Here is another way to get a RotationMatrix which transforms x to xp ( and vice versa)
rotM // MatrixForm RotationMatrix[Pi, (xx + xxp)/2] // MatrixForm RotationMatrix[Pi, (xx + xxp)/2].xx RotationMatrix[Pi, (xx + xxp)/2].xxp
Specifying the image of one vector only removes two degrees of freedom from the three present in the general 3D rotation, so your system of equations is underconstrained and does not have a unique solution.
Thanks, but these are huge expressions. I need short simple answers in degrees. Can you convert these answers in degrees?
For me works fine.
