## Quaternion explanation

A wonderful explanation of why $q p q^{-1}$ is rotation of $p$, where $q$ is most easily interpreted with the complex-number like structure of $\cos(\theta/2) + \sin(\theta/2)(a i + b j + c k)$ with $a i + bj +ck$ normalized (i.e., $a^2+b^2+c^2=1$). Note that $q^{-1}$ is then just $latex \cos(\theta/2) - \sin(\theta/2)(a i + b j + c k)$ (since $(a i + b j + c k)^2=-a^2 - b^2 -c^2 = -1$ just like the imaginary $i$ for complex number).

Basically quarternion is 4D stereographic projection back to 3D space. When multiply by a quarternion $\cos(\alpha) + \sin(\alpha) {\bf v}$ from the left, points parallel to $\bf v$ will be translate along $\bf v$ and points penpendicular to $\bf v$ will rotate around $\bf v$ (following the right hand rule). Similarly, when multiplying from the right by the same quarternion, points parallel to $\bf v$ will be translated in the same direction, but points perpendicular to $\bf v$ will be rotated in the opposite direction (following left hand rule). So if we multiple $q$ and $q^{-1}$ from both left and right hand sides, the translation effect will be exactly cancelled, and the the rotation will be doubled. That is why the angle in $q$ should be halved.