Given two orthogonal unit vectors, A and B, and two different orthogonal unit vectors C and D, I need the 3×3 direction cosine matrix or a quaternion which will rotate A to align with C AND will rotate B to align with D.
The vectors are all 3-vectors (x, y, z).
I have a brute force algorithm, but I am almost certain there is a much simpler solution. And my web searches have not yet revealed one.
I am using C#, but if you have something in C, or FORTRAN, or Basic (etc.) I can convert it. Or, I can use the terms written out mathematically.
The application is determining the orientation needed for a spacecraft so an optical device attached rigidly to it will be properly aligned to take a picture. (Both the bore sight direction and the appropriate rotation of the optic device about bore sight must be attained, thus the need to align TWO vectors at once.) The computation may be used in a loop fed by real-time sensed data, and the brute force solution is too slow.
I reread your question and the answer below (while right) doesn’t give you what you want. This link is about constructing a 3×3 rotation matrix.
since they are both orthogonal unit vectors, you just need to add one more too each to construct a basis (use the cross product for this). So now you have two basis {A, B, AxB} and {C, D, CxD}. A rotation that moves {A, B} onto {C, D} will re-express a vector a1A + a2B + a3(AXB) as b1C + b2D + b3(CxD). because it’s linear you only need to know how it behaves on the basis (This uniquely determines the linear transformation). So, taking {A, B, ..} as our basis and letting the transformation be T, we see that T(1, 0, 0) = C, T(0, 1, 0) = D and T(0, 0, 1) = CxD. Remember A = (1, 0, 0) etc. But the columns of this matrix are just M =(C, D, CxD)
To use this matrix as it stands, you have to express every vector in the basis {A, B, CxD} before you left-multiply it by M. You do this in the same way. In fact, it N is the matrix which translates from your normal basis to {A, B, ..} and M above translates that to {C, D…}, then MN (left multiplication here) will translate from your basis to {C, D, ..} and provide the rotation you want.
So now, all of your vectors are expressed in the basis {C, D, ..} 🙁
The solution is yet another transformation matrix. This one maps from {A, B, ..} to your primary basis and undoes N, also known as the inverse and denoted N^-1. So your final matrix is (N^-1)MN. The good news is that because N is orthogonal, you just need it’s transpose.
The trick is to choose your primary basis so that the matrices you deal with most are pretty.