# Matrix transform question about angle of rotation

1. Jun 15, 2009

### Gregg

$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)$ 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?

2. Jun 15, 2009

### 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.

Last edited: Jun 15, 2009
3. Jun 15, 2009

### Gregg

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

$\Rightarrow x+y+z=0$

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

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

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

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

$\Rightarrow \theta = 120$

Last edited: Jun 15, 2009
4. Jun 15, 2009

### Dick

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.

5. Jun 15, 2009

### Dick

Ah, I see you fixed it. Much better.

6. Jun 15, 2009

### Gregg

Yep made an error.Thanks for the help.

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