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Homework Help: Linear Transformation

  1. May 29, 2010 #1
    Hi,

    I am trying to find an orthogonal transformation that maps the point (0,5) to the point (3,4).
    Now, I found that the transformation matrix M for a reflection in the line y=mx is as follows:

    [tex] M = \left(
    \begin{array}{cc}
    cos(2\theta) & sin(2\theta)\\
    sin(2\theta) & -cos(2\theta)
    \end{array}
    \right)
    [/tex]

    [tex] \therefore \left(
    \begin{array}{cc}
    cos(2\theta) & sin(2\theta)\\
    sin(2\theta) & -cos(2\theta)
    \end{array}
    \right)
    \left(
    \begin{array}{c} 0 \\
    5 \\
    \end{array}
    \right)=
    \left(
    \begin{array}{c} 3 \\
    4\\
    \end{array}
    \right)

    [/tex]

    However this means that
    [tex] 5sin(2\theta)=3 [/tex]
    [tex] -5cos(2\theta)=4 [/tex]

    [tex] \frac{5sin2\theta}{-5cos2\theta} = 3/4 [/tex]
    [tex] \theta = arctan(-\frac{3}{4}) [/tex]
    [tex]\therefore m=\frac{3}{4}
    [/tex]

    I noticed that if instead I find the angle by taking [tex] 5sin(2\theta)=3 [/tex]
    [tex] \theta = 18.43 and tan(\theta)=0.333 [/tex]
    Or [tex] -5cos(2\theta)=4 [/tex]
    [tex] \theta = 71.56 [/tex] and [tex] tan(71.56)=3[/tex]

    Why don't they all yield the same result? isn't this like solving a system of linear equations?

    Having said this, I tried to derive the matrix of the transformation myself. I drew the basis vectors i(1,0) and j(0,1) and checked what their new coordinates would be when reflected in a line that makes an angle [tex] \theta [/tex] with the x-axis.

    When considering the j(0,1) vector, the angle between j and j' & that between i and i' is [tex] 2\theta [/tex].

    The new coordinates for i' would be:
    [tex] x = cos(2\theta)
    y = sin(2\theta) [/tex]

    and those for j' would be:

    [tex] x = sin(2\theta)
    y = cos(2\theta).
    [/tex]

    Why would you say that for j' [tex] y=-cos(2\theta)? [/tex]
    The only way j' will have negative y coordinates is if the gradient of the line is >45, and if this happens, cos(2x) will be negative (since 90 < 2x < 180, cos(2x) is negative).

    Sorry for the very long post, I just wanted to show what I tried before asking any questions. With this being said, could someone please tell me what's wrong with the derivation I attempted? And most of all, why the first one doesn't work?

    Thank you!
     
    Last edited: May 29, 2010
  2. jcsd
  3. May 29, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    NO, this is 2[itex]\theta[/itex]

    [itex]2\theta= 36.86[/itex] and [itex]tan(36.86)= 0.75[/itex]

    NO, cos(71.56)= 0.3163, nowhere near -.8. I don't know how you got this. If [itex]cos(\theta)= -.8[/itex] then [itex]\theta= 143[/itex] or [itex]\theta= -37[/itex]. tan(143)= -.75 but tan(-37)= .75.

     
  4. May 29, 2010 #3
    The gradient of the line is [tex] tan(\theta) [/itex] not [tex] tan(2\theta) [/itex], or am I wrong?

    [tex]2\theta [/itex] is the angle with which the vectors i and j will be rotated when they're reflected in the line [tex] y=(tan\theta)x [/itex].
     
  5. May 29, 2010 #4
    Ok fair enough, I meant [tex] tan(2\theta) = -\frac{3}{4} [/itex]
    [tex] \therefore \theta = \frac{arctan(-\frac{3}{4})}{2} [/itex].

    [tex] -5cos(2\theta)=4 [/itex]
    then [tex] cos(2\theta) = -4/5 = -0.8 [/itex]
    therefore [tex] \theta = \frac{arccos(-0.8)}{2} = 71.56 [/itex]
    and hence [tex] cos(2 * 71.56) = -0.8 [/itex]
     
    Last edited: May 29, 2010
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