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Homework Help: Linear transformations and rotations

  1. Nov 1, 2004 #1

    phy

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    Linear transformations and rotations....

    Hi everyone. I need some help getting started on this question.

    Let R: R3 ---> R3 be a rotation of pi/4 around the axix in R3. Find the matrix [R]E that defines the linear transformation R in the standard basis E={e1, e2, e3} of R3. Find R(1,2,1)

    The problem I'm having is just I don't know how to handle the question since I'm not given an equation for R nor am I given some sort of vector to start with. Or am I supposed to put vectors e1, e2, and e3 as the colums of a matrix and do something like that? I'm confused so any help would be greatly appreciated. Thanks.
     
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  3. Nov 1, 2004 #2

    Dr Transport

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    Since R3 is most likely the z-axis, the rotation is in the x-y plane. Think about the rotation of a vector in that plane, the rows of the transformation matrix would correspond to the coefficients of the transformation [tex] \vec{R}' = A \vec{R} [/tex]. A hint, the 3rd row of the matrix will be (0 0 1).
     
  4. Nov 1, 2004 #3

    phy

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    Hmmmmm, I'm not quite sure I understand. Would the first row be (0 0 1) and the second (0 1 0)?
     
  5. Nov 1, 2004 #4

    phy

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    Ooops I meant (1 0 0) and (0 1 0)
     
  6. Nov 2, 2004 #5

    Dr Transport

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    No, you are rotating about the z-axis, [tex] x' = x \cos(\pi/4) + y \sin(\pi/4) and y' = -x \sin(\pi/4) + y \sin(\pi/4) [/tex] check my signs, but I think thay may be correct. the 3rd row is as above. The initial vector is (1,2,1).
     
  7. Nov 2, 2004 #6

    Fredrik

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    A rotation is a linear transformation that doesn't change the length of any vector. This means that

    [tex]x^tx=(Rx)^t(Rx)[/tex]

    for all x. This fact, together with the condition that any vector in the 3 direction is left unchanged by left action of R, is enough to completely determine the components of R.
     
  8. Nov 2, 2004 #7

    Dr Transport

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    This is a rotation, not magnitude change in the vector, only direction.
     
  9. Nov 3, 2004 #8

    Fredrik

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    That's what I said.
     
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