Matrix transform question about angle of rotation

Gregg

[latex] \left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}
\right) [/latex] represents a rotation.

(a) find the axis of the rotation

[latex]
\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)
[/latex]

[latex]
\Rightarrow y=x=z
[/latex]
(b) what is the angle of rotation

I found a perpendicular vector.

[latex]
\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
[/latex]

Transform the perpendicular vector.


[latex] \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) [/latex]

Product of the perpendicular and transformed perpendicular

[latex]
\left(
\begin{array}{c}
-1 \\
1 \\
1
\end{array}
\right) \times
\left(
\begin{array}{c}
1 \\
1 \\
-1
\end{array}
\right) = -2i-2k [/latex]

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

[latex]
\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[/latex]

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

[latex]
\left(
\begin{array}{c}
1 \\
1 \\
1
\end{array}
\right).\left(
\begin{array}{c}
x \\
y \\
z
\end{array}
\right)=0[/latex]

[latex]

\Rightarrow x+y+z=0

[/latex]


[latex] b=\left(
\begin{array}{c}
-1 \\
0 \\
1
\end{array}
\right)[/latex]

[latex]

\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)[/latex]


[latex]
\left(
\begin{array}{c}
-1 \\
0 \\
1
\end{array}
\right).\left(
\begin{array}{c}
0 \\
1 \\
-1
\end{array}
\right) = -1[/latex]


[latex] -1 = \sqrt{2}\sqrt{2}\cos \theta[/latex]

[latex] \Rightarrow \theta = 120 [/latex]

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

Ah, I see you fixed it. Much better.

Gregg

Yep made an error.Thanks for the help.