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Matrix transform question about angle of rotation

  1. Jun 15, 2009 #1
    [itex] \left(
    \begin{array}{ccc}
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    1 & 0 & 0
    \end{array}
    \right) [/itex] represents a rotation.

    (a) find the axis of the rotation

    [itex]
    \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)
    [/itex]

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

    I found a perpendicular vector.

    [itex]
    \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
    [/itex]

    Transform the perpendicular vector.


    [itex] \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) [/itex]

    Product of the perpendicular and transformed perpendicular

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

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

    [itex]
    \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[/itex]

    Is the perpendicular vector wrong? Am I trying to solve this correctly?
     
  2. jcsd
  3. Jun 15, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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
  4. Jun 15, 2009 #3
    [itex]
    \left(
    \begin{array}{c}
    1 \\
    1 \\
    1
    \end{array}
    \right).\left(
    \begin{array}{c}
    x \\
    y \\
    z
    \end{array}
    \right)=0[/itex]

    [itex]

    \Rightarrow x+y+z=0

    [/itex]


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

    [itex]

    \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)[/itex]


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


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

    [itex] \Rightarrow \theta = 120 [/itex]
     
    Last edited: Jun 15, 2009
  5. Jun 15, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  6. Jun 15, 2009 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ah, I see you fixed it. Much better.
     
  7. Jun 15, 2009 #6
    Yep made an error.Thanks for the help.
     
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