Rotation matrices describe rotations about the origin.
The angle of rotation \(\theta\) is counterclockwise off of the positive x-axis (due east). It follows the same convention as the unit circle and the direction of vectors.
Rotation matrices are square matrices, orthogonal matrices and have a determinant of 1.
Notes
Rotation Matrix
\(R(\theta)=\left[{\begin{array}{cc}\cos{\theta} & -\sin{\theta} \\\sin{\theta} & \cos{\theta} \\\end{array} } \right]\)
To rotate \(\left[{\begin{array}{c}x \\y\\\end{array} } \right]\) by \(\theta\) degrees
\( \left[{\begin{array}{cc}\cos{\theta} & -\sin{\theta} \\\sin{\theta} & \cos{\theta} \\\end{array} } \right] \left[{\begin{array}{c}x \\y\\\end{array} } \right] = \left[{\begin{array}{c}x’ \\y’\\\end{array} } \right] \)
Practice Problems
\(\textbf{1)}\) \( \text{Find the rotation matrix for } 30^{\circ} \)
The rotation matrix is \(\left[{\begin{array}{cc}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\\ \frac{1}{2} & \frac{\sqrt{3}}{2} \\\end{array} } \right]\)
\(\textbf{2)}\) \( \text{Find the rotation matrix for } 45^{\circ} \)
The rotation matrix is \(\left[{\begin{array}{cc}\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\\end{array} } \right]\)
\(\textbf{3)}\) \( \text{Find the rotation matrix for } 90^{\circ} \)
The rotation matrix is \(\left[{\begin{array}{cc}0 & -1 \\\ 1 & 0 \\\end{array} } \right]\)
\(\textbf{4)}\) \( \text{Find the rotation matrix for } 135^{\circ} \)
The rotation matrix is \(\left[{\begin{array}{cc}-\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\\end{array} } \right]\)
\(\textbf{5)}\) \( \text{Find the rotation matrix for } 300^{\circ} \)
The rotation matrix is \(\left[{\begin{array}{cc}\frac{1}{2} & \frac{\sqrt{3}}{2} \\\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \\\end{array} } \right]\)
Perform the stated rotation
\(\textbf{6)}\) \( \text{Rotate the vector }\left[{\begin{array}{c}1 \\1\\\end{array} } \right] \text{ by } 30^{\circ} \)
The answer is \(\left[{\begin{array}{c}\frac{-1+\sqrt{3}}{2} \\\frac{1+\sqrt{3}}{2}\\\end{array} } \right] \approx \left[{\begin{array}{c}.366 \\1.366\\\end{array} } \right]\)
\(\textbf{Step 1: Rotation Matrix Notes}\)
\(\,\,\,\,\, \left[{\begin{array}{cc}\cos{\theta} & -\sin{\theta} \\\sin{\theta} & \cos{\theta} \\\end{array} } \right] \left[{\begin{array}{c}x \\y\\\end{array} } \right] \)
\(\,\,\,\,\, \left[{\begin{array}{cc}\cos{30} & -\sin{30} \\\sin{30} & \cos{30} \\\end{array} } \right] \left[{\begin{array}{c}1 \\1\\\end{array} } \right]\)
\(\,\,\,\,\, \left[{\begin{array}{cc}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\\ \frac{1}{2} & \frac{\sqrt{3}}{2} \\\end{array} } \right] \left[{\begin{array}{c}1 \\1\\\end{array} } \right]\)
\(\textbf{Step 2: Matrix Multiplication}\)
