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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 62

Hi all,

I have been dealing with a problem, but I did not get any solution so far. I am looking for a rotation operator in 60°-coordination. A plane of a 60°-coordination is shown below.

I am familiar with rotation in 90° coordination. it is (cos(θ)+jsin(θ)). I tried to modify this formula to use in 60°-coordination, but no success so far.

some examples: 60° rotation of (1,0), (-1,2), (1,-2) in 60°-coordination should yields (0,1), (-2,1), (2,-1), respectively.

I would be grateful if you share your ideas with me.

Thank you in advance.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,524

hi kappa_am

Here's a rotation of 60 anticlockwise.

Providing the origin is an invariant point (ie. the centre) you can work out the matrix by considering where (1,0) and (0,1) move to. As these form the identity matrix when written as vectors, the place they move to gives the required matrix. ie If they move to (a,b) and (c,d) then the matrix is

Hope that helps,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 62

Thank you Bob. Do you have any proof or find it by considering vectors movements?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,524

hi kappa_am

It's a result of vector theory. If a coordinate (in any system of axes) is written as a vector then it is a 2x1 matrix. After a transformation the result is too. So only a 2x2 matrix may be a candidate for the transformation … 2x2 multiplied by 2x1 gives 2x1. If you try multiplying the origin (0,0) by such a matrix you will get (0,0) again, so this will only work if the origin is an invariant point. So what follows can be applied to rotations around the origin, reflections in a line that goes through the origin and shears, where the invariant line goes through the origin. (You could also make up fancy transformations that would also work.)

(1,0) and (0,1) are used as a basis for the coordinate system. When you assemble a 2x2 matrix from these vectors you get the identity matrix. That's why this method works for those two points. Those transformations listed above are all 'linear' meaning that straight lines are transformed to straight lines. That's important too, because it means, once you have a matrix that works for the base vectors it will work for all other points too.

So decide where you want (1,0) and (0,1) to transform to. Let's say (a,b) and (c,d)

Then the transformation is:

That is (transformation matrix M) x (vectors) = (transformed vectors)

As 1, 0, 0, 1 is the identity matrix that means that:

So, to get the matrix for a rotation of 60 degrees I wrote down the vectors for four points in a parallelogram shape, before and after the rotation, and wrote out the matrix multiplication. To get M, I realised that I only needed two of the four vectors so the correct matrix was easy to write down. The other two points in my parallelogram were the origin (obviously stays the same) and the point (1,1) which maps to (-1,2) which helps to confirm that I've got the right matrix.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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