[Math] rotate vector around a point using matrices

linear algebramatrices

Im trying to rotate a vector around a point.
For this example lets say i want to rotate 30 degrees around point 2,3
I learned that it takes 3 steps

1. move vector using a "move matrix".

so the matrix would look like:

(matrix 1)

$$
\begin{bmatrix}1&0&2\\ 0&1&3\\ 0&0&1\end{bmatrix}.
$$

2. rotate using a rotation matrix

$$
\begin{bmatrix}
\cos 30& -\sin 30& 0\\
\sin 30& \cos 30& 0\\
0& 0& 1
\end{bmatrix}=\begin{bmatrix}
.866& -.5& 0\\
.5& .866& 0\\
0& 0& 1
\end{bmatrix}
$$
3. move back using a "move matrix" (same as matrix one using negative values for x and y):

(matrix 3)

$$
\begin{bmatrix}1&0&-2\\ 0&1&-3\\ 0&0&1\end{bmatrix}.
$$

But I don't understand how I should do these steps.
Should I multipy matrix one with matrix two, the result of this with matrix 3 and the result of that with my input, or the input with the result?

Or should I take my input, multiply with matrix 1, multiply the result with matrix 2 and after that multiply that result with matrix 3? I understand the steps but im not sure how to use these steps (if that makes sense).

edit:

lets call my input i

$$
\begin{bmatrix}x\\ y\\ 1\end{bmatrix}.
$$

should I do

$$
m1 * m2 * m3* i
$$

$$
i * m1 * m2 * m3
$$
or
$$
(i * m1) * (i * m2) * (i * m3)
$$

If anyone is able to find a link to a solved transformation that would be incredible helpfull, that way I can check if im doing it correctly.

Best Answer

$\begin{bmatrix}1&0&2\\ 0&1&3\\ 0&0&1\end{bmatrix} \begin{bmatrix} \cos 30& -\sin 30& 0\\ \sin 30& \cos 30& 0\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}1&0&-2\\ 0&1&-3\\ 0&0&1\end{bmatrix} \begin{bmatrix}x\\ y\\ 1\end{bmatrix}$

$\begin{bmatrix}1&0&2\\ 0&1&3\\ 0&0&1\end{bmatrix} \begin{bmatrix} \cos 30& -\sin 30& 0\\ \sin 30& \cos 30& 0\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}x-2\\ y-3\\ 1\end{bmatrix}$

$\begin{bmatrix}1&0&2\\ 0&1&3\\ 0&0&1\end{bmatrix} \begin{bmatrix}(x-2)\cos 30 - (y-3)\sin 30\\ (x-2)\sin 30+ (y-3)\cos30\\ 1\end{bmatrix}$

$\begin{bmatrix}(x-2)\cos 30 - (y-3)\sin 30+2\\ (x-2)\sin 30+ (y-3)\cos30+3\\ 1\end{bmatrix}$

Alternatively,

$\begin{bmatrix} \cos 30& -\sin 30& 2\\ \sin 30& \cos 30& 3\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}1&0&-2\\ 0&1&-3\\ 0&0&1\end{bmatrix} \begin{bmatrix}x\\ y\\ 1\end{bmatrix}$

$\begin{bmatrix} \cos 30& -\sin 30& -2\cos 30 + 3\sin30+2\\ \sin 30& \cos 30& -2\sin 30 - 3\cos 30 +3\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}x\\ y\\ 1\end{bmatrix}$

$\begin{bmatrix} x\cos 30 -y\sin 30 -2\cos 30 + 3\sin30+2\\ x\sin 30 +y\cos 30 -2\sin 30 - 3\cos 30 +3\\ 1 \end{bmatrix}=\begin{bmatrix}(x-2)\cos 30 - (y-3)\sin 30+2\\ (x-2)\sin 30+ (y-3)\cos30+3\\ 1\end{bmatrix} $

Related Solutions

[Math] Java Tetris – Using rotation matrix math to rotate piece

The operations you are wanting to perform can be easily implemented without the need of rotation matrices since the transforms you desire are limited to right angle rotations and mirroring.

Assuming $B$ is a two dimensional array representing a block and $B[i,j]$ is the $i^{\text{th}}$ row and $j^{\text{th}}$ column of the block taking on values of 1 for a single block segment and 0 for an empty space.

The first operation you need is a transpose which is effectively a 90 degree clockwise rotation:

B T(B b)
    B b' <- new B[b.columns, b.rows]

    for i = 0 to b.rows - 1
        for j = 0 to b.columns -1
            b'[j,i] <- b[i, j]

    return b'

The second operation is mirroring a block about the y-axis:

B Y(B b)
    B b' <- new B[b.rows, B.columns]

    for i = 0 to b.rows - 1
        for j = 0 to b.columns - 1
            b'[i,j] <- b[i, b.columns - 1 - j]

    return b'

And finally, mirroring a block about the x-axis:

B X(B b)
    B b' <- new B[b.rows, B.columns]

    for i = 0 to b.rows - 1
        for j = 0 to b.columns - 1
            b'[i,j] <- b[b.rows - 1 - i, j]

    return b'

Given a block $B$ in a neutral position, a 90 degree clockwise rotation is $T(B)$, 180 is $T(X(T(B)))$, 270 is $Y(T(B))$ and 360 is simply $B$. counter clockwise rotations are just 360 minus the complementing clockwise rotations.

Also, as @StevenStadnicki pointed out in the comments, you can hard code much of this and you will find that if you are replicating the original Tetris, you'll encounter more that requires a hard coded solution than not.

[Math] Rotation Matrices – Rotating a point on a graph

The matrix $$ \begin{bmatrix} -1 & 0\\ 0 & 1 \end{bmatrix}$$ is not the matrix of a rotation, but rather a reflection about the $y$-axis. The rotation through 180 degrees is given by the matrix $$ \begin{bmatrix} -1 & 0\\ 0 & -1 \end{bmatrix}.$$ The second matrix $$ \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}$$ is indeed the matrix of a rotation through 90 degrees in the positive (counterclockwise) direction, with center of rotation at the origin $(0,0)$. So it's not surprising that your red and green points moved while the blue point (origin) remained fixed!

In general, to compute the matrix of a linear transformation $T$ (such as a reflection or rotation fixing the origin), you compute its effect on the basis vectors---the matrix of $T$ has columns $T(e_1)$ and $T(e_2)$ (or, in higher dimensions, $T(e_1),...,T(e_n)$). Using a little geometry and the definitions of sine and cosine, a rotation by $\theta$ degrees fixing the origin maps the vectors $$ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$ to the vectors $$ \begin{bmatrix} \mathrm{cos}(\theta) \\ \mathrm{sin}(\theta) \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} -\mathrm{sin}(\theta) \\ \mathrm{cos}(\theta) \end{bmatrix}.$$ So the formula you give in the question is off by a little (and doesn't match the correct matrix you gave for rotation through 90 degrees).

Best Answer

Related Solutions

[Math] Java Tetris – Using rotation matrix math to rotate piece

[Math] Rotation Matrices – Rotating a point on a graph

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