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Transformation Matrices

  1. Aug 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the components of A after a rotation of -45 degrees about X3.

    A=(1,1,2)

    2. Relevant equations

    [tex]\lambda[/tex]=

    (cos[tex]\theta[/tex] 0 -sin[tex]\theta[/tex])
    ( 0 1 0 )
    (sin[tex]\theta[/tex] 0 cos[tex]\theta[/tex])


    3. The attempt at a solution

    Above is my attempt to show you guys the rotation matrix when rotated about the X3 axis. So my strategy was to plug in -45 degrees into the thetas. Once I got a value for each element in the matrix, I checked to see if A' was equal to A. However, I ended up getting A'=(0,SQRT(2),2). This does not equal A.

    Where did I go wrong? Any help would be much appreciated.
     
  2. jcsd
  3. Aug 26, 2010 #2

    lanedance

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    shoudn't the compononents be different after a rotatino?

    now a couple of points on what you;ve done anyway...

    if you're rotating around X3, the X3 component should be unchanged, so the matrix you have isn't correct.

    even if it were correct, the multiplication doesn't look correct, the 1st component should be
    [tex]1.cos(\theta) + 1.0 2sin(\theta)[/tex]
     
  4. Aug 26, 2010 #3
    No. My X3 component did remain unchanged. It started out as 2 and after the transform it remained 2.

    I am confused as to where you got 1.cos(theta) + 1.02sin(theta).
     
  5. Aug 26, 2010 #4

    vela

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    What you've written so far doesn't make sense. You said you calculated

    [tex]\mathbf{A}' = \begin{bmatrix}\cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & \cos\theta\end{bmatrix} \begin{pmatrix} 1 \\ 1 \\ 2\end{pmatrix}
    = \begin{pmatrix}\cos\theta-2\sin\theta\\1\\\sin\theta+2\cos\theta\end{pmatrix}[/tex]

    when [itex]\theta=-45^\circ[/itex], right? How did you get

    [tex]\mathbf{A}'=\begin{pmatrix}0 \\ \sqrt{2} \\ 2\end{pmatrix}[/tex]

    from that?
     
  6. Aug 26, 2010 #5
    @vela

    That is not the transformation matrix I used. When rotating about X3, the transformation matrix should look like:


    (cos(theta) sin(theta) 0)
    (-sin(theta) cos(theta) 0)
    (0 0 1)


    Then, after multiplying this with A, it would look like:

    (cos(-45) + sin(-45))
    (-sin(-45) + cos(-45))
    (2 )

    Which equals:

    (1/(sqrt(2)) + -1/(sqrt(2))
    (1/(sqrt(2)) + 1/(sqrt(2)))
    (2 )


    Which equals:

    (0)
    (2/sqrt(2))
    (2)


    That's how I got that answer.
     
    Last edited: Aug 26, 2010
  7. Aug 26, 2010 #6

    lanedance

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    ok that looks better... you need to be clear in communicating what you've what you've done, we can't guess...

    so is there an issue?
     
  8. Aug 26, 2010 #7

    vela

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    OK, but that's the matrix you gave in your original post.
    You have the minus sign in the wrong place in your rotation matrix. Your answer is actually A rotated by +45 degrees.
     
  9. Aug 26, 2010 #8
    Okay. Then all that would do is switch the top and middle value of my A' matrix.
     
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