IBDP Maths AHL 3.11 Relationships in trigonometric functions AA HL Paper 1- Exam Style Questions- New Syllabus

(a)
In parallelogram \( ABCD \), opposite sides are equal and parallel: \( \overrightarrow{AD} = \overrightarrow{BC} \).
\( \overrightarrow{BC} = C – B = \begin{pmatrix} -1 – (-3) \\ -2 – (-6) \\ 4 – 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix} \).
\( D = A + \overrightarrow{AD} = \begin{pmatrix} 1 \\ -4 \\ 0 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix} \).
Answer: \( \boxed{D = (3,0,2)} \)

(b)
The diagonals of a parallelogram bisect each other, so \( E \) is the midpoint of \( AC \).
\( E = \left( \frac{1-1}{2}, \frac{-4-2}{2}, \frac{0+4}{2} \right) = (0, -3, 2) \).
Answer: \( \boxed{E = (0,-3,2)} \)

(c)(i)
\( \overrightarrow{AB} = B – A = \begin{pmatrix} -4 \\ -2 \\ 2 \end{pmatrix} \); \( \overrightarrow{AD} = D – A = \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix} \).
\( \overrightarrow{AB} \times \overrightarrow{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & -2 & 2 \\ 2 & 4 & 2 \end{vmatrix} = \begin{pmatrix} -4-8 \\ -(-8-4) \\ -16+4 \end{pmatrix} = \begin{pmatrix} -12 \\ 12 \\ -12 \end{pmatrix} = 12\begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix} \).
Answer: \( \boxed{m = 12} \)

(c)(ii)
Area \( = |\overrightarrow{AB} \times \overrightarrow{AD}| = \sqrt{(-12)^2 + 12^2 + (-12)^2} = \sqrt{432} = 12\sqrt{3} \).
Answer: \( \boxed{12\sqrt{3}} \)

(d)
The normal vector to \( \Pi_1 \) is \( \mathbf{n}_1 = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix} \).
Using \( \mathbf{n} \cdot \mathbf{r} = \mathbf{n} \cdot \mathbf{a} \):
\( -1(x) + 1(y) – 1(z) = -1(1) + 1(-4) – 1(0) \implies -x + y – z = -5 \).
Answer: \( \boxed{x – y + z = 5} \)

(e)
\( \mathbf{n}_1 = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix} \), \( \mathbf{n}_2 = \begin{pmatrix} 5 \\ 1 \\ -7 \end{pmatrix} \).
\( \cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|-5 + 1 + 7|}{\sqrt{3}\sqrt{25+1+49}} = \frac{3}{\sqrt{3}\sqrt{75}} = \frac{3}{\sqrt{3} \cdot 5\sqrt{3}} = \frac{3}{15} = \frac{1}{5} \). (Shown)

(f)
Line \( L \) through \( E(0,-3,2) \) with direction \( \mathbf{n}_1 = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix} \):
\( L: \mathbf{r} = \begin{pmatrix} 0 \\ -3 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix} \).
Substitute \( x = -\lambda, y = -3 + \lambda, z = 2 – \lambda \) into \( \Pi_2 \):
\( 5(-\lambda) + (-3 + \lambda) – 7(2 – \lambda) = 1 \)
\( -5\lambda – 3 + \lambda – 14 + 7\lambda = 1 \implies 3\lambda = 18 \implies \lambda = 6 \).
\( F = \begin{pmatrix} -6 \\ -3+6 \\ 2-6 \end{pmatrix} = \begin{pmatrix} -6 \\ 3 \\ -4 \end{pmatrix} \).
Answer: \( \boxed{F = (-6, 3, -4)} \)