around an axis $r$. There are two alternative ways to arrive to that relation: i) composing rotation matrices, ii) constructing the rotated vector geometrically.

i) From: Siciliano, Sciavicco, Villani, Oriolo - Robotics, Modelling, Planning and Control, Sec.2.5:

"

1. Align $r$ with $z$, which is obtained as the sequence of a rotation by $−\alpha$ about $z$ and a rotation by $−\beta$ about $y$.
2. Rotate by $\theta$ about $z$.
3. Realign with the initial direction of $r$, which is obtained as the sequence of a rotation by $\beta$ about $y$ and a rotation by $\alpha$ about $z$.

In sum, the resulting rotation matrix is $R(\theta,r) = R_z(\alpha)R_y(\beta)R_z(\theta)R_y(−\beta)R_z(−\alpha)$.

"

where $R_y(\beta) = \begin{bmatrix} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{bmatrix}$ and $R_z(\alpha) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

2) Is weel explained here https://en.wikipedia.org/wiki/Rodrigues%27_rotation_formula.

To help you directly with your specific example, you should say about which axes the rotations you want to do are, and in which order you want to perform them.

Answer 2

First express the axes of the rotated coordinate system in terms of the original one. The rotation is equivalent to two consecutive rotations: rotation about the $z$ axis by an angle $\phi$ followed by a rotation about the $y'$ axis by $\theta$, hence the rotation matrix representing the final $(x',y',z')$ is given by

$ R = \begin{bmatrix} \cos \phi && -\sin \phi && 0 \\ \sin \phi && \cos \phi && 0 \\ 0 && 0 && 1 \end{bmatrix} \begin{bmatrix} \cos \theta && 0 && \sin \theta \\ 0 && 1 && 0 \\ -\sin \theta && 0 && \cos \theta \end{bmatrix} $

This evaluates to

$ R = \begin{bmatrix}\cos \theta \cos \phi && -\sin \phi && \sin \theta \cos \phi \\ \cos \theta \sin \phi && \cos \phi && \sin \theta \sin \phi \\ -\sin \theta && 0 && \cos \theta \end{bmatrix}$

The coordinates in this rotated frame of a vector $\vec{v} = (x,y,z)$ is the vector $\vec{u} = (x',y', z') $ given by

$ \vec{u} = (x',y', z') = R^T (x, y, z) $

That is,

$ \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \cos \theta \cos \phi && \cos \theta \sin \phi && - \sin \theta \\ - \sin \phi && \cos \phi && 0 \\ \sin \theta \cos \phi && \sin \theta \sin \phi && \cos \theta \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}$