\(\,\,\,\,\, \left[{\begin{array}{c}\frac{\sqrt{3}}{2}(1) -\frac{1}{2}(1) \\\ \frac{1}{2}(1) + \frac{\sqrt{3}}{2}(1) \\\end{array} } \right] \)
\(\left[{\begin{array}{c}\frac{-1+\sqrt{3}}{2} \\\frac{1+\sqrt{3}}{2}\\\end{array} } \right] \approx \left[{\begin{array}{c}.366 \\1.366\\\end{array} } \right]\)
\(\textbf{7)}\) \( \text{Rotate the vector }\left[{\begin{array}{c}5 \\5\\\end{array} } \right] \text{ by } 30^{\circ} \)
The answer is \(\left[{\begin{array}{c}\frac{-5+5\sqrt{3}}{2} \\\frac{5+5\sqrt{3}}{2}\\\end{array} } \right] \approx \left[{\begin{array}{c}1.83 \\6.83\\\end{array} } \right]\)
\(\,\,\,\,\, \left[{\begin{array}{cc}\cos{\theta} & -\sin{\theta} \\\sin{\theta} & \cos{\theta} \\\end{array} } \right] \left[{\begin{array}{c}x \\y\\\end{array} } \right] \)
\(\,\,\,\,\, \left[{\begin{array}{cc}\cos{30} & -\sin{30} \\\sin{30} & \cos{30} \\\end{array} } \right] \left[{\begin{array}{c}5 \\5\\\end{array} } \right]\)
\(\,\,\,\,\, \left[{\begin{array}{cc}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\\ \frac{1}{2} & \frac{\sqrt{3}}{2} \\\end{array} } \right] \left[{\begin{array}{c}5 \\5\\\end{array} } \right]\)
\(\,\,\,\,\, \left[{\begin{array}{c}\frac{\sqrt{3}}{2}(5) -\frac{1}{2}(5) \\\ \frac{1}{2}(5) + \frac{\sqrt{3}}{2}(5) \\\end{array} } \right] \)
\(\left[{\begin{array}{c}\frac{-5+5\sqrt{3}}{2} \\\frac{5+5\sqrt{3}}{2}\\\end{array} } \right] \approx \left[{\begin{array}{c}1.83 \\6.83\\\end{array} } \right]\)
\(\textbf{8)}\) \( \text{Rotate the vector }\left[{\begin{array}{c}5 \\5\\\end{array} } \right] \text{ by } 45^{\circ} \)
The answer is \(\left[{\begin{array}{c}0 \\5\sqrt{2}\\\end{array} } \right] \approx \left[{\begin{array}{c}0 \\7.07\\\end{array} } \right]\)
\(\textbf{9)}\) \( \text{Rotate the vector }\left[{\begin{array}{c}1 \\-4\\\end{array} } \right] \text{ by } 45^{\circ} \)
The answer is \(\left[{\begin{array}{c}\frac{5\sqrt{2}}{2} \\\frac{-3\sqrt{2}}{2}\\\end{array} } \right] \approx \left[{\begin{array}{c}3.54 \\-2.12\\\end{array} } \right]\)
\(\textbf{10)}\) \( \text{Rotate the vector }\left[{\begin{array}{c}2 \\3\\\end{array} } \right] \text{ by } 90^{\circ} \)
The answer is \(\left[{\begin{array}{c}-3 \\2\\\end{array} } \right]\)
\(\textbf{11)}\) \( \text{Rotate the vector }\left[{\begin{array}{c}-8 \\3\\\end{array} } \right] \text{ by } 135^{\circ} \)
The answer is \(\left[{\begin{array}{c}\frac{5\sqrt{2}}{2} \\\frac{-11\sqrt{2}}{2}\\\end{array} } \right] \approx \left[{\begin{array}{c}3.54 \\-7.78\\\end{array} } \right]\)
\(\textbf{12)}\) \( \text{Rotate the vector }\left[{\begin{array}{c}-5 \\-5\\\end{array} } \right] \text{ by } 300^{\circ} \)
The answer is \(\left[{\begin{array}{c}\frac{-5-5\sqrt{3}}{2} \\\frac{-5+5\sqrt{3}}{2}\\\end{array} } \right]\approx \left[{\begin{array}{c}-6.83 \\1.83\\\end{array} } \right]\)
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