0
$\begingroup$

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 ?
$\endgroup$
  • $\begingroup$ $\sum\mathbf q_i = \mathbf{Q1}$ where $\mathbf1$ is the vector of all ones, so multiplying $-\mathbf Q$ with any orthogonal matrix that fixes $\mathbf1$ will do. $\endgroup$ – Rahul Nov 29 '16 at 18:01
  • $\begingroup$ @Rahul Are you sure? I have doubts... I have to check...and $R$ is given, $Q$ is searched for ... if $R$ is rotation matrix $(-R)$ is not a rotation matrix.. $\endgroup$ – Widawensen Nov 29 '16 at 18:06
  • 2
    $\begingroup$ $\sum \mathbf q_i = \mathbf{Q1}$ where $\mathbf 1$ is the vector of all ones, so any rotation matrix $\mathbf Q$ that maps $\mathbf 1$ to $-\mathbf{R1}$ will do. $\endgroup$ – Rahul Nov 29 '16 at 18:21
  • $\begingroup$ @Rahul Ok. We have $R1+ Q1=0$ then $ - Q1=R1$ ...so maps $1$ to $-R1$ ? but how to achieve it? $\endgroup$ – Widawensen Nov 29 '16 at 18:31
  • $\begingroup$ @Rahul So this $Q$ rotation is probably rotation about $-1\times{(R1)}$ axis. I see that the most important in a problem is to start with a good representation of data as you have shown. If you would like, Rahul, you can put your solution from comments as the answer.. $\endgroup$ – Widawensen Nov 29 '16 at 19:27
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ A rotation about the $\mathbf 1$ axis preserves the $\mathbf 1$ vector. So if $\mathbf A$ is such a rotation, then $\mathbf{A1}=\mathbf 1$ and so $\mathbf{QA1}=\mathbf{Q1}=-\mathbf{R1}$. Thus $\mathbf{QA}$ is also a solution. $\endgroup$ – Rahul Nov 30 '16 at 15:20
  • $\begingroup$ ok, I see. Maybe this comment would be also good to insert in the answer.. $\endgroup$ – Widawensen Nov 30 '16 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.