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Affine transformations problem

  1. Jun 5, 2009 #1
    1. The problem statement, all variables and given/known data
    In R3:
    T1 symmetry with respect to x -√3y = 0 & z = 0
    T2 symmetry with respect to the X axis

    Find:
    The matrices for T1 and T2, T1(T2) and check that T1(T2) is a rotation around a line.


    2. Relevant equations


    3. The attempt at a solution
    T2 is:
    [tex]\begin{pmatrix}
    {1}&{0}&{0}&{0}\\
    {0}&{1}&{0}&{0}\\
    {0}&{0}&{-1}&{0}\\
    {0}&{0}&{0}&{-1}
    \end{pmatrix}[/tex]

    The line in T1 belongs to z = 0 and the image of (0,0,0) is (0,0,0), therefore the image of (0,0,1) is (0,0,-1).

    In the plane z = 0, with the basis {O, (√3/2, 1/2), (1/2, -√3/2)} the transformation's matrix is:
    [tex]\begin{pmatrix}
    {1}&{0}\\
    {0}&{-1}
    \end{pmatrix}[/tex]

    So putting this in the canonical basis and all together in one matrix:
    [tex]\begin{pmatrix}
    {1}&{0}&{0}&{0}\\
    {0}&{\frac{\sqrt{3}}{2}}&{-\frac{1}{2}}&{0}\\
    {0}&{\frac{1}{2}}&{\frac{\sqrt{3}}{2}}&{0}\\
    {0}&{0}&{0}&{-1}
    \end{pmatrix}[/tex]

    Then for T1(T2) = T2·T1 which yields:
    [tex]\begin{pmatrix}
    {1}&{0}&{0}&{0}\\
    {0}&{\frac{\sqrt{3}}{2}}&{-\frac{1}{2}}&{0}\\
    {0}&{-\frac{1}{2}}&{-\frac{\sqrt{3}}{2}}&{0}\\
    {0}&{0}&{0}&{1}
    \end{pmatrix}[/tex]

    I don't know if I'm doing fine so far, and I don't know how to do the end of the problem.

    Thanks for your help.
     
  2. jcsd
  3. Jun 6, 2009 #2
    (Gentle) bump :)
     
    Last edited: Jun 6, 2009
  4. Jun 6, 2009 #3

    Mark44

    Staff: Mentor

    Run a few vectors through your matrix, and see if the output vectors are rotated around a line. The line x = sqrt(3)y seems a likely candidate for the line. The endpoint of each input vector should be the same distance away from the line as the endpoint of the output vector, and the angle of rotation of an input/output vector pair should be the same as that of any other input/output pair.

    That's the way I would go. Hope that helps.
    Mark
     
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