I need to determine the matrix that represents the following rotation of $\mathbb{R}^3$.
(a) angle $\theta$, the axis $e_2$
(b) angle $2\pi/3$, axis contains the vector $(1,1,1)^t$
(c) angle $\pi/2$, axis contains the vector $(1,1,0)^t$
Now, I would like to check if I got the right answers because this problem has been quite difficult for me. Any help is greatly appreciated.
Please forgive me for skipping the work because formatting matrices is a real pain. Especially when I have a lot of them.
For part $(a)$, I got that $(e_2,e_3,e_1)$ is an orthonormal basis of $\mathbb{R}^3$. Then after simplification, the matrix is
$$\left( \begin{matrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \\ \end{matrix}\right) $$
For part $(b)$, I got an orthonormal basis as $\{\left[1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}\right]^t, \left[1/\sqrt{2},-1/\sqrt{2},0\right]^t,\left[1/\sqrt{6},1/\sqrt{6},-2/\sqrt{6}\right]^t\}$.
Then after simplification, the matrix is
$$\left( \begin{matrix} -\frac{\sqrt{3}}{2} & 0 & -\frac12 \\ 0 & 1 & 0 \\ \frac12 & 0 & -\frac{\sqrt{3}}{2} \\ \end{matrix}\right) $$
Is what I have done so far correct such that I can proceed with part $(c)$?
If you mean to rotate by $\theta$ around an axis $u$, and $u$ is a unit vector, then the general formula is:
$R = \begin{bmatrix} \cos \theta +u_x^2 \left(1-\cos \theta\right) & u_x u_y \left(1-\cos \theta\right) - u_z \sin \theta & u_x u_z \left(1-\cos \theta\right) + u_y \sin \theta \\ u_y u_x \left(1-\cos \theta\right) + u_z \sin \theta & \cos \theta + u_y^2\left(1-\cos \theta\right) & u_y u_z \left(1-\cos \theta\right) - u_x \sin \theta \\ u_z u_x \left(1-\cos \theta\right) - u_y \sin \theta & u_z u_y \left(1-\cos \theta\right) + u_x \sin \theta & \cos \theta + u_z^2\left(1-\cos \theta\right) \end{bmatrix}$