I have an equation which is the Rodrigues Rotation formula but rewritten different.

$$R=p \cos(nw)+\sin(nw)(tr-vq)+(1-\cos(nw))(ps+qt+rv)s$$

Where $p,w,t,r,v,q,s$ are all the unknown variables and $R,n$ are known. For every positive value $n$ there is a known value for $R$, therefore one can obtain the 7 equations needed to solve for the 7 unknowns. But of course, due to the similarities between these 7 equations I wonder if there’s is a better and easier way to solve for the unknowns. Also, what if instead of 7 variables there are 14, or 21, or any multiple of 7 variables in the same conditions, where for any positive value $n$ there is a known value $R$.

Another case I need to solve is for 8 unknowns, where the only difference is the following

$$R=p \cos(nw)+\sin(nw)(tr-vq)+(1-\cos(nw))(ps+qt+rv)s+x$$

Where the same question applies to this case, and what would happen if the unknowns would double or triple in this case ?

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  • $\begingroup$ So it appears you are calculating a rotation of a vector of some kind, and you then want to do the process in reverse. But with a geometric interpretation, you would just have to make the inverse rotation to arrive at the original vector. Am I all wrong here, or how does this sound like to you? $\endgroup$ – Matti P. Aug 14 at 9:28
  • $\begingroup$ But if I don’t know the values for the matrix I initially applied to an unknown vector, how can I compute the inverse matrix if I only know the output ? $\endgroup$ – ijk-1 Aug 14 at 9:44

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