# 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.

Thank you for your attention.

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:

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

• So, you get uu = -u.u from that identity and the fact that uxu=0(zero vector), but should not the result of a cross product be a vector instead of a scalar(zero)? So, how could uu=-u.u ? Sep 9 '20 at 14:58
• I know it will not change anything here in these equations when 0 or 0 vector used but actually i want to know we could always use zero scalar and zero vector interchangeably? Sep 9 '20 at 15:08
• Here math.stackexchange.com/questions/2456909/… they said in the answers, they are usually not equal or equal, check especially the last one "So the zero vector quantity is the zero scalar quantity (in whatever dimension)." Sep 9 '20 at 15:14
• @lockedscope You’re forgetting that the vectors and scalars are embedded in the algebra of the quaternions, and that we don’t need to distinguish vectors and scalars any longer. The operations are just quaternions addition and multiplication. The real parts of quaternions are interpreted as scalars, and the pure complex parts are interpreted as vectors. Everything is consistent. Sep 9 '20 at 15:19
• You could say that the quaternions zero is the pair consisting of the scalar zero and the vector zero. Sep 9 '20 at 15:22

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 $uu=-u \cdot u$.

• 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... Feb 4 '13 at 18:42
• Ha! That should teach you not to leave your answers over lunch. ;-) Feb 4 '13 at 18:44
• Thank you both. Now I will have problems with choosing the answer to accept. :) Feb 4 '13 at 19:01
• @Pippo: Toss a coin. :-) Feb 4 '13 at 19:04
• Sorry Hans, a random number generator has decided. ;) Feb 4 '13 at 19:08