The matrix of a proper rotation $R$ by angle $θ$ around the axis u = $(u_x, u_y, u_z)$, a unit vector with $u_x^2 + u_y^2 + u_z^2=1$ is given by:(as per wikipedia {https://en.wikipedia.org/wiki/Rotation_matrix})

$R=\begin{bmatrix}\cos \theta +u_{x}^{2}\left(1-\cos \theta \right)&u_{x}u_{y}\left(1-\cos \theta \right)-u_{z}\sin \theta &u_{x}u_{z}\left(1-\cos \theta \right)+u_{y}\sin \theta \\u_{y}u_{x}\left(1-\cos \theta \right)+u_{z}\sin \theta &\cos \theta +u_{y}^{2}\left(1-\cos \theta \right)&u_{y}u_{z}\left(1-\cos \theta \right)-u_{x}\sin \theta \\u_{z}u_{x}\left(1-\cos \theta \right)-u_{y}\sin \theta &u_{z}u_{y}\left(1-\cos \theta \right)+u_{x}\sin \theta &\cos \theta +u_{z}^{2}\left(1-\cos \theta \right)\end{bmatrix}$

What I want to show is, if $P \in S^2$ i.e if $P = (p_1, p_2, p_3)$ then $p_1^2 + p_2^2 +p_3^2 = 1$ and if $R( \theta_1 )*P = R( \theta_2 )*P$ and if $P\neq (u_x, u_y, u_z)$ and $P\neq (-u_x, -u_y, -u_z)$ then, $\theta_1 = \theta_2 + 2*k*\pi$ such that $k \in Z$.

As per my understanding: $(u_x, u_y, u_z), (-u_x, -u_y, -u_z) \in S^2$ and $P$ does not lie on the axis of the rotation.

My idea: If $R( \theta_1 )*P = R( \theta_2 )*P$, then

$R(- \theta_1 ) R( \theta_1 )*P = R(- \theta_1 )R( \theta_2 )*P$

=> $R(- \theta_1 )R( \theta_2 )*P = P$

=> $R(\theta_2 - \theta_1) *P = P$

I have the below information (let's assume $\theta_2 - \theta_1 = \theta_3)$:

  1. $p_1^2 + p_2^2 +p_3^2 = 1$
  2. $u_x^2 + u_y^2 + u_z^2=1$
  3. $P\neq (u_x, u_y, u_z)$ and $P\neq (-u_x, -u_y, -u_z)$
  4. $R(\theta_3) *P = P$ (we get 3 equations)
  5. Rotations preserve distances:

$(((\cos \theta_3 +u_{x}^{2}\left(1-\cos \theta_3 \right))*p_1)$ + $((u_{x}u_{y}\left(1-\cos \theta_3 \right)-u_{z}\sin \theta_3) * p_2)$ +$((u_{x}u_{z}\left(1-\cos \theta_3 \right)+u_{y}\sin \theta_3)*p_3))^2$ +

$(((u_{y}u_{x}\left(1-\cos \theta_3 \right)+u_{z}\sin \theta_3)*p_1)$ + $((\cos \theta_3 +u_{y}^{2}\left(1-\cos \theta_3 \right))*p_2)$ + $((u_{y}u_{z}\left(1-\cos \theta_3 \right)-u_{x}\sin \theta_3)*p_3))^2$ +

$(((u_{z}u_{x}\left(1-\cos \theta_3 \right)-u_{y}\sin \theta_3)*p_1)$+ $(u_{z}u_{y}\left(1-\cos \theta_3 \right)+u_{x}\sin \theta_3)*p_2)$+ $((\cos \theta_3 +u_{z}^{2}\left(1-\cos \theta_3 \right)*p_3))^2$ = 1

How to show that $\theta_3=2*k*\pi$ from the above information?

Also, $R(\theta_3) *P = P$ => $R(- \theta_3) *P = P$

If there is anything wrong or a mistake in my question, please let me know.


1 Answer 1


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I have proved if $u = (u_x, 0, u_z)$ then $\sin\theta_3 = 0$.

Maybe this will give some idea on how to prove if $u=(u_x, u_y, u_z)$.


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