4.4: Rotation Matrices in 2-Dimensions

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Page ID: 125046

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The Rotation Matrix

To this point, we worked with vectors and with matrices. Now, we will put them together to see how to use a matrix multiplication to rotate a vector in the counterclockwise direction through some angle $\theta$ in 2-dimensions.

This image shows two vectors on the xy plane: vector v and vector v'. Vector v = [x,y] and vector v' = [x',y']. Vector v and vector v' are separated by theta degrees.

Diagram of vector v and its rotational vector v'. Vector v is rotated counterclockwise through an angle. The matrix values of vector v and vector v' have not been changed.

Our plan is to rotate the vector $v=\left[ \begin{array}{c} x \\ y \end{array} \right]$ counterclockwise through some angle $\theta$ to the new position given by the vector $v^{\prime}=\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]$. To do so, we use the rotation matrix, a matrix that rotates points in the $xy$-plane counterclockwise through an angle $\theta$ relative to the $x$-axis.

\[\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right] \nonumber \]

The Rotation Process

To get the coordinates of the new vector $\left[{ \begin{array}{c} x \\ y^{\mathrm{'}} \end{array} }^{\mathrm{'}}\right]\mathrm{,\ }$perform the matrix multiplication

\[\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right] \nonumber \]

Example $\PageIndex{1}$

Find the vector $\left[{ \begin{array}{c} x \\ y^{\mathrm{'}} \end{array} }^{\mathrm{'}}\right]\mathrm{\ }$that results when the vector $\left[ \begin{array}{c} x \\ y \end{array} \right]\mathrm{=}\left[ \begin{array}{c} \mathrm{1} \\ --\mathrm{1} \end{array} \right]$ is rotated 90$\mathrm{{}^\circ}$ counterclockwise.

Solution

Using the rotation formula $\left[{ \begin{array}{c} x \\ y^{\mathrm{'}} \end{array} }^{\mathrm{'}}\right]\mathrm{=}\left[ \begin{array}{cc} \mathrm{cos}\theta & \mathrm{-}\mathrm{sin}\theta \\ \mathrm{sin}\theta & \mathrm{cos}\theta \end{array} \right]\left[ \begin{array}{c} x \\ y \end{array} \right]$ with $\left[ \begin{array}{c} x \\ y \end{array} \right]\mathrm{=}\left[ \begin{array}{c} \mathrm{1} \\ --\mathrm{1} \end{array} \right]$

and $\theta \mathrm{=90{}^\circ ,}$ we get

\[\begin{aligned}
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{cc}
\cos 90^{\circ} & -\sin 90^{\circ} \\
\sin 90^{\circ} & \cos 90^{\circ}
\end{array}\right]\left[\begin{array}{c}
1 \\
-1
\end{array}\right]} \\
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\left[\begin{array}{c}
1 \\
-1
\end{array}\right]=\left[\begin{array}{c}
0 \cdot 1+(-1) \cdot(-1) \\
1 \cdot 1+0 \cdot(-1)
\end{array}\right]} \\
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{l}
1 \\
1
\end{array}\right]}
\end{aligned} \nonumber \]

When rotated counterclockwise 90°, the vector $\left[\begin{array}{c}
1 \\
-1
\end{array}\right] $ becomes $\left[\begin{array}{c}
1 \\
1
\end{array}\right]$

This image shows a vector v = [1,negative 1] in the xy plane.

This images shows that vector v = [1,negative 1] has been rotated counterclockwise through 90 degrees into v' = [1,1].

Example $\PageIndex{2}$

Find the vector $\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]$ that results when the vector $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
2 \\
6
\end{array}\right]$ is rotated 60° counterclockwise.

