Rotation Matrix

Finding the Rotation Matrix about the Origin

The image of a point $(x,y)$ rotated about the origin $(0,0)$ by an angle $\theta$ is .

The image of a point $(x,y)$ rotated about the origin $(0,0)$ by an angle $\theta$ is .

(x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)

We want to find a $2 \times 2$ matrix, say $\begin{pmatrix} r & s \\ t & u \end{pmatrix}$ , that represents this rotation transformation.

This matrix must satisfy:

\begin{pmatrix} r & s \\ t & u \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \cos \theta - y \sin \theta \\ x \sin \theta + y \cos \theta \end{pmatrix}

From the matrix multiplication on the left side, we get:

\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} x \cos \theta - y \sin \theta \\ x \sin \theta + y \cos \theta \end{pmatrix}

By equating the corresponding components:

First row: $rx + sy = x \cos \theta - y \sin \theta$ .

For this equation to hold for all $x$ and $y$ , the coefficients of $x$ must be equal and the coefficients of $y$ must be equal. Thus, $r = \cos \theta$ and $s = -\sin \theta$ .
Second row: $tx + uy = x \sin \theta + y \cos \theta$ .

Similarly, $t = \sin \theta$ and .

The matrix associated with a rotation by an angle $\theta$ radians (or degrees) about the origin $O(0,0)$ is:

R_\theta = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}

To rotate a point $(x,y)$ about an arbitrary point $P(a,b)$ by an angle $\theta$ , we perform three steps:

Translate the point $(x,y)$ so that $P(a,b)$ becomes the origin: $(x-a, y-b)$ .
Rotate the translated point about the origin by $\theta$ using the matrix $R_\theta$ .
Translate the rotated point back by adding $(a,b)$ .

The operation associated with rotation by an angle $\theta$ radians about the point $(a,b)$ is:

\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x-a \\ y-b \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix}

The matrix associated with a rotation by $\theta = \frac{1}{4}\pi$ radians ( $45^\circ$ ) about the origin is:

We know $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ .

R_{\frac{\pi}{4}} = \begin{pmatrix} \cos(\frac{\pi}{4}) & -\sin(\frac{\pi}{4}) \\ \sin(\frac{\pi}{4}) & \cos(\frac{\pi}{4}) \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}

This is the required matrix.

Visualization of Rotating Point

(2,0)

by

45^\circ

about the Origin

Point

A(2,0)

is rotated by

45^\circ

to become

A'(\sqrt{2}, \sqrt{2})

.

\sqrt{2} \approx 1.414

.

Determine the matrices associated with a rotation about the origin $O(0,0)$ by $\frac{1}{6}\pi$ radians.
Determine the image of point $P(4,2)$ if it is rotated about the origin $O(0,0)$ by $60^\circ$ .
Determine the image of point $Q(3,1)$ if it is rotated about the point $C(1,-2)$ by $90^\circ$ .

Given $\theta = \frac{\pi}{6}$ or $30^\circ$ :
$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$
Rotation matrix:

$\begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}$
Point $P(4,2)$ , $\theta = 60^\circ$ . $\cos 60^\circ = \frac{1}{2}$ , .
Given point $Q(3,1)$ , center $C(1,-2)$ , $\theta = 90^\circ$ . $(a,b)=(1,-2)$ .

u = \cos \theta

\sin 60^\circ = \frac{\sqrt{3}}{2}

\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} (4)(\frac{1}{2}) - (2)(\frac{\sqrt{3}}{2}) \\ (4)(\frac{\sqrt{3}}{2}) + (2)(\frac{1}{2}) \end{pmatrix} = \begin{pmatrix} 2 - \sqrt{3} \\ 2\sqrt{3} + 1 \end{pmatrix}

Image: $P'(2-\sqrt{3}, 2\sqrt{3}+1)$ .

\cos 90^\circ = 0

Image: $Q'(-2,0)$ .