Angle between two quaternions

[Fidelis]

Hello guys!

I have two quaternions with norm equal to 1. Both are represented in the angle-phase form, i.e, I have q=exp(i*\phi)exp(k*\psi)exp(j*\theta) and p=exp(i*\phi')exp(k*\psi')exp(j*\theta'). Let \alpha be the angle between q and p. I need to write \alpha in function of \phi-\phi', \psi-\psi' and \theta-\theta' in a simple way. Could anyone give me some idea?

 

[elfmotat]

Given two unit quaternions, [math]p[/math] and [math]q[/math], you can always find another unit quaternion [math]r[/math] such that [math]rq=p[/math] which represents the amount you need to move [math]q[/math] to match [math]p[/math]. Solving for [math]r[/math] gives:

 

[math]r = pq^{-1} = p~ \textup{Conj}(q)[/math]

 

The angle between [math]p[/math] and [math]q[/math] is the angle of [math]r[/math], which is given by:

 

[math]\alpha =2 \textup{cos}^{-1}(\textup{Re}(r )) =2 \textup{cos}^{-1}(\textup{Re}(p~ \textup{Conj}(q)))[/math]

Edited April 13, 2016 by elfmotat