Solution

Using the rotation formula $\left[\begin{array}{c}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]$ with $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
2 \\
6
\end{array}\right]$ and $\theta =60\mathrm{{}^\circ },$ we get

\[\begin{aligned}
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{cc}
\cos 60^{\circ} & -\sin 60^{\circ} \\
\sin 60^{\circ} & \cos 60^{\circ}
\end{array}\right]\left[\begin{array}{l}
2 \\
6
\end{array}\right]} \\
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
1 / 2 & -\sqrt{3} / 2 \\
\sqrt{3} / 2 & 1 / 2
\end{array}\right]\left[\begin{array}{l}
2 \\
6
\end{array}\right]=\left[\begin{array}{c}
1 / 2 \cdot 2+(-\sqrt{3} / 2) \cdot 6 \\
\sqrt{3} / 2 \cdot 2+1 / 2 \cdot 6
\end{array}\right]} \\
& {\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{l}
1-3 \sqrt{3} \\
3+\sqrt{3}
\end{array}\right]}
\end{aligned} \nonumber \]

When rotated counterclockwise 60$\mathrm{{}^\circ}$, the vector $\left[\begin{array}{l}
2 \\
6
\end{array}\right]$ becomes $\left[\begin{array}{l}
1-3 \sqrt{3} \\
3+\sqrt{3}
\end{array}\right]$.

This image shows the vector v = [2,6] being rotated counterclockwise 60 degrees to become v' = [1 - (3 multiplied by square root of 3), 3 + square root of 3].

Using Technology

We can use technology to help us find the rotation. WolframAlpha evaluates the trig functions for us.

Go to www.wolframalpha.com.

We can check the above problem from Example 2 by using WolframAlpha. Find the vector $\left[\begin{array}{c}
x^{\prime} \\
y^{\prime}
\end{array}\right]$ that results when the vector $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
2 \\
6
\end{array}\right]$ is rotated 60° counterclockwise. To find rotation of the vector enter evaluate $[[\cos (60),-\sin (60)],[\sin (60), \cos (60)]] *[2,6]$ into the entry field.

$This screenshot from Wolfram Alpha shows [[2],[6]] being rotated 60 degrees. This is done by [[cos(60), –sin(60)], [sin(60), cos(60)]] * [2,6]. The result is [1 minus (3 multiplied by square root of 3), 3 + square root of 3].$

When rotated counterclockwise 60$\mathrm{{}^\circ}$, the vector $\left[ \begin{array}{c} 2 \\ 6 \end{array} \right]$ becomes $\left[ \begin{array}{c} 1--3\sqrt{3} \\ 3+\sqrt{3} \end{array} \right]$.

Try these

Exercise $\PageIndex{1}$

Find the vector $\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]$ that results when $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
1 \\
1
\end{array}\right]$ is rotated 90° counterclockwise.

Answer: $\left[ \begin{array}{c} -1 \\ 1 \end{array} \right]$

Exercise $\PageIndex{2}$

Answer: $\left[ \begin{array}{c} -1 \\ -1 \end{array} \right]$

Exercise $\PageIndex{3}$

Answer: $\left[ \begin{array}{c} 1 \\ -1 \end{array} \right]$

Exercise $\PageIndex{4}$

Find the vector $\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]$ that results when $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
0 \\
1
\end{array}\right]$ is rotated 90° counterclockwise.

Answer: $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$

Exercise $\PageIndex{5}$

Answer: $\left[ \begin{array}{c} 0 \\ \sqrt{2} \end{array} \right]$

Exercise $\PageIndex{6}$

Find the vector $\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]$ that results when $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
-1 \\
1
\end{array}\right]$ is rotated 45° counterclockwise.

Answer: $\left[ \begin{array}{c} -\sqrt{2} \\ 0 \end{array} \right]$

Exercise $\PageIndex{7}$

Find the vector $\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]$ that results when $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
– 2.20205 \\
4.48898
\end{array}\right]$ is rotated -63° counterclockwise.

Answer: $\left[ \begin{array}{c} 3 \\ 4 \end{array} \right]$

Exercise $\PageIndex{8}$

Find the vector $\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]$ that results when $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
-3 \\
-3
\end{array}\right]$ is rotated -90° counterclockwise.

Answer: $\left[ \begin{array}{c} -3 \\ 3 \end{array} \right]$

Exercise $\PageIndex{9}$

Approximate, to five decimal places, the coordinates of the vector $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
-1 \\
1
\end{array}\right]$ when it is rotated counterclockwise 30°.

Answer: $\left[ \begin{array}{c} -1.36603 \\ 0.36603 \end{array} \right]$

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