Let $\mathbf{R}=[r_1 \ r_2 \ r_3]$, $\mathbf{Q}=[q_1 \ q_2 \ q_3]$
be two different 3-dimensional rotation matrices built from orthonormal column vectors $r_i$ and $q_i$ appropriately.
Question:
- Could be obtained general form for a matrix $\mathbf{ Q}$ given the matrix $\mathbf{R}$ that it would be satisfied $${\sum_{i=1}^3}q_i + {\sum_{i=1}^3}r_i =0$$ Obvious solution is $q_i=-r_j$ and its "orthogonal" permutations.
- Are other solutions possible also? If not how to prove it ?
If $\mathbf Q=\begin{bmatrix}\mathbf q_1 & \mathbf q_2 & \mathbf q_3\end{bmatrix}$ and $\mathbf 1 = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$, then $\mathbf{Q1} = \mathbf q_11 + \mathbf q_21 + \mathbf q_31 = \sum_{i=1}^3\mathbf q_i.$ Similarly, $\mathbf{R1} = \sum_i\mathbf r_i$. So you want to find a rotation matrix $\mathbf Q$ such that $\mathbf{Q1} = -\mathbf{R1}$, that is, a rotation that maps $\mathbf 1$ to $-\mathbf{R1}$. One solution is a rotation about the axis $\mathbf 1\times(-\mathbf{R1})$ by the angle $\cos^{-1}\Bigl(\frac{\mathbf 1\cdot(-\mathbf{R1})}{\|\mathbf 1\|\|\mathbf{-R1}\|}\Bigr)$. Infinitely many other solutions can be obtained by right-multiplying this matrix by an arbitrary rotation about the $\mathbf 1$ axis.