In a Solar Power Tower, mirrors in a field, re-direct the sunlight to a central reciever.  The problem is to find the Azimuth and Elevation of the mirror, to do so. The sun is moving and the Target is stationary. The givens are the Azimuth and Elevation of the Sun, and the coordinates of the Tower. These are 2 vectors - sort of. Where I am going with this is, that the vector that bisects these two 3 dimensional vectors, is the sum of the unitized Sun and Target vector.  This is easy to prove. For a parallelogram, the diagonal does not bisect the angle. But for a rhombus, it does. A rhombus is a parallelogram with equal sides. Hence, unitizing the Sun vector and the Target vector. 


    Let AZ(imuth)sun and EL(evation) sun be given. These 2 values, in degrees, need to be turned into the x,y,z components of the SUN's vector. The conversion (courtesy of is Sz = Sin(Elevation); Hypotenuse = Cos(Elevatio) ; Sy = Hyp * Cos(Azimuth); Sx = Hyp * Sin(Azimuth). So the Sun vector is Sx*i+Sy*j+Sz*k. This needs to be a unit vector. So it is divided by its magnitude, or by Smag = sqrt(Sx^2+Sy^2+Sz^2). The Sun vector is Sx/Smag*i + Sy/Smag*j + Sz/Smag*k   The target vector is Tx*i+Ty*j+Tz*k. This also needs to be a unit vector.  Tmag = sqrt(Tx^2+Ty^2+Tz^2). So the Target vector is i*(Tx/Tmag ) + j*(Ty/Tmag  ) + k*(Tz/Tmag ). The Mirror vector is obtained as the sum of the Sun Unit Vector and the Target Unit Vector. It is i*(Tx/Tmag + Sx/Smag) + j*(Ty/Tmag + Sy/Smag) + k*(Tz/Tmag + Sz/Smag). This sum is not a unit vector. The sum needs to be converted to AZimuth_mirror and ELevation_mirror. I am working on this.