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Vector Question

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that the coordinates of the point (x',y') where the counter-clockwise rotation through the angle @ around (0,0) brings the given point (x,y) are

    x' = xcos@ - ysin@
    y' = xsin@ + ycos@

    Hint: show that for the points (x,y) = (1,0) and (x,y) = (0,1) directly,
    and use the fact that the vector (x,y) is equal to the combination
    x.(1,0) + y.(0,1)

    2. Relevant equations
    For vectors u and v, angle @ between them
    u.v = |u||v|Cos@

    3. The attempt at a solution

    I don't want to be told how to do it, I would prefer if someone would kind of tease the solution out of me, if you know what i mean..

    I've included a diagram, showing my interpretation of the question.

    I've tried a few different approaches for the question.
    I used the fact that tan@ = (m1 - m2)/(1 +m1m2).
    I got the slopes of the lines being y/x and y'/x'. When I plugged everything in and rewrote tan as sin/cos, I got the required formulae, but they were both being divided by each other.

    I also used the dot product, put this just resulted with a lot of squares which doesn't help.

    I don't entirely understand the hint also.

    Attached Files:

  2. jcsd
  3. Sep 29, 2010 #2
    Use the hint that was given.

    First draw a rectangle with corners (0,0) and (x,y) (with sides parallel to x- and y-axis) and then draw the rectangle that you get when the whole plane is rotated by an angle @.

    Then you can see in your drawing that the new x' and y' are just sums of the height and width of the rectangle multiplied by sines and/or cosines of @
  4. Sep 30, 2010 #3
    I don't really understand what the hint means.

    Okay, I'll try that.
  5. Sep 30, 2010 #4
    The hint is basically saying that if your starting vectors are either

    [tex]\left(\begin{array}{c}1\\0\end{array}\right)\mbox{ or }\left(\begin{array}{c}0\\1\end{array}\right)[/tex]

    then it's pretty easy to figure out what the new coordinates will be if you rotate it through an angle [tex]\theta[/tex].

    You'll need the sum of angles formulas for cos and sin, [tex]\cos(\theta_{0}+\theta)[/tex] and [tex]\sin(\theta_{0}+\theta)[/tex].
    ([tex]\theta_{0}[/tex] is the initial trig angle for the vector).

    Does this make sense?
  6. Sep 30, 2010 #5
    It makes sense, but I have no idea where to start.
  7. Sep 30, 2010 #6
    You original vector is:

    [tex]v = a\cdot \left(\begin{array}{c}1\\0\end{array}\right)
    + b\cdot \left(\begin{array}{c}0\\1\end{array}\right)[/tex]

    so the rotated vector will be:

    [tex]rot_\theta(v) = a \cdot rot_\theta (\left(\begin{array}{c}1\\0\end{array}\right) )
    + b\cdot rot_\theta ( \left(\begin{array}{c}0\\1\end{array}\right) )[/tex]

    You can first determine c and d in:

    [tex] rot_\theta (\left(\begin{array}{c}1\\0\end{array}\right) ) =
    c\cdot \left(\begin{array}{c}1\\0\end{array}\right)
    d\cdot \left(\begin{array}{c}0\\1\end{array}\right)[/tex]

    and e and f in:

    [tex]rot_\theta (\left(\begin{array}{c}0\\1\end{array}\right) ) =
    e\cdot \left(\begin{array}{c}1\\0\end{array}\right)
    f\cdot \left(\begin{array}{c}0\\1\end{array}\right)[/tex]

    and then substitute those in your expression for [tex]rot_\theta(v) [/tex].
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