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I have the situation depicted in the figure below. Horizontal and vertical axes are $(x,y)$ respectively. The axes $(x',y')$ are denoted by dotted lines and they are rotated by $-31^\circ$ and their origin is shifted from the center point $C$ to the point $A$. The vectors $\vec{r0}=\vec{CA}$, $\vec{r}=\vec{CB}$, $\vec{\rho}=\vec{AB}$ are given in the picture. I want to find the coordinates of the vector $\rho$ with respect to the primed axes. This should simply be $\rho_A = \mathbb{R}(-\theta)\rho_C = \mathbb{R}(-\theta)(\vec{r}-\vec{r0})$ where $\mathbb{R}(\theta)$ is the rotation matrix

$$ \begin{bmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{bmatrix} $$

and the minus sign is because we rotate the system clockwise. And $\cos{\theta} = \frac{-\vec{r0}\cdot(1,0)}{|\vec{r0}|} = \cos{31^\circ}$. The problem is that it does not work. The only way, it works is that $\rho_A = \mathbb{R}(\theta)\rho_C$ that is if I rotate counterclockwise. Why?

enter image description here

Just to show in Mathematica that clockwise rotation will not lead to the result:

r0 = {-5, 3};
r = {7, 7};
\[Theta] = ArcCos[-r0.{1, 0}/Norm[r0]]
rot = RotationMatrix[-\[Theta]];
rot.(r - r0)
N@%
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1 Answer 1

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Good news is that all your computations are correct; but the correct rotation matrix to apply is indeed $R(\theta)$, because since you are rotating the coordinate axes an angle $\theta$ *clockwise*, the vector $\rho$ is now rotated by $\theta$ *counterclockwise*, relative to the rotated axes. The problem can be boiled down to: given a vector $\rho$ in one coordinate basis, what is $\rho$ expressed in rotated coordinate basis?

$\rho$ in original/rotated coordinate system

For simplicity, assume $\rho$ is simply a vector in the $x$ direction. Do you see that whereas it forms an angle of $0$ with the $x$-axis, it forms an angle of $\theta$ with the $x'$-axis?

base example

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