How to rotate a whole rectangle by an arbitrary angle around the origin using a transformation matrix?

Question

Suppose, I have a 2D rectangle ABCD like the following:

$A(0,0)$, $B(140,0)$, $C(140,100)$, $D(0,100)$.

I want to rotate the whole rectangle by $\theta = 50°$.

I want to rotate it around the Z-axis by an arbitrary angle using a rotation transformation matrix.

How to do that?

I know that, $$ A = \begin{bmatrix} \ 0 & 0 & 1 \\ \end{bmatrix}; B = \begin{bmatrix} \ 140 & 0 & 1 \\ \end{bmatrix}; C = \begin{bmatrix} \ 140 & 100 & 1 \\ \end{bmatrix}; D = \begin{bmatrix} \ 0 & 100 & 1 \\ \end{bmatrix}. $$

And, I know that the rotation matrix is, $$R = \begin{bmatrix} \ cos \theta & -sin \theta & 0 \\ sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

Now, what is the calculation?

I have tried the following ways,

$$Rotation = \begin{bmatrix} \ 0 & 0 & 1 \\ 140 & 0 & 1 \\ 140 & 100 & 1 \\ 0 & 100 & 1 \\ \end{bmatrix}.\begin{bmatrix} \ cos \theta & -sin \theta & 0 \\ sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

And, $$ A' = \begin{bmatrix} \ 0 & 0 & 1 \\ \end{bmatrix} . \begin{bmatrix} \ cos \theta & -sin \theta & 0 \\ sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\\ B' = \begin{bmatrix} \ 140 & 0 & 1 \\ \end{bmatrix} . \begin{bmatrix} \ cos \theta & -sin \theta & 0 \\ sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\\ C' = \begin{bmatrix} \ 140 & 100 & 1 \\ \end{bmatrix} . \begin{bmatrix} \ cos \theta & -sin \theta & 0 \\ sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\\ D' = \begin{bmatrix} \ 0 & 100 & 1 \\ \end{bmatrix} . \begin{bmatrix} \ cos \theta & -sin \theta & 0 \\ sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$

For example,

$$B' = \begin{bmatrix} \ 140 & 0 & 1 \\ \end{bmatrix} . \begin{bmatrix} \ cos 50^\circ & -sin 50^\circ & 0 \\ sin 50^\circ & cos 50^\circ & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} \ 89.99 & -107.24 & 1 \\ \end{bmatrix}\\ C' = \begin{bmatrix} \ 140 & 100 & 1 \\ \end{bmatrix} . \begin{bmatrix} \ cos 50^\circ & -sin 50^\circ & 0 \\ sin 50^\circ & cos 50^\circ & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} \ 166.59 & -42.96 & 1 \\ \end{bmatrix}$$ What is the right way to work with?

Answer 1

I have solved my problem.

$$Rotation = \begin{bmatrix} \ 0 & 0 & 0 \\ 140 & 0 & 0 \\ 140 & 100 & 0 \\ 0 & 100 & 0 \\ \end{bmatrix}.\begin{bmatrix} \ cos \theta & -sin \theta & 0 \\ sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

Alternatively,

$$ A' = \begin{bmatrix} \ 0 & 0 & 0 \\ \end{bmatrix} . \begin{bmatrix} \ cos \theta & sin \theta & 0 \\ -sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix}0 & 0 & 0\end{bmatrix}\\ B' = \begin{bmatrix} \ 140 & 0 & 0 \\ \end{bmatrix} . \begin{bmatrix} \ cos \theta & sin \theta & 0 \\ -sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix}98.9949 & 98.9949 & 0\end{bmatrix}\\ C' = \begin{bmatrix} \ 140 & 100 & 0 \\ \end{bmatrix} . \begin{bmatrix} \ cos \theta & sin \theta & 0 \\ -sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix}28.2843 & 169.706 & 0\end{bmatrix}\\ D' = \begin{bmatrix} \ 0 & 100 & 0 \\ \end{bmatrix} . \begin{bmatrix} \ cos \theta & sin \theta & 0 \\ -sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix}-70.7107 & 70.7107 & 0\end{bmatrix}\\ $$

Answer 2

When you show $B'$ as $(5,7)$ it is rotated by about 54.5^\circ and the length of the side is now $\sqrt{74} \approx 8.6$, not $7$ This is a problem with your expectation, not the code. You also have the sign of the sine backwards if you are using row vectors-rotate $(1,0,1)$ by $45^\circ$ and it should be $(\frac 12\sqrt 2, \frac 12\sqrt 2,1)$ but you put a minus sign on the second coordinate. A proper rotation of $B$ by $50^\circ$ is $(7 \cos 50^\circ, 7 \sin 50^\circ)\approx (4.5,5.36)$. $C'=(7 \cos 50^\circ-5 \sin 50^\circ, 7 \sin 50^\circ + 5 \cos 50^\circ)\approx(0.6693,8.576)$