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Rotation matrix in 3D

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi

    I have a coordinate system as attached, and I want to rotate it along the y-axis in the clockwise direction as shown. For this purpose I use
    [tex]
    R = \left( {\begin{array}{*{20}c}
    {\cos \theta } & 0 & {\sin \theta } \\
    0 & 1 & 0 \\
    { - \sin \theta } & 0 & {\cos \theta } \\

    \end{array} } \right)
    [/tex]
    where θ<0 since I rotate CW. Say I now look at the point (x, y, z) = (1, 0, 1). This I want to rotate by θ=-45 degrees, which then becomes in the new system (x', y', z') = (0, 0, 1/√2). But that is wrong, according to my figure (which satisfies the right-hand rule) it should give (x', y', z') = (1/√2, 0, 0).

    I thought I had followed the convention with the right-hand rule, but I must have made an error somewhere. I can't figure out where it is though, it seems consistent. I would be happy to get some feedback.

    Best,
    Niles.
     

    Attached Files:

    Last edited: Oct 16, 2012
  2. jcsd
  3. Oct 16, 2012 #2
    Hello Niles

    I don't know how you got (0, 0, 1/√2).

    I multiplied your rotation matrix with (1,0,1) & got the right answer, which is (√2, 0, 0)

    cos(-45) = 0.707
    sin(-45) = -0.707
     
  4. Oct 16, 2012 #3
    Hi, and thanks

    I think you must have made an error, it does give (0, 0, 1/√2). I just tried it out in Mathematica.
     
  5. Oct 16, 2012 #4
    Hello Niles,

    I think it should give (0,0,√2)

    The point (1,0,1) lies on XZ plane. So its distance from origin is √ (1 + 1) = √2
    As per the image you have attached.
    It makings an angle of 45° CCW with Z-axis.
    Now you are rotating the point CW 45°.
    i.e you are moving the point about Y-axis, towards Z-axis.
    So with distance from origin (or Y-axis) remaining constant, the point should now lie on Z-axis.
    So the final point is (0,0,√2)
     
  6. Oct 16, 2012 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    IF you are doing the calculation as "Av" so that v is a column vector, then it should be
    [tex]R= \begin{pmatrix}cos(\theta) & 0 & -sin(\theta) \\ 0 & 1 & 0 \\ sin(\theta) & 0 & cos(\theta)\end{pmatrix}[/tex]

    If you do the calculation as vA, with v a row vector, then what you give would be correct.

     
  7. Oct 16, 2012 #6
    I'm rotating the coordinate system, not the point as such. So it should give (√2,0,0) -- and it does when I follow HallsOfIvy's approach.

    Thanks to both of you.
     
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