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.

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