How to find Rotation Matrix with 6 known parameters?

[ercandu...@gmail.com]

Dear people,

Assume that we have a R rotation matrix with first two columns are known. By using the orthoganility, how can I find the whole rotation matrix? Actually I follow these steps below however I would receive multiple solutions. (2 solutions for each, that makes 2*2*2=8 eight solutions.) What I am doing is correct or I miss something? Is there one solution or multiple solutions?

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=1

r21^2+r22^2+r23^2=1

r31^2+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?

[Erik van Oene]

The columns are unit length and orthogonal. So you could use the cross product of the first 2 columns to get the last column. Please use the correct order such that you end up with a right handed coordinate frame (I expect that is your goal).

Erik

Op maandag 1 februari 2021 om 09:07:22 UTC+1 schreef ercandu...@gmail.com: