I'm trying to figure out the general form for the matrix (let's say in $\mathbb R^3$ for simplicity) of a rotation of $\theta$ around an arbitrary vector $v$ passing through the origin (look towards the origin and rotate counterclockwise). This is inspired by a similar problem which asked me to find the matrix for a rotation of $120^\circ$ around the vector $v=\begin{bmatrix}1&1&1\end{bmatrix}^\top$. However, in this case I was able to cheat a little since the transformation corresponds to a rotation of the vertices. So even though I found a solution, I'm not satisfied with my methodology. Is there a general form for rotation around an arbitrary vector in $\mathbb R^3$? A reference would be perfectly acceptable. Thanks.
Linear Algebra – Matrix for Rotation Around a Vector
geometrylinear algebramatricestransformation
Related Solutions
Stick your thumb of your right hand along the direction of the rotation axis. A positive angle is along the way your fingers point.
Like this
http://electron9.phys.utk.edu/Collisions/rotational_motiondetails.htm
You can find $R$ as $PSP^{-1}$ where $P$ is a rotation (orthonormal transformation) that maps the $x$-axis to the axial vector $(1,2,3)$, and $S$ is a rotation of angle $\alpha$ around the $x$-axis.
Finding $S$ is easy: $$S = \begin{bmatrix}1&0&0\\0&\cos\alpha&-\sin\alpha\\0&\sin\alpha&\cos\alpha\end{bmatrix}$$
Finding $P$ is more complex. Firstly, $Pe_1={(1,2,3)^T\over\sqrt{1^2+2^2+3^2}}$. Next, we meed to make sure that $Pe_i \perp Pe_j$ for all $i \neq j$, and $|Pe_1|=|Pe_2|=|Pe_3|=1$; for this we can use the cross product: Make $Pe_2={(1,2,3)^T\times e_1 \over |(1,2,3)^T \times e_1|}$, which is the unit vector perpendicular to $e_1$ and $(1,2,3)^T$, and similarly make $Pe_3 = {((1,2,3)^T\times e_1) \times (1,2,3)^T \over |(((1,2,3)^T\times e_1) \times (1,2,3)^T|}$, which is perpendicular to both $Pe_1$ and $Pe_2$ and of unit length. This satisfies all the conditions we need for $P$.
A remark: A key step in finding $P$ is finding a pair of vectors perpendicular to the axial vector and to each other. There is no continuous mapping that does this because of the hairy ball theorem. In spite of that, there are algorithms for doing this (see Yves Daoust's answer).
Best Answer
To settle this question: one can use the Rodrigues rotation formula to construct the rotation matrix that rotates by an angle $\varphi$ about the unit vector $\mathbf{\hat u}=\langle u_x,u_y,u_z\rangle$ (and if your vector is not a unit vector, normalization does the trick). Letting
$$\mathbf W=\begin{pmatrix}0&-u_z&u_y\\u_z&0&-u_x\\-u_y&u_x&0\end{pmatrix}$$
the Rodrigues rotation matrix is constructed as
$$\mathbf I+\left(\sin\,\varphi\right)\mathbf W+\left(2\sin^2\frac{\varphi}{2}\right)\mathbf W^2$$
where $\mathbf I$ is an identity matrix.
(For those of an advanced bent, one constructs $\mathbf W$ from $\mathbf{\hat u}$ through a premultiplication with the Levi-Civita tensor.)
Conventionally, the scalar multiplying the $\mathbf W^2$ term above is written as $1-\cos\,\varphi$, but this version is more prone to subtractive cancellation when $\varphi$ is near $2k\pi$ ($k$ is an integer), so the expression with the sine is more numerically sound.