How can I calculate a rotation using geometric algebra?

[JonnyMaddox]

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.

 

[Pond Dragon]

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.

Would you please clarify?

 

[JonnyMaddox]

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 !
Thx for reply.

 

[Pond Dragon]

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 !
Thx for reply.

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

 

[JonnyMaddox]

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.