4.4: Rotation Matrices in 2-Dimensions
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.
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
\[\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right] \nonumber \]
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 \(\left[\begin{array}{c}
1 \\
1
\end{array}\right]\)
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 \(\left[\begin{array}{l}
1-3 \sqrt{3} \\
3+\sqrt{3}
\end{array}\right]\).
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.
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
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
-
\(\left[ \begin{array}{c} -1 \\ 1 \end{array} \right]\)
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 180° counterclockwise.
- Answer
-
\(\left[ \begin{array}{c} -1 \\ -1 \end{array} \right]\)
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 270° counterclockwise.
- Answer
-
\(\left[ \begin{array}{c} 1 \\ -1 \end{array} \right]\)
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
-
\(\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]\)
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
-
\(\left[ \begin{array}{c} 0 \\ \sqrt{2} \end{array} \right]\)
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
-
\(\left[ \begin{array}{c} -\sqrt{2} \\ 0 \end{array} \right]\)
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
-
\(\left[ \begin{array}{c} 3 \\ 4 \end{array} \right]\)
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
-
\(\left[ \begin{array}{c} -3 \\ 3 \end{array} \right]\)
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
-
\(\left[ \begin{array}{c} -1.36603 \\ 0.36603 \end{array} \right]\)