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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 ?
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  • $\begingroup$ $\sum\mathbf q_i = \mathbf{Q1}$ 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
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    $\begingroup$ $\sum \mathbf q_i = \mathbf{Q1}$ 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
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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.

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  • $\begingroup$ 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

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