Solve Rotation Matrix Homework: 60,892857 & 12,496875 -4,837500

[skrat]

Homework Statement

A have two points
60,892857 12,496875 -4,837500
70,714286 14,915625 -5,240625
I have to rotate for -0,067195795 radians.

Homework Equations

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

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?

 

[vela]

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?

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.

 

[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!

 

[skrat]

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

 

[Fredrik]

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.

 

[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!