# Rotation matrix

1. Jul 13, 2014

### skrat

1. The problem statement, all variables and given/known data
A have two points
60,892857 12,496875 -4,837500
70,714286 14,915625 -5,240625
I have to rotate for -0,067195795 radians.

2. Relevant equations

$\begin{bmatrix} {x}'\\ {y}' \end{bmatrix}=\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}$

3. The attempt at a solution

Just to explain my data:
60,892857 12,496875 -4,837500
70,714286 14,915625 -5,240625

The first column is x value, the second one is y and the third one is also y. So the idea is, that these points represent some kind of a body, and the problem says I have to rotate for given angle and plot both bodies; rotated and original.

My problem here is very basic and it may sound stupid but:

After calculating the new x and y value using the rotation matrix for first and second column I get
61,59454052 8,380006791
71,55621175 10,13383602

BUT calculating the new x and y value for the first and third column gives me
60,43061978 -8,915248198
70,20281554 -9,976925593

HOW ON EARTH IS THAT POSSIBLE?

Both y values were at the same x before rotation, and they should also be after the rotation. So my question to you is, why the hell aren't they?

2. Jul 13, 2014

### vela

Staff Emeritus
That's not true. For example, take the points (1,1) and (1,-1) and rotate by 45 degrees counterclockwise. One ends up on the x-axis, and the other ends up on the y-axis.

3. Jul 13, 2014

### skrat

Ammmm.. Still I have to disagree... :D

Basically, this should be rotation of the coordinate system for given angle.
Now to simplify it, let's say that one origin system has horizontal x axis and that the coordinates I have are given in the other coordinate system. Now both y are at the same x, meaning there is a line that goes through both of the points and is parallel to the system's y axis.
What I am trying to say is that rotating these two points is identical to rotating that line, which means, that the $x^{'}$ in the new (the horizontal one) coordinate system should be the same for both!

4. Jul 13, 2014

### skrat

If you say no to that, than I should really start thinking about going back to primary school. Please don't do that.

5. Jul 13, 2014

### Fredrik

Staff Emeritus
I don't understand what you're saying in these posts. What exactly did you plug into the transformation equation
$$\begin{bmatrix} x'\\ y' \end{bmatrix}=\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}$$ and why? Can you post the full problem statement?

I also don't understand statements like "both y are at the same x".

One thing that you should think about is that if a transformation changes (x,y) to (x',y'), and we have x'=x, then if |y'|≠|y|, the points (x,y) and (x',y') aren't at the same distance from (0,0). This would mean that the transformation isn't a rotation.

6. Jul 13, 2014

### skrat

aaaaargh!

Ok, I see it now. I tried to draw everything thing to explain to you what my problem is and by doing that I realized that yes math doesn't lie and that I was wrong too say that both y will still be at the same x.

Ah, Thanks!