What is the possible solutions for an $3x3$ 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:
$R= \begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}$
The parameters of $r_{11},r_{21},r_{31}, r_{12},r_{22},r_{32}$ are known values. And it is desired to find the last (third) column with the parameters of $r_{13},r_{23},r_{33}$.
Is there a unique solution? What are/is the possible solution(s)?
I am confused with possible solutions, therefore I wanted to ask to be sure, how to find it.
Thanks in advance.
Edit: I take the square root of each raw which is equivalent to 1 unit vector.
$r_{11}^2+r_{12}^2+r_{13}^2=1$
$r_{21}^2+r_{22}^2+r_{23}^2=1$
$r_{31}^2+r_{32}^2+r_{33}^2=1$
Therefore we can get to the equations of
$r_{13}= \pm \sqrt[]{1-r_{11}^2-r_{12}^2}$
$r_{23}= \pm \sqrt[]{1-r_{21}^2-r_{22}^2}$
$r_{33}= \pm \sqrt[]{1-r_{31}^2-r_{32}^2}$
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 $r_{13},r_{23},r_{33}$) I would take, it would be still an orthogonal matrix:
$r_{13}^2+r_{23}^2+r_{33}^2=1$
An ortogonal matrix is a matrix whose columns are an orthonormal basis of vectors. So, if you are given the first two columns, $(r_{11},r_{21},r_{31})$ and $(r_{12},r_{22},r_{32})$, the third column must ve a vector which forms an orthonormal base with the other two. That is, it is a vector perpendicular to the other two columns and of norm 1. If you are told it is a rotation matrix, there will be just 1 possible vector. To calculate it you just have to take the cross product of the two other columns.
The cross product of two perpendicular unit vectors $v_1$, $v_2$ is the only vector $v_3$ such that $v_1,v_2,v_3$ form a orthonormal basis and the determinant of the matrix with columns $v_1,v_2,v_3$, in that order, is positive. So you have to take the third column to be the cross product of the first column and the second column.