4.5: Finding the Angle of Rotation Between Two Rotated Vectors in 2-Dimensions

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Given the Rotated Vector, Find the Angle of Rotation

Suppose we did not know the angle $\theta$ of rotation. We can get it by working backwards and solving a system of equations. The rotation formula

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

produces the system of equations

\[\left\{\begin{array}{c}
x^{\prime}=x \cdot \cos \theta+y \cdot(-\sin \theta) \\
y^{\prime}=x \cdot \sin \theta+y \cdot \cos \theta
\end{array}\right. \nonumber \]

Example $\PageIndex{1}$

In Example 4.4.1 of Section 4.4, we found that when the vector $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
1 \\
-1
\end{array}\right]$ was rotated counterclockwise by 90°, it became the vector $\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{l}
1 \\
1
\end{array}\right]$. We got this rotated vector by applying the rotation formula $\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]$.

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

Since two vectors are equal only if their corresponding components are equal, we have the system of two equations

\[\left\{\begin{array}{c}
1=1 \cdot \cos \theta+(-1) \cdot(-\sin \theta) \\
1=1 \cdot \sin \theta+(-1) \cdot \cos \theta
\end{array}\right. \nonumber \]

Using Technology

We can use WolframAlpha to help us solve above system for the angle of rotation, $\theta .$

Go to www.wolframalpha.com.

Since we want to rotate only one time around the coordinate system, we want to instruct WA to give us solutions only where the angle $\theta$ is between 0 and 2$\pi$.

Using the English letter $x$ in place of the Greek letter $\theta$, enter

Solve $1=1^* \cos (x)+(-1)^*(-\sin (x)), 1=1^* \sin (x)+(-1)^* \cos (x), 0<=x<=2^* \text { pi }$ in the entry field.

$This screenshot from WolframAlpha shows how many degrees [[1][-1]] was rotated to get [[1],[1]]. To do this we input Solve 1 = 1*cos(x) + (-1)*(-sin(x)), 1 = 1*sin(x) + (-1)*cos(x), 0 <= x <= 2*pi. The result is pi over 2.$

WA shows the angle of rotation is $\theta =\frac{\pi }{2}$, which is 90°. We conclude that the angle of rotation is 90°.

Example $\PageIndex{2}$

In Example 4.4.2 of Section 4.4, we found that when the vector $\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
2 \\
6
\end{array}\right]$ was rotated counterclockwise by 60°, it became the vector $\left[\begin{array}{l}
x^{\prime} \\
y^{\prime}
\end{array}\right]=\left[\begin{array}{c}
1-3 \sqrt{3} \\
3+\sqrt{3}
\end{array}\right]$. We got this rotated vector by applying the rotation formula $\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]$.

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

Since two vectors are equal only if their corresponding components are equal, we have the system of two equations

\[\left\{\begin{array}{c}
1-3 \sqrt{3}=2 \cdot \cos \theta+6 \cdot(-\sin \theta) \\
3+\sqrt{3}=2 \cdot \sin \theta+6 \cdot \cos \theta
\end{array}\right. \nonumber \]

We will use WolframAlpha to help us solve this system for the angle of rotation, $\theta$.

Using the English letter x in place of the Greek letter $\theta ,$ enter

Solve $1-3 \operatorname{sqrt}(3)=2^* \cos (x)+6^*(-\sin (x)), 3+\operatorname{sqrt}(3)=2^* \sin (x)+6^* \cos (x), 0<=x<=2^* \text { pi }$

in the entry field. Separate the two equations with a comma.

$This screenshot from WolframAlpha shows how many degrees [2,6] was rotated to [1 – (3 multiplied by square root of 3)],[3 + square root of 3]. To do this, we input Solve 1 – 3sqrt(3) = 2*cos(x) + 6*(–sin(x)), 3 + sqrt(3) = 2*sin(x) + 6*cos(x), 0 <= x <= 2*pi. The result is pi over 3.$

WA shows the angle of rotation is $\theta =\frac{\pi }{3}$, which is 60°. We conclude that the angle of rotation is 60°.

Try these

Exercise $\PageIndex{1}$

Find the angle $\theta$ through which the vector $\left[\begin{array}{l}
3 \\
3
\end{array}\right]$ is rotated to become $\left[\begin{array}{c}
0 \\
3 \sqrt{2}
\end{array}\right]$.

Answer: $\theta =\frac{\pi }{4}=45{}^\circ$

Exercise $\PageIndex{2}$

Find the angle $\theta$ through which the vector $\left[\begin{array}{l}
2 \\
-2
\end{array}\right]$ is rotated to become $\left[\begin{array}{c}
1+\sqrt{3} \\
-1+\sqrt{3}
\end{array}\right]$.

Answer: $\theta =\frac{\pi }{3}=60{}^\circ$

Exercise $\PageIndex{3}$

Find the angle $\theta$ through which the vector $\left[\begin{array}{l}
2 \\
0
\end{array}\right]$ is rotated to become $\left[\begin{array}{c}
0 \\
-2
\end{array}\right]$.

Answer: $\theta =\frac{3\pi }{2}=270{}^\circ$

Exercise $\PageIndex{4}$

Find the angle $\theta$ through which the vector $\left[\begin{array}{l}
-2 \\
-2
\end{array}\right]$ is rotated to become $\left[\begin{array}{l}
-1+\sqrt{3} \\
-1-\sqrt{3}
\end{array}\right]$.

Answer: $\theta =\frac{\pi }{3}=60{}^\circ$

Exercise $\PageIndex{5}$

Find the angle $\theta$ through which the vector $\left[\begin{array}{l}
1 \\
1
\end{array}\right]$ is rotated to become $\left[\begin{array}{c}
\sqrt{2} \\
0
\end{array}\right]$.

Answer: $\theta =\frac{7\pi }{4}=315{}^\circ = -45{}^\circ$

Search

Example \(\PageIndex{1}\)

Example \(\PageIndex{2}\)

Exercise \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)

Exercise \(\PageIndex{3}\)

Exercise \(\PageIndex{4}\)

Exercise \(\PageIndex{5}\)