I have something starting at (50, 10) it then rotates counter clockwise by 30 degrees, around the point at (50, 0), essentially mapping out an arc of a circle. How do I find the point it now lies on?
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You apply a rotation matrix to the vector from the center of rotation to the point that is rotating. Here the vector is $(50-50,10-0)=(0,10)$. Then you multiply that by $\begin {bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end {bmatrix}$, getting $$\begin {bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end {bmatrix}\begin {bmatrix} 0\\10 \end {bmatrix}=\begin {bmatrix} 10\sin 30^\circ \\10\cos 30^\circ \end {bmatrix}=\begin {bmatrix} 5 \\5 \sqrt 3 \end {bmatrix}$$ and add that to the center, getting $(50+5,0+5\sqrt 3)=(55,5\sqrt 3)$
Ross Millikan
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sorry I meant 50, 0 – EasilyBaffled Oct 01 '13 at 16:28
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What is the angle in 10 cos 0, and where did the 30 in the 10 sin 30 come from? – EasilyBaffled Oct 01 '13 at 22:54
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The $\theta$ should have been $30^\circ$, which you specified in your question. The $10$ comes from the radius-the distance from the center of rotation to the point of interest. – Ross Millikan Oct 01 '13 at 23:03