Specimen 2017 H2 Mathematics Paper 2 Question 3

Vectors I: Basics, Dot and Cross Products

Answers

{\overrightarrow{OE} = \frac{18}{23} \mathbf{a} + \frac{20}{23} \mathbf{b}.}

(bii)

{OE:OF = 8:23}

Full solutions

(a)

\begin{gather*} \mathbf{a} \cdot \mathbf{b} = | \mathbf{a} | | \mathbf{b} | \cos \theta \\ \begin{pmatrix} 3 \\ - 2 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ d \\ - \sqrt{7} \end{pmatrix} = \sqrt{13} \sqrt{36+d^2+7} \left( \frac{6}{13} \right) \\ 18 - 2 d = \frac{6\sqrt{13}}{13} \sqrt{36+d^2+7} \\ 9 - d = \frac{3\sqrt{13}}{13} \sqrt{43+d^2} \\ 13^2 (9 - d)^2 = 9 (13) (43 + d^2) \\ 13(81 - 18 d + d^2) = 9(43 + d^2) \\ 4 d^2 - 234 d + 666 = 0 \\ 2 d^2 - 117 d + 333 = 0 \; \blacksquare \end{gather*}

(bi)

{\overrightarrow{OC}=6 \mathbf{a},}

{\overrightarrow{OD}=4 \mathbf{b}}

\begin{align*} \overrightarrow{AD} &= \overrightarrow{OD} - \overrightarrow{OA} \\ &= 4 \mathbf{b} - \mathbf{a} \\ \overrightarrow{BC} &= \overrightarrow{OC} - \overrightarrow{OB} \\ &= 6 \mathbf{a} - \mathbf{b} \end{align*}

\begin{align} && \quad l_{AD}: \mathbf{r} = \mathbf{a} + \lambda (4 \mathbf{b} - \mathbf{a}) \\ && \quad l_{BC}: \mathbf{r} = \mathbf{b} + \mu (6 \mathbf{a} - \mathbf{b}) \\ \end{align}

At point

{E,}

\mathbf{a} + \lambda (4 \mathbf{b} - \mathbf{a}) = \mathbf{b} + \mu (6 \mathbf{a} - \mathbf{b})

Comparing,

\begin{align} && \quad 1 - \lambda &= 6 \mu \\ && \quad 4\lambda &= 1 - \mu \\ \end{align}

Solving with a GC,

\lambda = \frac{5}{23}, \quad \mu = \frac{3}{23}

\begin{align*} \overrightarrow{OE} &= \mathbf{a} + \frac{5}{23} (4 \mathbf{b} - \mathbf{a}) \\ &= \frac{18}{23} \mathbf{a} + \frac{20}{23} \mathbf{b} \; \blacksquare \\ &= \frac{2}{23} (9 \mathbf{a} + 10 \mathbf{b}) \end{align*}

(bii)

\begin{align*} \overrightarrow{OF} &= \frac{5 \overrightarrow{OD} + 3 \overrightarrow{OC}}{5+3} \\ &= \frac{18 \mathbf{a} + 20 \mathbf{b}}{8} \\ &= \frac{1}{4} (9 \mathbf{a} + 10 \mathbf{b}) \end{align*}

Hence

{\overrightarrow{OE} = \frac{8}{23} \overrightarrow{OF}}

Hence ${O, E}$ and ${F}$ are collinear since ${\overrightarrow{OE}}$ and ${\overrightarrow{OF}}$ are parallel and they share a common point ${O. \; \blacksquare}$

${OE:OF = 8:23 \; \blacksquare}$