What is the possible solutions for an $3×3$ size orthogonal matrix with the known parameters of the first two columns which makes 6 parameters known?
Let's think of a rotation matrix which is an orthogonal matrix:
R=[r11 r12 r13; r21 r22 r23; r31 r32 r33]
The parameters of r11, r21, r31, r12, r22, r32 are known values. And it is desired to find the last (third) column with the parameters of r13,r23,r33.
I take the square root of each raw which is equivalent to 1 unit vector.\
r11^2+r12^2+r13^2=1r21^2+r22^2+r23^2=1r31^2+r32^2+r33^2=1
Therefore we can get to the equations of
r13= sqrt(1-r11^2-r12^2)r23= sqrt(1-r21^2-r22^2)r33= sqrt(1-r31^2-r32^2)
However, if there is one solution for the rotation matrix of R, how can I find that unique solution. Square root of any raw gives me two solutions.
Whichever (positive/negative square root of r13,r23,r33 I would take, it would be still an orthogonal matrix:
r1^2+r23^2+r33^2=1
So is there one unique solution or multiple solutions?
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