Matrix transform question about angle of rotation

[Gregg]

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

(a) find the axis of the rotation

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

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

I found a perpendicular vector.

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

Transform the perpendicular vector.


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

Product of the perpendicular and transformed perpendicular

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

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

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

Is the perpendicular vector wrong? Am I trying to solve this correctly?

 

[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.

 

[Gregg]

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.

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

$<br /> \Rightarrow x+y+z=0<br />$


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

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


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


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

$\Rightarrow \theta = 120$

 

[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.

 

[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.

Ah, I see you fixed it. Much better.

 

[Gregg]

Ah, I see you fixed it. Much better.

Yep made an error.Thanks for the help.