A2 June 2025 Paper 1 Q9

9 The vectors \(\mathbf{a}\) and \(\mathbf{b}\) are such that

\[\begin{aligned} &|\mathbf{a} \times \mathbf{b}| = 12 \\ &\mathbf{a} \bullet \mathbf{b} = -\sqrt{3} \\ &|\mathbf{a}| = 7 \end{aligned}\]

Find the value of \(|\mathbf{b}|\) [4 marks]

Mark scheme

Scheme	Marks	AO
Obtains \(-\sqrt{3} = 7\|\mathbf{b}\|\cos\theta\) OE or \(-\sqrt{3} = \|\mathbf{a}\|\|\mathbf{b}\|\cos\theta\) OE	B1	1.2
Obtains \(12 = 7\|\mathbf{b}\|\sin\theta\) OE or \(12 = \|\mathbf{a}\|\|\mathbf{b}\|\sin\theta\) OE	B1	1.1b
Obtains an equation in only \(\|\mathbf{b}\|^2\) or \(\|\mathbf{b}\|\) from a valid method	M1	3.1a
Deduces that \(\|\mathbf{b}\| = \sqrt{3}\) Accept AWRT 1.73	A1	2.2a
	(4 marks)

Typical solution

\[-\sqrt{3} = 7|\mathbf{b}|\cos\theta\]\[\cos\theta = \frac{-\sqrt{3}}{7|\mathbf{b}|}\]\[12 = 7|\mathbf{b}|\sin\theta\]\[\sin\theta = \frac{12}{7|\mathbf{b}|}\]\[\begin{aligned} 1 &= \cos^2\theta + \sin^2\theta \\ &= \frac{3}{49|\mathbf{b}|^2} + \frac{144}{49|\mathbf{b}|^2} \\ &= \frac{3}{|\mathbf{b}|^2} \end{aligned}\]\[|\mathbf{b}|^2 = 3\]\[|\mathbf{b}| = \sqrt{3}\]