(a)(i)
Vectors: \(\mathbf{a} = \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix} 0 \\ -3 \\ 4 \end{pmatrix}\)
Dot product: \(\mathbf{a} \cdot \mathbf{b} = (6)(0) + (3)(-3) + (2)(4) = -1\)
Magnitudes: \(|\mathbf{a}| = \sqrt{6^2 + 3^2 + 2^2} = 7\), \(|\mathbf{b}| = \sqrt{0^2 + (-3)^2 + 4^2} = 5\)
Cosine: \(\cos \theta = \frac{-1}{7 \times 5} = -\frac{1}{35}\)
Marks: (M1) for dot product formula, (A1) for dot product, (A1) for magnitudes, (A1) for cosine
Result: \(\cos \theta = -\frac{1}{35}\) [4]

(a)(ii)
Cross product: \(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & 3 & 2 \\ 0 & -3 & 4 \end{vmatrix}\)
\(\mathbf{i}\): \(3 \times 4 – 2 \times (-3) = 18\)
\(\mathbf{j}\): \(-(6 \times 4 – 2 \times 0) = -24\)
\(\mathbf{k}\): \(6 \times (-3) – 3 \times 0 = -18\)
Result: \(\mathbf{a} \times \mathbf{b} = 18\mathbf{i} – 24\mathbf{j} – 18\mathbf{k}\)
Marks: (M1) for cross product setup, (A1) for result
Result: \(18\mathbf{i} – 24\mathbf{j} – 18\mathbf{k}\) [2]

(a)(iii)
Normal: \(\mathbf{n} = \mathbf{a} \times \mathbf{b} = \begin{pmatrix} 18 \\ -24 \\ -18 \end{pmatrix}\)
Point: \(\mathbf{p} = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\)
Plane: \(\mathbf{r} \cdot \mathbf{n} = \mathbf{p} \cdot \mathbf{n}\)
Compute: \(\mathbf{p} \cdot \mathbf{n} = (1)(18) + (1)(-24) + (-1)(-18) = 12\)
Equation: \(18x – 24y – 18z = 12 \implies 3x – 4y – 3z = 2\)
Marks: (M1) for plane equation setup, (A1) for equation
Result: \(3x – 4y – 3z = 2\) [2]

(a)(iv)
Set \(z = 0\): \(3x – 4y = 2\)
Intercepts: \(x = \frac{2}{3}\), \(y = -\frac{1}{2}\)
Area: \(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{6}\)
Marks: (M1) for \(z = 0\), (A1) for intercepts, (A1) for area
Result: Area = \(\frac{1}{6}\) [3]

(b)(i)
Dot product: \(\mathbf{p} \cdot \mathbf{p} = |\mathbf{p}| |\mathbf{p}| \cos 0 = |\mathbf{p}|^2\)
Marks: (M1) for dot product, (A1) for result
Result: \(\mathbf{p} \cdot \mathbf{p} = |\mathbf{p}|^2\) [2]

(b)(ii)
Compute: \(|\mathbf{p} + \mathbf{q}|^2 = (\mathbf{p} + \mathbf{q}) \cdot (\mathbf{p} + \mathbf{q})\)
Expand: \(|\mathbf{p}|^2 + \mathbf{p} \cdot \mathbf{q} + \mathbf{q} \cdot \mathbf{p} + |\mathbf{q}|^2\)
Simplify: \(|\mathbf{p}|^2 + 2\mathbf{p} \cdot \mathbf{q} + |\mathbf{q}|^2\)
Marks: (M1) for expansion, (A1) for terms, (A1) for result
Result: \(|\mathbf{p} + \mathbf{q}|^2 = |\mathbf{p}|^2 + 2\mathbf{p} \cdot \mathbf{q} + |\mathbf{q}|^2\) [3]

(b)(iii) METHOD 1
Use: \(\mathbf{p} \cdot \mathbf{q} \leq |\mathbf{p}| |\mathbf{q}|\)
From (ii): \(|\mathbf{p} + \mathbf{q}|^2 \leq |\mathbf{p}|^2 + 2|\mathbf{p}| |\mathbf{q}| + |\mathbf{q}|^2\)
Take square root: \(|\mathbf{p} + \mathbf{q}| \leq |\mathbf{p}| + |\mathbf{q}|\)

(b)(iii) METHOD 2

A triangle with vectors  \(\mathbf{p}\), \(\mathbf{q}\), \(\mathbf{p} + \mathbf{q}\))

since the sum of any two sides of a triangle is greater than the third side, Triangle inequality: \(|\mathbf{p}| + |\mathbf{q}| \geq |\mathbf{p} + \mathbf{q}|\)
Equality when collinear
Conclusion: \(|\mathbf{p} + \mathbf{q}| \leq |\mathbf{p}| + |\mathbf{q}|\)
Marks: (M1) for inequality, (A1) for application, (A1) for deduction
Result: \(|\mathbf{p} + \mathbf{q}| \leq |\mathbf{p}| + |\mathbf{q}|\) [3]