Quaternion - trigonometric form - $q=\cos \theta +u \sin \theta$ Components for $u$?

Question

It is proven that a quaternion has the following trigonometric form: $$q=\cos \theta +u \sin \theta.$$

My question is: Which are the components of the $u$?

Thanks!

Answer 1

Just take the pure imaginary part of $\bf q$ and normalize it to get $\bf u$.

The pure imaginary part of ${\bf q}=a+b{\bf i}+c{\bf j}+d{\bf k}$ is the $b{\bf i}+c{\bf j}+d{\bf k}$ part.

Normalizing a vector means dividing it by its length, i.e. ${\bf v}\mapsto {\bf v}/\|{\bf v}\|$.

So $\displaystyle {\bf u}=\frac{b}{\sqrt{b^2+c^2+d^2}}{\bf i}+\frac{c}{\sqrt{b^2+c^2+d^2}}{\bf j}+\frac{d}{\sqrt{b^2+c^2+d^2}}{\bf k}$.

(Note there is some sign ambiguity, since ${\bf q}=\cos(-\theta)+(-{\bf u})\sin(-\theta)$ too.)