2018 H2 Mathematics Paper 1 Question 6

Vectors I: Basics, Dot and Cross Products

Answers

(ii)

λ=±231{\lambda = \pm 2 \sqrt{31}}

Full solutions

(i)

a×3b=2a×ca×3b−2a×c=0a×(3b−2c)=0\begin{gather*} \mathbf{a} \times 3 \mathbf{b} = 2 \mathbf{a} \times \mathbf{c} \\ \mathbf{a} \times 3 \mathbf{b} - 2 \mathbf{a} \times \mathbf{c} = 0 \\ \mathbf{a} \times \left( 3\mathbf{b}-2\mathbf{c}\right) = 0 \end{gather*}
Hence a{\mathbf{a}} and 3b−2c{3\mathbf{b}-2\mathbf{c}} are parallel (or one/both are 0{\mathbf{0}})
∴3b−2c=λa  ■\therefore 3\mathbf{b}-2\mathbf{c} = \lambda \mathbf{a} \; \blacksquare

(ii)

∣a∣=∣c∣=1{|\mathbf{a}|=|\mathbf{c}|=1} and ∣b∣=4{|\mathbf{b}|=4}
From (i), taking the dot product with itself,
(3b−2c)⋅(3b−2c)=(λa)⋅(λa)9b⋅b−6c⋅b−6b⋅c+4c⋅c=λ2a⋅a\begin{gather*} (3\mathbf{b}-2\mathbf{c})\cdot(3\mathbf{b}-2\mathbf{c}) = (\lambda \mathbf{a})\cdot(\lambda \mathbf{a}) \\ 9\mathbf{b}\cdot\mathbf{b}-6\mathbf{c}\cdot\mathbf{b}-6\mathbf{b}\cdot\mathbf{c}+4\mathbf{c}\cdot\mathbf{c} = \lambda^2 \mathbf{a}\cdot\mathbf{a} \\ \end{gather*}
λ2∣a∣2=9∣b∣2−12b⋅c+4∣c∣2λ2(1)2=9(4)2−12∣b∣∣c∣cos⁡θ+4(1)2λ2=148−12(4)(1)cos⁡60∘=124λ=±231  ■\begin{align*} \lambda^2 \left|\mathbf{a}\right|^2 &= 9\left|\mathbf{b}\right|^2 - 12\mathbf{b}\cdot\mathbf{c} + 4\left|\mathbf{c}\right|^2 \\ \lambda^2 (1)^2 &= 9(4)^2 - 12 \left|\mathbf{b}\right|\left|\mathbf{c}\right|\cos\theta + 4(1)^2 \\ \lambda^2 &= 148 - 12(4)(1)\cos 60^\circ \\ &= 124 \\ \lambda &= \pm 2 \sqrt{31} \; \blacksquare \end{align*}