## Matrix transform question about angle of rotation

$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)$ represents a rotation.

(a) find the axis of the rotation

$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} y \\ z \\ x \end{array} \right)$

$\Rightarrow y=x=z$
(b) what is the angle of rotation

I found a perpendicular vector.

$\left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right) \times \left( \begin{array}{c} -1 \\ 1 \\ 1 \end{array} \right) = 0 \Rightarrow \theta = 90$

Transform the perpendicular vector.

$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)\left( \begin{array}{c} -1 \\ 1 \\ 1 \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \\ -1 \end{array} \right)$

Product of the perpendicular and transformed perpendicular

$\left( \begin{array}{c} -1 \\ 1 \\ 1 \end{array} \right) \times \left( \begin{array}{c} 1 \\ 1 \\ -1 \end{array} \right) = -2i-2k$

this does not indicate the 120 degree rotation that i need.

$\left( \begin{array}{c} -1 \\ 1 \\ 1 \end{array} \right)\left( \begin{array}{c} 1 \\ 1 \\ -1 \end{array} \right) = -1 = \sqrt{3}\cos\theta \Rightarrow \theta = 109$

Is the perpendicular vector wrong? Am I trying to solve this correctly?
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 Recognitions: Homework Help Science Advisor The perpendicular vector is wrong. You want to solve (1,1,1).(a,b,c)=0 to find a perpendicular. The DOT product. Then you want to find the dot product of the perpendicular with the transformed perpendicular. No cross products necessary.

 Quote by Dick The perpendicular vector is wrong. You want to solve (1,1,1).(a,b,c)=0 to find a perpendicular. The DOT product. Then you want to find the dot product of the perpendicular with the transformed perpendicular. No cross products necessary.
$\left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right).\left( \begin{array}{c} x \\ y \\ z \end{array} \right)=0$

$\Rightarrow x+y+z=0$

$b=\left( \begin{array}{c} -1 \\ 0 \\ 1 \end{array} \right)$

$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)\left( \begin{array}{c} -1 \\ 0 \\ 1 \end{array} \right)=\left( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right)$

$\left( \begin{array}{c} -1 \\ 0 \\ 1 \end{array} \right).\left( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right) = -1$

$-1 = \sqrt{2}\sqrt{2}\cos \theta$

$\Rightarrow \theta = 120$

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## Matrix transform question about angle of rotation

b=(-1,0,1) is a good perpendicular. But the transformed perpendicular isn't (1,0,-1) is it? There's another mistake in the dot product that makes me think you just copied it wrong.

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