a. [2 marks]

Velocity vector of airplane B: \(\mathbf{v}_B = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix}\).
Bearing uses horizontal components (east, north): \(\begin{pmatrix} 4 \\ 2 \end{pmatrix}\).
Angle from north: \(\tan \theta = \frac{\text{east}}{\text{north}} = \frac{4}{2} = 2 \implies \theta = \arctan(2) \approx 63.4349^\circ\) (M1).
Three-figure bearing: \(063^\circ\) (A1).
Answer: \(063^\circ\).

b. [2 marks]

Speed is magnitude of velocity vector.
Airplane A: \(\mathbf{v}_A = \begin{pmatrix} -6 \\ 2 \\ 4 \end{pmatrix}\), \(|\mathbf{v}_A| = \sqrt{(-6)^2 + 2^2 + 4^2} = \sqrt{36 + 4 + 16} = \sqrt{56} \approx 7.4833 \, \text{km/min}\) (M1).
Airplane B: \(\mathbf{v}_B = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix}\), \(|\mathbf{v}_B| = \sqrt{4^2 + 2^2 + (-2)^2} = \sqrt{16 + 4 + 4} = \sqrt{24} \approx 4.8990 \, \text{km/min}\) (A1).
Compare: \(7.4833 > 4.8990\).
Answer: Airplane A travels faster than airplane B.

c. [3 marks]

Angle between flight paths is angle between velocity vectors: \(\mathbf{v}_A = \begin{pmatrix} -6 \\ 2 \\ 4 \end{pmatrix}\), \(\mathbf{v}_B = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix}\).
Dot product: \(\mathbf{v}_A \cdot \mathbf{v}_B = (-6) \cdot 4 + 2 \cdot 2 + 4 \cdot (-2) = -24 + 4 – 8 = -28\) (M1).
Magnitudes: \(|\mathbf{v}_A| = \sqrt{56} \approx 7.4833\), \(|\mathbf{v}_B| = \sqrt{24} \approx 4.8990\).
\(\cos \theta = \frac{\mathbf{v}_A \cdot \mathbf{v}_B}{|\mathbf{v}_A| |\mathbf{v}_B|} = \frac{-28}{\sqrt{56} \cdot \sqrt{24}} = \frac{-28}{\sqrt{1344}} \approx \frac{-28}{36.6606} \approx -0.7638\).
Acute angle: \(\theta = \arccos(0.7638) \approx 40.236^\circ \approx 40.2^\circ\) (A1)(A1).
Answer: \(40.2^\circ\).

d. [3 + 2 marks]

(i) Intersection at \( P \): Set \(\mathbf{r}_A(t_1) = \mathbf{r}_B(t_2)\).
\(\mathbf{r}_A(t_1) = \begin{pmatrix} 19 – 6t_1 \\ -1 + 2t_1 \\ 1 + 4t_1 \end{pmatrix}\), \(\mathbf{r}_B(t_2) = \begin{pmatrix} 1 + 4t_2 \\ 2t_2 \\ 12 – 2t_2 \end{pmatrix}\).
Equate components:
(1) \(19 – 6t_1 = 1 + 4t_2\),
(2) \(-1 + 2t_1 = 2t_2\),
(3) \(1 + 4t_1 = 12 – 2t_2\).
From (2): \(t_2 = t_1 – 0.5\) (M1).
Substitute into (1): \(19 – 6t_1 = 1 + 4(t_1 – 0.5) \implies 19 – 6t_1 = 1 + 4t_1 – 2 \implies 20 = 10t_1 \implies t_1 = 2\).
Then: \(t_2 = 2 – 0.5 = 1.5\) (A1).
Verify (3): \(1 + 4 \cdot 2 = 9\), \(12 – 2 \cdot 1.5 = 9\).
Coordinates: \(\mathbf{r}_A(2) = \begin{pmatrix} 19 – 12 \\ -1 + 4 \\ 1 + 8 \end{pmatrix} = \begin{pmatrix} 7 \\ 3 \\ 9 \end{pmatrix}\).
Confirm: \(\mathbf{r}_B(1.5) = \begin{pmatrix} 1 + 4 \cdot 1.5 \\ 2 \cdot 1.5 \\ 12 – 2 \cdot 1.5 \end{pmatrix} = \begin{pmatrix} 7 \\ 3 \\ 9 \end{pmatrix}\) (A1).
Answer: \(\begin{pmatrix} 7 \\ 3 \\ 9 \end{pmatrix} \, \text{km}\).
(ii) Time difference: \(|t_1 – t_2| = |2 – 1.5| = 0.5 \, \text{minutes}\) (A1)(A1).
Answer: \(0.5 \, \text{minutes}\).

e. [3 marks]

Distance \( D(t) \): \(\mathbf{r}_B(t) – \mathbf{r}_A(t) = \begin{pmatrix} (1 + 4t) – (19 – 6t) \\ 2t – (-1 + 2t) \\ (12 – 2t) – (1 + 4t) \end{pmatrix} = \begin{pmatrix} 10t – 18 \\ 1 \\ -6t + 11 \end{pmatrix}\).
\( D(t) = \sqrt{(10t – 18)^2 + 1 + (-6t + 11)^2} \).
Minimize \( f(t) = D^2(t) = (10t – 18)^2 + 1 + (6t – 11)^2 = 100t^2 – 360t + 324 + 1 + 36t^2 – 132t + 121 = 136t^2 – 492t + 446 \) (M1).
Derivative: \( f'(t) = 272t – 492 = 0 \implies t = \frac{492}{272} = \frac{123}{68} \approx 1.8088 \).
Second derivative: \( f”(t) = 272 > 0 \), confirms minimum (A1).
Evaluate: \( 10t – 18 = 10 \cdot \frac{123}{68} – \frac{1224}{68} = \frac{6}{68} = \frac{3}{34} \), \( 6t – 11 = 6 \cdot \frac{123}{68} – \frac{748}{68} = -\frac{10}{68} = -\frac{5}{34} \).
\( D^2 = \left( \frac{3}{34} \right)^2 + 1 + \left( -\frac{5}{34} \right)^2 = \frac{9}{1156} + \frac{1156}{1156} + \frac{25}{1156} = \frac{1190}{1156} = \frac{595}{578} \).
\( D = \sqrt{\frac{595}{578}} \approx 1.0146 \, \text{km} \) (A1).
Answer: \( D \approx 1.01 \, \text{km} \).