# Detail in the proof of the quaternion rotation identity

I am trying to understand the proof of the quaternion rotation identity illustrated in wikipedia (http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation#Proof_of_the_quaternion_rotation_identity). I cannot understand the development of the last term in the first passage, i.e. why this should be true:

$\vec{u}\vec{v}\vec{u}=\vec{v}(\vec{u}\cdot\vec{u})-2\vec{u}(\vec{u}\cdot\vec{v})$

$\vec{u}$ and $\vec{v}$ are pure imaginary quaternions.

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Further up the page, there is the identity: $uv=u\times v-u\cdot v$. Using this and the fact that $uu=-u\cdot u$, we have:

$$uv=u\times v-u\cdot v\\=-v\times u +u\cdot v -2(u\cdot v)\\=-vu-2(u\cdot v)$$

Mulitplying on the right by $u$, you have the identity:

$$uvu=(-vu-2(u\cdot v))u=-vuu-2u(u\cdot v)=v(u\cdot u)-2u(u\cdot v)$$

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The rule $uv = -u \cdot v + u \times v$ gives $vu = -v \cdot u + v \times u = -u \cdot v - u \times v$, and hence $uv+vu = -2 u \cdot v$. In other words, $uv = -vu - 2 u \cdot v$. Now multiply this by $u$ from the right and use $uu=-u \cdot u$.

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I had left my answer on my machine over lunch and was surprised no solutions had arrived... but then seconds before I hit submit you answered too! Strange timing on both our parts... – rschwieb Feb 4 '13 at 18:42
Ha! That should teach you not to leave your answers over lunch. ;-) – Hans Lundmark Feb 4 '13 at 18:44
Thank you both. Now I will have problems with choosing the answer to accept. :) – Pippo Feb 4 '13 at 19:01
@Pippo: Toss a coin. :-) – Hans Lundmark Feb 4 '13 at 19:04
Sorry Hans, a random number generator has decided. ;) – Pippo Feb 4 '13 at 19:08