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Matrix transformations

  1. Mar 20, 2009 #1
    1. The problem statement, all variables and given/known data

    The matrix

    [itex] \left[ \begin{array}{ccc} 0 &1 &0 \\ 0 &0 &1 \\ 1 &0 &0 \end{array} \right] [/itex]

    represents a rotation.

    (a) Find the equation of the axis of this rotation.
    (b) What is the angle of the rotation?

    2. Relevant equations

    [itex]\left[ \begin{array}{ccc} 1 &0 &0 \\ 0 &\cos\theta &-\sin\theta \\ 0 &\sin\theta &\cos\theta \end{array} \right] [/itex]

    [itex] \left[ \begin{array}{ccc} \cos\theta &0 &\sin\theta \\ 0 &1 &0 \\ -\sin\theta & 0 &\cos\theta \end{array} \right] [/itex]

    [itex]\left[ \begin{array}{ccc} \cos\theta &-\sin\theta &0 \\ \sin\theta &\cos\theta &0 \\ 0 &0 &1 \end{array} \right] [/itex]

    Rotations of \theta about x, y and z axes respectively.

    3. The attempt at a solution

    I thought this would just be a case of looking at the matrix and deciding whether it was a rotation about the x,y or z. I'm not sure how to determine the equation for the axis of rotation.

    I discovered that:

    [itex] \left[ \begin{array}{ccc} 0 &1 &0 \\ 0 &0 &1 \\ 1 &0 &0 \end{array} \right] \left[ \begin{array}{ccc} x_1 &x_2 &x_3 \\ y_1 &y_2 &y_3 \\ z_1 &z_2 &z_3 \end{array} \right] = \left[ \begin{array}{ccc} y_1 &y_2 &y_3 \\ z_1 &z_2 &z_3 \\ x_1 &x_2 &x_3 \end{array} \right] [/itex]

    But can't get close to the answer.
     
  2. jcsd
  3. Mar 20, 2009 #2

    alphysicist

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    Homework Helper

    Hi Gregg,

    I believe here you want to know how your rotation matrix affects a vector (not multiply it with another matrix).

    So for a general vector

    [tex]
    \vec r=x\hat i+y\hat j+z\hat k
    [/tex]

    you might try examining what happens to that vector when you apply your rotation matrix to it. What do you get? (Are you familiar with writing vectors as a matrix?)
     
  4. Mar 23, 2009 #3
    I'm new to matrices and such and I don't really understand it well. [itex]
    \vec r=x\hat i+y\hat j+z\hat k
    [/itex]

    I'm not sure how to find the axis of rotation since the matrix that is in the problem does not look similar to any of the standard results I'm given as in my relevant equations. Would you write the vector like this:

    [itex] \vec r=\begin{bmatrix} x &y &z \end{bmatrix} [/itex]

    and then

    [itex] \begin{bmatrix} x &y &z \end{bmatrix} \begin{bmatrix} 0 &1 &0 \\ 0 &0 &1 \\ 1 &0 &0 \end{bmatrix} = \begin{bmatrix} z &x &y \end{bmatrix} [/itex] ?

    then maybe...

    [itex] \vec r = z \hat i + x \hat j + y \hat k [/itex]?
     
    Last edited: Mar 23, 2009
  5. Mar 23, 2009 #4

    gabbagabbahey

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    Gold Member

    No, vectors are usually written as columns when they are being operated on:

    [tex]\vec r=\begin{bmatrix} x \\ y \\ z \end{bmatrix}[/tex]



    No, the rotation matrix operates on the vector not vice versa...the operator is always written to the left of what it operates on:

    [tex]\vec{r}'=\begin{bmatrix} 0 &1 &0 \\ 0 &0 &1 \\ 1 &0 &0 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} ? \\ ? \\ ? \end{bmatrix}[/tex]

    (I used a prime to denote the rotated vector)

    Now, suppose you chose a vector that was parallel to the axis of rotation; what could you say about r and r'?:wink:
     
  6. Mar 23, 2009 #5
    [itex]\vec r=\begin{bmatrix} x \\ y \\ z \end{bmatrix}[/itex]

    [itex] \vec r' = \begin{bmatrix} 0 &1 &0 \\ 0 &0 &1 \\ 1 &0 &0 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} y \\ z \\ x \end{bmatrix}[/itex]

    So

    [itex]\vec r=\begin{bmatrix} x \\ y \\ z \end{bmatrix}[/itex]

    and

    [itex] \vec r' = \begin{bmatrix} y \\ z \\ x \end{bmatrix}[/itex]

    I'm not sure if I could immediately take x = y = z though? and as for the angle of rotation...
     
  7. Mar 23, 2009 #6

    lanedance

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    Homework Helper

    hi gregg

    say we say

    [itex] \textbf{M} = \begin{bmatrix} 0 &1 &0 \\ 0 &0 &1 \\ 1 &0 &0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} [/itex]

    then if a is a vector parallel to your rotation axis, then

    [itex] \textbf{M} \textbf{a} = a [/tex] ie it is unchanged by the rotation

    this is a good check to see if you have the correct axis of rotation

    to work out the angle of rotation find a vector perpindicular to a, and apply the rotation, the angle will be given by teh angel between the initial and final vectors

    (also i found it helpful to draw how each axis x,y,z is rotated)
     
  8. Mar 23, 2009 #7

    lanedance

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    Homework Helper

    hi gregg

    say we say

    [itex] \textbf{M} = \begin{bmatrix} 0 &1 &0 \\ 0 &0 &1 \\ 1 &0 &0 \end{bmatrix} [/itex]

    then if a is a vector parallel to your rotation axis, then

    [itex] \textbf{M} \textbf{a} = \textbf{a} [/tex] ie it is unchanged by the rotation

    this is a good check to see if you have the correct axis of rotation

    to work out the angle of rotation find a vector perpindicular to a, and apply the rotation, the angle will be given by the angle between the initial and final vectors

    (also i found it helpful to draw how each axis x,y,z is rotated)
     
    Last edited: Mar 23, 2009
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