Rotation Matrix

Matching Matrix Entries to Rotated Coordinates

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

Start with a general $2 \times 2$ matrix:

It represents this rotation only if it produces the same coordinates for every $x$ and $y$.

This matrix must satisfy:

Expanding the multiplication on the left gives:

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 $u = \cos \theta$.

Rotation Matrix about the Origin

Using one consistent angle unit, the rotation matrix about the origin $O(0,0)$ is:

Rotating about an Arbitrary Center

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

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

Centered Matrix Formula

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

Finding a Specific Rotation Matrix

For $\theta = \frac{1}{4}\pi$ radians, or $45^\circ$, both trigonometric values equal $\frac{1}{\sqrt{2}}$:

This matrix rotates every position vector by $45^\circ$ counter-clockwise.

Exercises

The problems move from building the rotation matrix to applying it. The last one rotates about a center that is not the origin.

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

Solutions

1. For $\theta = \frac{\pi}{6}$, or $30^\circ$, the sine and cosine are:

   $$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$$

   $$\sin(\frac{\pi}{6}) = \frac{1}{2}$$

   Substituting these values gives:

   $$\begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}$$

2. For point $P(4,2)$ and $\theta = 60^\circ$, use $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$.

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

   The image is $P'(2-\sqrt{3}, 2\sqrt{3}+1)$.

3. The point being rotated is $Q(3,1)$, the center of rotation is $C(1,-2)$, and $\theta = 90^\circ$. Thus, $(a,b)=(1,-2)$.

   $$\cos 90^\circ = 0$$

   $$\sin 90^\circ = 1$$

   $$\begin{aligned} \begin{pmatrix} x' \\ y' \end{pmatrix} &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 3-1 \\ 1-(-2) \end{pmatrix} \\ &\quad{}+ \begin{pmatrix} 1 \\ -2 \end{pmatrix} \end{aligned}$$

   $$\begin{aligned} &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 1 \\ -2 \end{pmatrix} \\ &= \begin{pmatrix} -3 \\ 2 \end{pmatrix} + \begin{pmatrix} 1 \\ -2 \end{pmatrix} \\ &= \begin{pmatrix} -2 \\ 0 \end{pmatrix} \end{aligned}$$

   The image is $Q'(-2,0)$.