IB Mathematics AHL 3.9 Geometric transformations AI HL Paper 2- Exam Style Questions- New Syllabus

(a)(i)
    Rotation anticlockwise \( \frac{\pi}{6} \) radians:
    \( \begin{pmatrix} \cos\left(\frac{\pi}{6}\right) & -\sin\left(\frac{\pi}{6}\right) \\ \sin\left(\frac{\pi}{6}\right) & \cos\left(\frac{\pi}{6}\right) \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \) (M1)(A1)
    Reflection in the line \( y = \frac{x}{\sqrt{3}} \) (angle \( \theta = \frac{\pi}{6} \)):
    \( \theta = \frac{\pi}{6} \), so \( 2\theta = \frac{\pi}{3} \).
    Matrix: \( \begin{pmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \end{pmatrix} \).
    Substitute: \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \), \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \).
    \( \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \) (M1)(A1)(A1)
    Rotation clockwise \( \frac{\pi}{3} \) radians (i.e., \( -\frac{\pi}{3} \)):
    \( \begin{pmatrix} \cos\left(-\frac{\pi}{3}\right) & -\sin\left(-\frac{\pi}{3}\right) \\ \sin\left(-\frac{\pi}{3}\right) & \cos\left(-\frac{\pi}{3}\right) \end{pmatrix} \).
    Substitute: \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \), \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \).
    \( \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \) (A1)
Result:
Correct matrices.

(a)(ii)
    Compute the overall transformation matrix \( P \) by multiplying the matrices in reverse order (right to left application):
    Start with Rotation \( \frac{\pi}{6} \), then Reflection, then Rotation \( -\frac{\pi}{3} \).
    \( P = \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \times \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \times \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \) (M1).
    First, \( \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \times \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \).
    Then, multiply the result by \( \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \).
    \( P = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{\sqrt{3}}{2} \end{pmatrix} \) (A1)(A1)
Result:
\( P \) as above.

(a)(iii)
    Compute \( P^2 \) by multiplying \( P \) with itself:
    \( P = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{\sqrt{3}}{2} \end{pmatrix} \).
    \( P^2 = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{\sqrt{3}}{2} \end{pmatrix} \).
    Element (1,1): \( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = 1 \).
    Element (1,2): \( \frac{\sqrt{3}}{2} \times \left(-\frac{1}{2}\right) + \left(-\frac{1}{2}\right) \times \left(-\frac{\sqrt{3}}{2}\right) = -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = 0 \).
    Element (2,1): \( -\frac{1}{2} \times \frac{\sqrt{3}}{2} + \left(-\frac{\sqrt{3}}{2}\right) \times \left(-\frac{1}{2}\right) = -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = 0 \).
    Element (2,2): \( -\frac{1}{2} \times \left(-\frac{1}{2}\right) + \left(-\frac{\sqrt{3}}{2}\right) \times \left(-\frac{\sqrt{3}}{2}\right) = \frac{1}{4} + \frac{3}{4} = 1 \).
    \( P^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) (M1)(A1)
Result:
\( P^2 = I \), the identity matrix.

(a)(iv)
    Since \( P^2 = I \), the identity matrix, applying the movement twice results in no net change.
    This indicates the drone returns to its original position after two applications.
    (R1)
Result:
The overall movement is an involution.

(b)
METHOD 1
    Calculate the determinant of \( P \):
    \( P = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{\sqrt{3}}{2} \end{pmatrix} \).
    \( \det(P) = \left( \frac{\sqrt{3}}{2} \times -\frac{\sqrt{3}}{2} \right) – \left( -\frac{1}{2} \times -\frac{1}{2} \right) \).
    \( = -\frac{3}{4} – \frac{1}{4} = -1 \).
    \( |\det(P)| = |-1| = 1 \) (M1).
    Since the absolute value of the determinant is \( 1 \), the area is preserved.
    (R1)
METHOD 2
    Rotations preserve area.
    Reflections preserve area.
    Thus, the combination of rotations and reflections preserves area.
    (R1)(R1)
Result:
Areas are equal.

(c)
    Compare \( P \) with a reflection matrix form:
    \( P = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{\sqrt{3}}{2} \end{pmatrix} \).
    This matches \( \begin{pmatrix} \cos(-\frac{\pi}{6}) & \sin(-\frac{\pi}{6}) \\ \sin(-\frac{\pi}{6}) & -\cos(-\frac{\pi}{6}) \end{pmatrix} \), where \( \cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \) and \( \sin(-\frac{\pi}{6}) = -\frac{1}{2} \) (M1).
    For a reflection matrix \( \begin{pmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \end{pmatrix} \), set \( 2\theta = -\frac{\pi}{6} \).
    Solve for \( \theta \): \( \theta = -\frac{\pi}{12} \) (M1)(A1).
    The line of reflection is \( y = \tan(\theta) x \), where \( \theta = -\frac{\pi}{12} \).
    \( \tan(-\frac{\pi}{12}) \approx -0.268 \).
    Thus, the line is \( y = -0.268x \) (A1).
Result:
Reflection in the line \( y = \tan(-\frac{\pi}{12}) x \).