a. [3 marks]

Given: \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta\), \(|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}| |\mathbf{v}| \sin \theta\) (M1).
Compute: \((\mathbf{u} \cdot \mathbf{v})^2 = (|\mathbf{u}| |\mathbf{v}| \cos \theta)^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 \cos^2 \theta\),
\(|\mathbf{u} \times \mathbf{v}|^2 = (|\mathbf{u}| |\mathbf{v}| \sin \theta)^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 \sin^2 \theta\) (A1).
Sum: \((\mathbf{u} \cdot \mathbf{v})^2 + |\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 (\cos^2 \theta + \sin^2 \theta) = |\mathbf{u}|^2 |\mathbf{v}|^2 \cdot 1 = |\mathbf{u}|^2 |\mathbf{v}|^2\) (A1).
Answer: \((\mathbf{u} \cdot \mathbf{v})^2 + |\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2\).

b. [2 + 3 + 4 marks]

Given: \(\mathbf{u} = \overrightarrow{AB} = \begin{pmatrix} p \\ q – 1 \\ 1 \end{pmatrix}\), \(\mathbf{v} = \overrightarrow{AC} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}\), \(\mathbf{u} \cdot \mathbf{v} = 3\), area of \(\triangle ABC = \sqrt{6}\).
(i) Area: \(\frac{1}{2} |\mathbf{u} \times \mathbf{v}| = \sqrt{6} \implies |\mathbf{u} \times \mathbf{v}| = 2 \sqrt{6}\) (M1)(A1).
Answer: \(|\mathbf{u} \times \mathbf{v}| = 2 \sqrt{6}\).
(ii) Use identity: \((\mathbf{u} \cdot \mathbf{v})^2 + |\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2\).
Compute: \(|\mathbf{v}|^2 = 3^2 + 1^2 + (-1)^2 = 11\), \((\mathbf{u} \cdot \mathbf{v})^2 = 3^2 = 9\), \(|\mathbf{u} \times \mathbf{v}|^2 = (2 \sqrt{6})^2 = 24\) (M1).
Substitute: \(9 + 24 = |\mathbf{u}|^2 \cdot 11 \implies 33 = 11 |\mathbf{u}|^2 \implies |\mathbf{u}|^2 = 3 \implies |\mathbf{u}| = \sqrt{3}\) (A1)(A1).
Answer: \(|\mathbf{u}| = \sqrt{3}\).
(iii) Dot product: \(\mathbf{u} \cdot \mathbf{v} = 3p + (q – 1) \cdot 1 + 1 \cdot (-1) = 3p + q – 2 = 3 \implies 3p + q = 5\) (M1).
Magnitude: \(|\mathbf{u}|^2 = p^2 + (q – 1)^2 + 1 = 3 \implies p^2 + (q – 1)^2 = 2\) (M1).
Substitute \(q = 5 – 3p\): \(p^2 + (4 – 3p)^2 = 2 \implies p^2 + 16 – 24p + 9p^2 = 2 \implies 10p^2 – 24p + 14 = 0 \implies 5p^2 – 12p + 7 = 0\) (A1).
Solve: \(p = \frac{12 \pm \sqrt{144 – 140}}{10} = \frac{12 \pm 2}{10} \implies p = 1, \frac{7}{5}\).
For \(p = 1\): \(q = 5 – 3 \cdot 1 = 2\). For \(p = \frac{7}{5}\): \(q = 5 – 3 \cdot \frac{7}{5} = \frac{4}{5}\) (A1).
Answer: \(p = 1, q = 2\); \(p = \frac{7}{5}, q = \frac{4}{5}\).

c. [5 marks]

