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If $g ∈ \mathrm{SO}(3)$ is the rotation about axis $p$ by angle $α$, and $h$ is a rotation mapping $p$ to another line $q$, then $g$ conjugated by $h$ is the rotation about $q$ by the same angle $α$. I've seen this fact presented as obvious, but I don't find 3D rotations trivial to visualize in general, and would like to see a formal proof. This could presumably be brute-forced in a number of ways, but I'm interested in an argument that avoids lengthy computation.

Since $hgh^{-1}$ obviously fixes $q$, that is indeed the axis of the resulting rotation, so it would suffice to prove it rotates $q^{\bot}$ by $α$.

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An easy way to see this is to use Rodrigues's rotation formula: $$ g{\bf v}=\cos \alpha \,{\bf v}+(1-\cos \alpha)({\bf u}\cdot {\bf v}){\bf u}+\sin \alpha \,{\bf u}\times {\bf v} $$

where ${\bf u}$ is the axis of rotation. Now it can be shown that conjugating $g$ by $h$ gives $$ hgh^{-1}{\bf v}=\cos \alpha \,{\bf v}+(1-\cos \alpha)((h{\bf u})\cdot {\bf v})h{\bf u}+\sin \alpha \,(h{\bf u})\times {\bf v}. $$

This is a rotation by $\alpha$ about axis $h{\bf u}$.

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  • $\begingroup$ Thanks for the answer. I was aware of the formula, but was hoping there was a more conceptual proof. I'll wait a bit more before accepting this just in case, but I doubt anyone else will comment. $\endgroup$ – user54748 Feb 8 '16 at 21:08

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