How to derive rotation matrix for quaternion

Question

I've been following the Wikipedia article on quaternions and spatial rotations and I've come across something I don't understand:

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Everything up to this point is clear yet I can't see how the first equation leads to the second. If someone could give a more clear explanation that would be greatly appreciated.

Thanks for any help

Answer 1

It comes down to straightforward manipulations, but a lot of them. First, simplify the scalar term and expand the cross product: \begin{align} s(-\mathbf v \cdot \mathbf p q_r + q_r \mathbf v\cdot \mathbf p&, \mathbf v(\mathbf v \cdot \mathbf p) + q_r^2 \mathbf p + q_r \mathbf v \times \mathbf p + \mathbf v \times (q_r \mathbf p + \mathbf v \times \mathbf p)) \\ & = s(0, \mathbf v(\mathbf v \cdot \mathbf p) + q_r^2 \mathbf p + q_r \mathbf v \times \mathbf p + \mathbf v \times (q_r \mathbf p + \mathbf v \times \mathbf p)) \\ & = s(0, \mathbf v(\mathbf v \cdot \mathbf p) + q_r^2 \mathbf p + q_r \mathbf v \times \mathbf p + q_r\mathbf v \times \mathbf p + \mathbf v \times (\mathbf v \times \mathbf p)) \\ & = s(0, \mathbf v(\mathbf v \cdot \mathbf p) + q_r^2 \mathbf p + 2q_r \mathbf v \times \mathbf p+ \mathbf v \times (\mathbf v \times \mathbf p)). \end{align} Now, rewrite each term as a matrix multiplied by $\mathbf p$:

• $\mathbf v(\mathbf v \cdot \mathbf p) = (\mathbf{v \otimes v})\mathbf p$
• $q_r^2\mathbf p = q_r^2 \mathbf I \mathbf p$
• $2q_r \mathbf v \times \mathbf p = 2q_r[\mathbf v]_\times \mathbf p$
• $\mathbf v \times (\mathbf v \times \mathbf p) = [\mathbf v]_\times^2 \mathbf p$.

From there, we simply factor the product.

$$ (\mathbf{v \otimes v})\mathbf p + q_r^2 \mathbf I \mathbf p + 2q_r[\mathbf v]_\times \mathbf p + [\mathbf v]_\times^2 \mathbf p = (\mathbf{v \otimes v} + q_r^2 \mathbf I + 2q_r[\mathbf v]_\times + [\mathbf v]_\times^2)\mathbf p. $$