# Geometric algebra, rotations

## Main Question or Discussion Point

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.

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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.
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.

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 !

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

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.