Rotation on a Plane | Using Rotation Matrix and Point Rotation

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The discussion focuses on rotating a square around a specific point on the X-Y plane, particularly point A(3,1), by 70 degrees clockwise. The user initially struggles with applying the rotation matrix to points not centered at the origin but realizes that by rotating the vector from point A to other corners and then adjusting with point A's coordinates, they can achieve the desired rotation. They also explore the relationship between the rotation matrix and its transpose, noting that the transpose corresponds to a reverse rotation. The conversation highlights the properties of the rotation matrix, including its orthogonality and the even and odd functions of sine and cosine. Ultimately, the user gains a clearer understanding of the rotation process.
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Homework Statement


On a X-Y plane we have a square with its 4 corners A(3,1) B(7,3) C(2,6) D(0,2).
We are to rotate the rest of the square around the point A clockwise by 70 degrees.



Homework Equations


(I am not sure how they are called in English)
The rotation matrix 2x2 1st row: cosa ,-sina 2nd row: sina, cosa - call it G(a) so that
X' = G(a) * X


The Attempt at a Solution


I know how to use this matrix transposition or conversion (not sure how it is called) when I am rotating a vector around the 0-point, but I don't know how to rotate a vector around a point on the plane.


EDIT: Just as I posted this I got a revelation - I will Rotate the vector AB using the rotation matrix and then add point A's x and y coordinate respectively to the product of the matrixes. And all the same with the other corners - construct vector AC AD and deja vu.

Now there is a question:
When I do the product of G(a) * A , where A is the vector matrix - the vector spins counterclockwise, but when I do the product of AT * G(a) - the vector spins clockwise. I don't understand why - does it mean that AT * G(a) = G(-a) * A?
 
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Yes, the rotation matrix,
\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}
is orthogonal- its columns or rows, considered as vectors, are "orthonormal", perpendicular and of length 1.

Also, the transpose is, as you say, equal to the reverse rotation:
\begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta)\end{bmatrix}= \begin{bmatrix}cos(-\theta) & -sin(-\theta) \\ sin(-\theta) & cos(-\theta)\end{bmatrix}
because cosine is an "even" function and sine is an "odd" function:
cos(-\theta)= cos(\theta) and sin(-\theta)= -sin(\theta).
 
lendav_rott said:
On a X-Y plane we have a square with its 4 corners A(3,1) B(7,3) C(2,6) D(0,2).
Funny looking square!
 
Yes I know it's not a square :D But that's what the assignment said, it's not important though. I guess you can call it a polygonia or however it is called in English :/

Also thanks HallsofIvy - read some about it and played around with the numbers a bit and I understand how it works :)
 

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