Given: \(p = 1\), \(q = 2\), so \(\mathbf{u} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\), \(\mathbf{v} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}\), \(\mathbf{w} = \overrightarrow{CD} = \begin{pmatrix} x – 3 \\ y – 2 \\ z – 1 \end{pmatrix}\), \(\mathbf{u} \cdot \mathbf{w} = 0\), \(\mathbf{v} \cdot \mathbf{w} = 0\), area of \(\triangle ACD = 5\).
Method 1:
Orthogonality: \(\mathbf{u} \cdot \mathbf{w} = x + y + z – 6 = 0 \implies x + y + z = 6\) (M1).
\(\mathbf{v} \cdot \mathbf{w} = 3x + y – z – 10 = 0\) (M1).
Solve: Subtract: \((x + y + z) – (3x + y – z) = 6 – 10 \implies -2x + 2z = -4 \implies z = x + 2\).
Substitute: \(x + y + (x + 2) = 6 \implies y = 4 – 2x\) (A1).
Area: \(\overrightarrow{CA} = -\mathbf{v} = \begin{pmatrix} -3 \\ -1 \\ 1 \end{pmatrix}\), area = \(\frac{1}{2} |(-\mathbf{v}) \times \mathbf{w}| = 5 \implies |(-\mathbf{v}) \times \mathbf{w}| = 10\).
Substitute \(\mathbf{w} = \begin{pmatrix} x \\ 4 – 2x \\ x + 2 \end{pmatrix}\), compute cross product: \(|(-\mathbf{v}) \times \mathbf{w}| = \sqrt{66} |x| = 10 \implies x^2 = \frac{100}{66} = \frac{50}{33} \implies x = \pm \frac{5\sqrt{66}}{33}\).
Then: \(y = 4 – 2 \left( \pm \frac{5\sqrt{66}}{33} \right) = \mp \frac{10\sqrt{66}}{33}\), \(z = \pm \frac{5\sqrt{66}}{33} + 2\) (A1).
Answer: \(\mathbf{w} = \pm \begin{pmatrix} \frac{5\sqrt{66}}{33} \\ -\frac{10\sqrt{66}}{33} \\ \frac{5\sqrt{66}}{33} \end{pmatrix}\).
Method 2:
Since \(\mathbf{w} \perp \mathbf{u}, \mathbf{v}\), \(\mathbf{w} = \lambda (\mathbf{u} \times \mathbf{v})\), where \(\mathbf{u} \times \mathbf{v} = \begin{pmatrix} -2 \\ 4 \\ -2 \end{pmatrix}\) (M1).
Area: \(|(-\mathbf{v}) \times \mathbf{w}| = |\lambda| \cdot |\begin{pmatrix} -3 \\ -1 \\ 1 \end{pmatrix} \times \begin{pmatrix} -2 \\ 4 \\ -2 \end{pmatrix}| = 10\). Compute cross product: \(\sqrt{140} = 2 \sqrt{35}\), so \(|\lambda| \cdot 2 \sqrt{35} = 10 \implies |\lambda| = \frac{5}{\sqrt{35}}\) (A1).
Thus: \(\mathbf{w} = \pm \frac{5}{\sqrt{35}} \begin{pmatrix} -2 \\ 4 \\ -2 \end{pmatrix} = \pm \begin{pmatrix} -\frac{10}{\sqrt{35}} \\ \frac{20}{\sqrt{35}} \\ -\frac{10}{\sqrt{35}} \end{pmatrix}\) (A1).
Answer: \(\mathbf{w} = \pm \begin{pmatrix} -\frac{10}{\sqrt{35}} \\ \frac{20}{\sqrt{35}} \\ -\frac{10}{\sqrt{35}} \end{pmatrix}\).
Method 3:
Area of \(\triangle ACD\): \(\frac{1}{2} |\mathbf{u} \times \mathbf{w}| = 5 \cdot \frac{|\mathbf{u}|}{|\mathbf{v}|} = 5 \cdot \frac{\sqrt{3}}{\sqrt{11}}\) (M1).
Since \(\mathbf{w} = \lambda \begin{pmatrix} -2 \\ 4 \\ -2 \end{pmatrix}\), compute \(\mathbf{u} \times \mathbf{w} = \lambda \begin{pmatrix} -6 \\ 0 \\ 6 \end{pmatrix}\), \(|\mathbf{u} \times \mathbf{w}| = |\lambda| \sqrt{72}\).
Set: \(\frac{1}{2} |\lambda| \sqrt{72} = 5 \cdot \frac{\sqrt{3}}{\sqrt{11}} \implies |\lambda| \sqrt{18} = \frac{5 \sqrt{33}}{\sqrt{11}} \implies |\lambda| = \frac{5}{\sqrt{66}}\) (A1).
Thus: \(\mathbf{w} = \pm \frac{5}{\sqrt{66}} \begin{pmatrix} -2 \\ 4 \\ -2 \end{pmatrix} = \pm \begin{pmatrix} -\frac{10}{\sqrt{66}} \\ \frac{20}{\sqrt{66}} \\ -\frac{10}{\sqrt{66}} \end{pmatrix}\) (A1).
Answer: \(\mathbf{w} = \pm \begin{pmatrix} -\frac{10}{\sqrt{66}} \\ \frac{20}{\sqrt{66}} \\ -\frac{10}{\sqrt{66}} \end{pmatrix}\).