# Geometric algebra, rotations

1. Sep 5, 2014

Hi, I want to calculate a rotation of a vector GA style with this formula $e^{-B \frac{\pi}{2}}(2e_{1}+3e_{2}+e_{3})e^{B\frac{\pi}{2}}$. Now since no book/pdf on GA exists where a calculation is explicitly done with numbers, I wounder how to calculate this. Should I substitude $e^{-B \frac{\pi}{2}}=cos (\frac{\pi}{2})-Bsin(\frac{\pi}{2})$(same for the other term) and see what I get, or should I convert the vector into a complex number and then convert it into the exponential form and then multiply it? Or how is this done? By the way, how do I convert a vector in $R^{3}$ into a complex number? Multiply on the left with $e_{1}e_{2}$? In $R^{2}$ it is just multiplication by $e_{1}$ on the left.

2. Sep 5, 2014

### Pond Dragon

Is $B$ the imaginary unit?

Your question is unclear. I am particularly confused by your conjugation by "$e^{-B\frac{\pi}{2}}$," since I'm not sure how it is acting.

3. Sep 5, 2014

Hi Pond Dragon. Sry, the general formula for a rotation is defined as $a^{,}= RaR^{\dagger}$ Where $R=nm$ and $R^{\dagger}=mn$. Now one uses $B= \frac{m\wedge n}{sin \phi}$, $B^{2}=-1$ (unit bivector) and rewrites $R=cos \frac{\phi}{2} -B sin \frac{\phi}{2}=e^{-B \frac{\phi}{2}}$ and $R^{\dagger}= cos \frac{\phi}{2} +B sin \frac{\phi}{2}=e^{B \frac{\phi}{2}}$. Hope that helps !

4. Sep 5, 2014

### Pond Dragon

So, you don't know how to compute the product? It looks like you could just apply definitions.

5. Sep 5, 2014

Hi again, so you think this is the right way

$(cos \frac{\pi}{2}-Bsin\frac{\pi}{2})(2e_{1}+3e_{2}+e_{3})=-e_{1}B2-e_{2}B3-e_{3}B$

then from the right
$(-e_{1}B2-e_{2}B3-e_{3}B)(cos\frac{\pi}{2}+Bsin\frac{\pi}{2})=-2e_{1}-3e_{2}+e_{3}$
It should be a rotation of 90 degrees, but the two vectors are not orthogonal hm.