# rotation matrix - why am I thinking this wrong?

The rotation given in Question 1 part ii) doesn't match with this wikipedia link http://en.wikipedia.org/wiki/Rotation_matrix.

$$\begin{array}{lcl} x' &=& x \cos\theta - y \sin\theta \\ y' &=& x \sin\theta + y \cos\theta \end{array}$$ What's the difference? Why are the two answers different? (I don't think it's a mistake on the previous link.)

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– user1831003 Feb 22 '13 at 5:38
You shouldn't post equations as images. This site supports MathJax for rendering LaTeX equations (here's a short guide, google for much more info). I'll fix this for you for now. – Michael Brown Feb 22 '13 at 6:14

You need to be careful to make the distinction between rotating the vector $\mathbf A$ counterclockwise in which case you would have $$A_x' = A_x\cos\theta - A_y\sin\theta, \qquad A_y' = A_x\sin\theta + A_y\cos\theta$$ and rotating the basis vectors counterclockwise and then finding the components of $\mathbf A$ in the resulting basis which results in the above expressions with the replacement $\theta\to-\theta$. To see what I mean, draw a vector $\mathbf A$ in the plane, then imagine (a) rotating the vector counterclockwise and then calculating the resulting components (b) rotating the basis vectors counterclockwise while keeping $\mathbf A$ fixed and then computing the components of $\mathbf A$ in the rotated basis.
Another way of seeing this is that for any rotation $R(\theta)$ counterclockwise by $\theta$ the situation (a) corresponds to (using the summation convention) $$A_i' = R(\theta)_{ij} A_j$$ while situation (b) corresponds to $$A_i' = \mathbf A\cdot (R(\theta)\mathbf{e}_i) = A_j(R(\theta) \mathbf e_i)_j = A_jR(\theta)_{jk}\delta_{ik} = A_jR(\theta)_{ji} = R(-\theta)_{ij} A_j$$ where the last equality follows from the fact that $R^{-1} = R^t$ for rotations. The situations differ by the replacement $\theta\to-\theta$ as claimed.