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I'm self learner and for some reason I can't wrap my head around quaternion multiplication. I just stumble upon one of equation in my text.
Can anyone show step-by-step workout for below: $$ \begin{equation} \begin{split} w =& qvq^* \\ =& (q_0+\vec{q})(0+\vec{v})(q_0-\vec{q}) \\ =&(2q_0^2-1)\vec{v}+2(\vec{q}\cdot\vec{v})\vec{q} + 2q_0(\vec{q}\times\vec{v}) \end{split} \end{equation} $$

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1 Answer 1

up vote 1 down vote accepted

Here's one way to do it. $\def\vq{{\bf q}} \def\vqb{{\bf q}^*} \def\vv{{\bf v}} \def\va{{\bf a}} \def\vb{{\bf b}} \def\vc{{\bf c}}$

$$\begin{eqnarray*} q v q^* &=& (v q + [q,v])q^* \\ &=& v \|q\|^2 + 2(\vq\times\vv) (q_0 - \vq) \\ &=& v \|q\|^2 + 2q_0(\vq\times\vv) - 2(\vq\times\vv)\vq \\ &=& v \|q\|^2 + 2q_0(\vq\times\vv) - 2\left[-(\vq\times\vv)\cdot\vq + (\vq\times\vv)\times\vq \right]\\ &=& v \|q\|^2 + 2q_0(\vq\times\vv) + 2\vq\times(\vq\times\vv) \\ &=& v \|q\|^2 + 2q_0(\vq\times\vv) + 2\vq(\vq\cdot\vv) - 2\vv(\vq\cdot\vq) \\ &=& v_0\|q\|^2 + (q_0^2-\vq\cdot\vq)\vv + 2(\vq\cdot\vv)\vq + 2q_0(\vq\times\vv) \\ &=& (2q_0^2-1)\vv + 2(\vq\cdot\vv)\vq + 2q_0(\vq\times\vv), \hspace{5ex} \textrm{if }v_0=0\textrm{ and }\|q\| = 1 \end{eqnarray*}$$

Some relevant links and formulas: $$\begin{eqnarray*} a b &=& (a_0+\va)(b_0+\vb) \\ &=& a_0b_0-\va\cdot\vb + a_0\vb+b_0\va+\va\times\vb \\ \hspace{0ex}[a,b] &=& ab-ba \\ &=& 2\va\times\vb \\ \|a\|^2 &=& a_0^2 + \va\cdot\va \\ &=& 1, \textrm{ for unit quaternion} \\ \va\cdot(\vb\times\vc) &=& \vb\cdot(\vc\times\va) \\ &=& \vc\cdot(\va\times\vb) \\ \va\times\va &=& 0 \\ \va\times(\vb\times\vc) &=& \vb(\va\cdot\vc) - \vc(\va\cdot\vb) \end{eqnarray*}$$

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thanks, I give up studying with my current quaternion book due to frustration.Now, I'm going to resume back, thanks once again. –  kypronite Oct 27 '12 at 4:22
    
@kypronite: Glad to help and good luck in your studies. –  user26872 Oct 27 '12 at 14:25

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