2007 H2 Mathematics Paper 1 Question 6

Vectors I: Basics, Dot and Cross Products

Answers

{\overrightarrow{OM}=\frac{1}{3} \begin{pmatrix} 4 \\ 2 \\ 5 \end{pmatrix}}

Area of

{\triangle OAC} \allowbreak {= \sqrt{35} \textrm{ units}^2}

\begin{align*} & \overrightarrow{OA} \cdot \overrightarrow{OB} \\ &= \begin{pmatrix} 1 \\ - 1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 4 \\ 1 \end{pmatrix} \\ &= 2 - 4 + 2 \\ &= 0 \end{align*}

Hence

{OA}

is perpendicular to

{OB \; \blacksquare}

By Ratio Theorem,

\begin{align*} \overrightarrow{OM} &= \frac{2\overrightarrow{OA}+\overrightarrow{OB}}{3} \\ &= \frac{2\begin{pmatrix} 1 \\ - 1 \\ 2 \end{pmatrix}+\begin{pmatrix} 2 \\ 4 \\ 1 \end{pmatrix}}{3} \\ &= \frac{1}{3} \begin{pmatrix} 4 \\ 2 \\ 5 \end{pmatrix} \; \blacksquare \end{align*}

\begin{align*} & \textrm{Area of } \triangle OAC \\ &= \frac{1}{2} \left| \overrightarrow{OA} \times \overrightarrow{OC} \right| \\ &= \frac{1}{2} \left|\begin{pmatrix} 1 \\ - 1 \\ 2 \end{pmatrix} \times \begin{pmatrix} - 4 \\ 2 \\ 2 \end{pmatrix}\right| \\ &= \frac{1}{2} \left|\begin{pmatrix} - 6 \\ - 10 \\ - 2 \end{pmatrix} \right| \\ &= \frac{1}{2} \sqrt{36 + 100 + 4} \\ &= \frac{1}{2} \sqrt{140} \\ &= \sqrt{35} \textrm{ units}^2 \; \blacksquare \end{align*}