Insight/Intuition into rotations in R²

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
STENDEC
Messages
21
Reaction score
0
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 \\<br /> 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:
Physics news on Phys.org
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?