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Rotation in R2, around a line?

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the points fixed by f, and show it is a line. We know that f is an isometry.

    0 [tex]\leq[/tex] [tex]\theta[/tex] < 2[tex]\Pi[/tex]
    f: R[tex]^{2}[/tex] [tex]\rightarrow[/tex] R[tex]^{2}[/tex]
    f(x) = Ax

    A = | cos [tex]\theta[/tex] sin [tex]\theta[/tex] |
    ......| sin [tex]\theta[/tex] -cos [tex]\theta[/tex]|

    2. Relevant equations
    fix(f) = {x | f(x) = x}


    3. The attempt at a solution
    fix(f) = | (x1)(cos [tex]\theta[/tex]) + (x2)(sin [tex]\theta[/tex]) = x1 |
    ...........| (x1)(sin [tex]\theta[/tex]) + (x2)(cos [tex]\theta[/tex]) = x2 |

    The thing that confuses me the most is rotating about a line in R[tex]^{2}[/tex]. I was under the impression you could only rotate around a point? Is there something I'm missing? I'm not looking for the answer so much as I'm looking to be pointed in the right direction.

    I apologize for my messy matrix formatting.
     
  2. jcsd
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