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Homework Help: Vector Proof

  1. Sep 7, 2010 #1
    1. The problem statement, all variables and given/known data
    An (x,y) coordinate system is rotated through an angle theta to produce an (x',y') system, see figure.
    A point with coordinates (x,y) will have coordinates (x',y') in the rotated system given by:
    x'1 = (x1 * cos theta) + (y1 * sin theta)
    y'1 = (-x1 * sin theta) + (y1 * cos theta)
    Show that the formula for the distance of the point from the origin is invariant, or unchanged, by the rotation. That is, show:
    sqrt (x12 + y12) = sqrt (x'12 + y'12)


    2. Relevant equations
    I don't know if these are really relevant, I just thought so:
    ax = a cos theta
    ay = a sin theta
    a = sqrt (ax2 + ay2)
    tan theta = ay/ax
    where a = the magnitude of vector a and theta = the angle vector a makes with the positive direction of the x axis

    3. The attempt at a solution
    So I thought this was asking, more or less, to prove that rotating the axes changes the components of the vector but not the vector itself.

    I set the two equations given equal to each other, subbing in the information given for x' and y', but I don't know how to proceed or even if this was a good place to start. Any ideas for starting off?

    sqrt (x12 + y12) = sqrt (((x1 * cos theta) + (y1 * sin theta))2 + ((-x1 * sin theta) + (y1 * cos theta))2)
     

    Attached Files:

  2. jcsd
  3. Sep 7, 2010 #2

    collinsmark

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    'Looks good to me so far. :approve:

    Leave the left side of the equation alone. Start expanding the right side. See if any terms cancel out as you go. :wink: Eventually you will find the trig identity,

    [tex] \cos^2 \theta + \sin^2 \theta = 1 [/tex]

    quite useful.
     
  4. Sep 8, 2010 #3
    So I worked on the given equation, simplifying until I got:

    (x12 + y12)1/2 = ( (x1 * cos theta)2 + (x1 sin theta)2 + 4(x1 * cos theta * y1 * sin theta) + (y1 * sin theta)2 + (y1 * cos theta)2 )1/2

    But I don't see where to go from here. I feel as if I should be able to see a2 + b2 (as in the Pythagorean theorem) in this mess, but I don't see it. Perhaps that's just off base anyway and I just haven't done enough algebra.
     
  5. Sep 8, 2010 #4

    collinsmark

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    The term in red involves a little mistake. But that little mistake makes a big difference.

    Here's a hint. In your original substitution,

    [tex] \sqrt{\left( x_1 \cos \theta + y_1 \sin \theta \right) ^2 + \left( {\color{red}{-}} x_1 \sin \theta + y_1 \cos \theta \right) ^2}, [/tex]

    the x1sinθ has a negative sign attached to it (in red, directly above). However, when you expanded all the terms, everything somehow ended up positive (which is where the mistake fits in). :wink:
     
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