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Insight/Intuition into rotations in R²

  1. Feb 24, 2016 #1
    I've been using rotation matrices for quite some time now without fully grasping them. Whenever I tried to develop an intuitive understanding of...[tex]
    x' = x\cos\theta - y\sin\theta \\
    y' = x\sin\theta + y \cos\theta
    [/tex]... I failed and gave up. I've looked at numerous online texts and videos, but following the step-by-step explanations didn't lead to me seeing the whole picture as I had hoped.

    Could someone explain to me (like I'm 5 years old), why [itex]-y\sin\theta[/itex] and [itex]x\sin\theta[/itex] are used to affect the value along the other axis?

    Looking at the following picture (pardon the quality):
    drawing.png
    Is the contribution of [itex]y[/itex] to [itex]x[/itex] and vice versa there, to ensure that [itex]P[/itex] maintains the correct distance to the origin, or is that a misguided simplification of mine? The yellow line cannot be [itex]sin + cos[/itex] (Pythagorean theorem) yet I may combine these two to get [itex]x'[/itex] and [itex]y'[/itex]. Do you see where my gap in understanding lies? Is there a drawing that could clarify how these terms combine to give the correct value we observe? Algebraic proofs don't work with me I'm afraid, I need a geometric/visual explanation.
     
    Last edited: Feb 24, 2016
  2. jcsd
  3. Feb 24, 2016 #2

    jedishrfu

    Staff: Mentor

    MAYBE THIS VIDEO WILL HELP:

     
  4. Feb 25, 2016 #3
  5. Feb 28, 2016 #4
    The component ##\left(\begin{smallmatrix}x\cos\theta\\y\cos\theta\end{smallmatrix}\right)## represents the projection of ##\left(\begin{smallmatrix}x'\\y'\end{smallmatrix}\right)## onto the original vector ##\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)##. The other component (the "rejection") should therefore have magnitude ##|\sqrt{x^2+y^2}\sin\theta|## and be orthogonal to the projection. There are two unique solutions for such a vector.

    Finally, when ##\theta## is nonzero but less than a straight angle, an increase in ##y'## corresponds to a decrease in ##x'##; i.e., the x-component of the rejection should be negative. (This condition is equivalent to a choice of orientation on ##\mathbb{R}^2##.) This tells us which solution is the desired one.

    As for your drawing, I don't quite understand what you mean. Could you elaborate?
     
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