1
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

enter image description here

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.