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Find $3 \times 3$ orthogonal matrix Q such that: $$Q \left[\begin{array}{r} 3\\ 0 \\ 4\end{array}\right] = \left[\begin{array}{r} 5\\ 0 \\ 0\end{array}\right]$$

How can I find matrix Q if I only know the info like above. Thank you for your help.

Alexander
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  • http://math.stackexchange.com/questions/180418/calculate-rotation-matrix-to-align-vector-a-to-vector-b-in-3d – Yimin May 11 '15 at 01:09
  • http://math.stackexchange.com/questions/114512/how-to-find-the-orthonormal-transformation-that-will-rotate-a-vector-to-the-x-ax – Yimin May 11 '15 at 01:11
  • I don't quite understand those posts, can you please explain to me ? – Alexander May 11 '15 at 01:16
  • since orthogonal matrix represents a rotation or refection, in 3D, there are two variables to control a rotate. they just did that. In you case, since the second term of vector does not change, you should only consider a matrix transferring $(3,4)'$ to $(5,0)'$. – Yimin May 11 '15 at 01:19

2 Answers2

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Since your second term of vector does not change. $A$ is in form of

$\begin{pmatrix}\cos\theta &0& \sin\theta\\ 0 & 1 & 0\\ -\sin\theta & 0 & \cos\theta\end{pmatrix}$

then just find out $\theta$.

$\cos \theta = 0.6$ and $\sin\theta = 0.8$

Yimin
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  • It would also be $\begin{pmatrix}\cos\theta &0& -\sin\theta\ 0 & 1 & 0\ -\sin\theta & 0 & -\cos\theta\end{pmatrix}$ –  May 11 '15 at 02:18
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Another answer: $$Q=I-\frac2{\bf d\cdot d}{\bf d}{\bf d}^T\quad\hbox{where}\quad {\bf d}=\pmatrix{3\cr0\cr4\cr}-\pmatrix{5\cr0\cr0\cr}\ .$$

David
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  • The form of $Q$ above makes it a Householder Reflector: http://en.wikipedia.org/wiki/Householder_transformation. Such matrices are nice because they are Hermitian, unitary, idempotent, and have unit magnitude eigenvalues. – Xoque55 May 11 '15 at 02:26