2018 H2 Maths Paper 1 Q6 — Worked Solution

Part (i)

2 marks

Since $\mathbf{a} \times 3\mathbf{b} = 2\mathbf{a} \times \mathbf{c}$ ,

3\left( \mathbf{a} \times \mathbf{b} \right) = 2\left( \mathbf{a} \times \mathbf{c} \right)

3\left( \mathbf{a} \times \mathbf{b} \right) - 2\left( \mathbf{a} \times \mathbf{c} \right) = \mathbf{0}

\mathbf{a} \times \left( 3\mathbf{b} - 2\mathbf{c} \right) = \mathbf{0}.

Since $\mathbf{a} \neq \mathbf{0}$ , a vector cross product $\mathbf{a} \times \mathbf{v} = \mathbf{0}$ means $\mathbf{v}$ is parallel to $\mathbf{a}$ (or $\mathbf{v} = \mathbf{0}$ , the degenerate case of parallel). So $3\mathbf{b} - 2\mathbf{c}$ is parallel to $\mathbf{a}$ , i.e.

3\mathbf{b} - 2\mathbf{c} = \lambda\mathbf{a}

for some constant $\lambda$ . (shown)

Part (ii)

5 marks

Approach

We are given magnitudes and an angle, which is scalar-product information, so we turn the vector equation from (i) into a scalar one by taking the modulus of both sides and squaring — this brings in $\left| \mathbf{b} \right|$ , $\left| \mathbf{c} \right|$ and $\mathbf{b} \cdot \mathbf{c}$ , all of which we know, and $\left| \mathbf{a} \right| = 1$ .

Since $\mathbf{a}$ and $\mathbf{c}$ are unit vectors, $\left| \mathbf{a} \right| = 1$ and $\left| \mathbf{c} \right| = 1$ ; also $\left| \mathbf{b} \right| = 4$ and the angle between $\mathbf{b}$ and $\mathbf{c}$ is $60^{\circ}$ , so

\mathbf{b} \cdot \mathbf{c} = \left| \mathbf{b} \right|\left| \mathbf{c} \right|\cos 60^{\circ} = 4(1)\left( \frac{1}{2} \right) = 2.

Taking the modulus squared of both sides of $3\mathbf{b} - 2\mathbf{c} = \lambda\mathbf{a}$ ,

\left| 3\mathbf{b} - 2\mathbf{c} \right|^{2} = \lambda^{2}\left| \mathbf{a} \right|^{2}

\left( 3\mathbf{b} - 2\mathbf{c} \right) \cdot \left( 3\mathbf{b} - 2\mathbf{c} \right) = \lambda^{2}

9\left| \mathbf{b} \right|^{2} - 12\left( \mathbf{b} \cdot \mathbf{c} \right) + 4\left| \mathbf{c} \right|^{2} = \lambda^{2}

9(16) - 12(2) + 4(1) = \lambda^{2}

124 = \lambda^{2}.

Solving,

\lambda = \pm \sqrt{124} = \pm 2\sqrt{31}.

$\therefore\lambda = 2\sqrt{31}$ or $\lambda = - 2\sqrt{31}$ .

2018 H2 Maths Paper 1, Question 6

Part (i)

Part (ii)

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