Rotation Matrix: Calculating Angle & Direction of Rotation

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breen155
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Homework Statement


Hey guys, I'm not sure if this bit is relevant but the first part of the question is... 'The diagram shows a triangle with vertices O, A(1,2) , B (0,2). The question I need help with is 'Each of the following matricies represents a rotation about the origin, Find the angle and direction of rotation in each case'

(1/2 -[tex]\sqrt{3}[/tex]/2)
([tex]\sqrt{3}[/tex]/2 1/2)

Imagine its one big set of brackets around the above matrix not 2 smaller ones :)

Homework Equations


I'm also not sure whether these are of relivance but x' = ax + cy and y' = bx + dy
also I have the matrix
cos[tex]\theta[/tex] -sin[tex]\theta[/tex]
sin[tex]\theta[/tex] cos[tex]\theta[/tex]


The Attempt at a Solution


I have been messing about with this for a while attemting to sub in co ordinates to the x' equations and y' equations but I feel I am getting nowhere. I would appreciate any help.
Thanks in advance guys
Breen155
 
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[tex]\begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end{pmatrix}[/tex]

[tex]a = \frac{1}{2}[/tex]
[tex]b = \frac{\sqrt{3}}{2}[/tex]Can you use the definition of equality of matrices to get two equations for theta?
 
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not too sure what you mean sorry I only started teaching this to myself this morning :S erm... is it costheta - sin theta = .5 and sintheta + costheta = sqt3/2 ?
 
I slipped up on my notation. I fixed my last post. Try to get the equations again because they aren't right. Remember, get them by using the definition of equality for matrices.
 
costheta = .5 and sintheta = sqrt3/2 ? :)
 
correct. Now you just need to find theta.
 
theta is 60 degrees but how do i tell the direction of rotation, clockwish or anticlockwise ? :) (thanks for the help so far btw)
 
Remember that a positive theta corresponds, by convention, to a counterclockwise rotation. If it were a clockwise rotation of 60 degrees, then theta would be -60 degrees. This is equivalent to a counterclockwise rotation of 300 degrees, and you see that the sine and cosine of 300 degrees is exactly equal to the sine and cosine of -60 degrees. Of course, this means the question is a bit vague, since you can change the direction of rotation just by changing the angle. If you have both a direction with a correct angle, however, it should be fine.
 
Ok I understand now. Thanks for all the help guys