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Rotation matrix

  1. Jul 13, 2014 #1
    1. The problem statement, all variables and given/known data
    A have two points
    60,892857 12,496875 -4,837500
    70,714286 14,915625 -5,240625
    I have to rotate for -0,067195795 radians.



    2. Relevant equations

    ##\begin{bmatrix}
    {x}'\\
    {y}'
    \end{bmatrix}=\begin{bmatrix}
    cos\alpha & -sin\alpha \\
    sin\alpha & cos\alpha
    \end{bmatrix}\begin{bmatrix}
    x\\
    y
    \end{bmatrix}##

    3. The attempt at a solution

    Just to explain my data:
    60,892857 12,496875 -4,837500
    70,714286 14,915625 -5,240625

    The first column is x value, the second one is y and the third one is also y. So the idea is, that these points represent some kind of a body, and the problem says I have to rotate for given angle and plot both bodies; rotated and original.

    My problem here is very basic and it may sound stupid but:

    After calculating the new x and y value using the rotation matrix for first and second column I get
    61,59454052 8,380006791
    71,55621175 10,13383602

    BUT calculating the new x and y value for the first and third column gives me
    60,43061978 -8,915248198
    70,20281554 -9,976925593

    HOW ON EARTH IS THAT POSSIBLE?

    Both y values were at the same x before rotation, and they should also be after the rotation. So my question to you is, why the hell aren't they?
     
  2. jcsd
  3. Jul 13, 2014 #2

    vela

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    That's not true. For example, take the points (1,1) and (1,-1) and rotate by 45 degrees counterclockwise. One ends up on the x-axis, and the other ends up on the y-axis.
     
  4. Jul 13, 2014 #3
    Ammmm.. Still I have to disagree... :D

    Basically, this should be rotation of the coordinate system for given angle.
    Now to simplify it, let's say that one origin system has horizontal x axis and that the coordinates I have are given in the other coordinate system. Now both y are at the same x, meaning there is a line that goes through both of the points and is parallel to the system's y axis.
    What I am trying to say is that rotating these two points is identical to rotating that line, which means, that the ##x^{'}## in the new (the horizontal one) coordinate system should be the same for both!
     
  5. Jul 13, 2014 #4
    If you say no to that, than I should really start thinking about going back to primary school. Please don't do that.
     
  6. Jul 13, 2014 #5

    Fredrik

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    I don't understand what you're saying in these posts. What exactly did you plug into the transformation equation
    $$\begin{bmatrix}
    x'\\
    y'
    \end{bmatrix}=\begin{bmatrix}
    \cos\alpha & -\sin\alpha \\
    \sin\alpha & \cos\alpha
    \end{bmatrix}\begin{bmatrix}
    x\\
    y
    \end{bmatrix}$$ and why? Can you post the full problem statement?

    I also don't understand statements like "both y are at the same x".

    One thing that you should think about is that if a transformation changes (x,y) to (x',y'), and we have x'=x, then if |y'|≠|y|, the points (x,y) and (x',y') aren't at the same distance from (0,0). This would mean that the transformation isn't a rotation.
     
  7. Jul 13, 2014 #6
    aaaaargh!

    Ok, I see it now. I tried to draw everything thing to explain to you what my problem is and by doing that I realized that yes math doesn't lie and that I was wrong too say that both y will still be at the same x.

    Ah, Thanks!
     
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