# Given a rotation matrix about a specific axis show that if $R(\theta_1)*p$ = $R(\theta_2)*p$ then $\theta_1=\theta_2+2*\pi*k$ where $k \in Z$.

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.

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 $$u=(u_x, u_y, u_z)$$.