A2 June 2024 Paper 2 Q8

8 The vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are such that \(\mathbf{a} \times \mathbf{b} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}\) and \(\mathbf{a} \times \mathbf{c} = \begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix}\)

Work out \((\mathbf{a} - 4\mathbf{b} + 3\mathbf{c}) \times (2\mathbf{a})\) [4 marks]

Mark scheme

Scheme	Marks	AO
Uses distributive property of cross-product. Condone one numerical error.	M1	3.1a
Uses \(\mathbf{a} \times \mathbf{a} = \mathbf{0}\) PI by only \(\mathbf{b} \times \mathbf{a}\) and \(\mathbf{c} \times \mathbf{a}\) terms remaining	M1	1.1a
Uses anti-commutative property of cross-product at least once.	M1	1.1a
Obtains \(\begin{bmatrix} 16 \\ 8 \\ -18 \end{bmatrix}\)	A1	1.1b
	(4 marks)

Typical solution

\[\begin{aligned} (\mathbf{a} - 4\mathbf{b} + 3\mathbf{c}) \times (2\mathbf{a}) &= \mathbf{a} \times 2\mathbf{a} - 4\mathbf{b} \times 2\mathbf{a} + 3\mathbf{c} \times 2\mathbf{a} \\ &= \mathbf{0} + 8\mathbf{a} \times \mathbf{b} - 6\mathbf{a} \times \mathbf{c} \\ &= \begin{bmatrix} 16 \\ 8 \\ -18 \end{bmatrix} \end{aligned}\]