2009 H2 Mathematics Paper 2 Question 2

Vectors I: Basics, Dot and Cross Products

Answers

{P \left( 12, - 4, 6 \right)}

{\displaystyle \mathbf{c}=\frac{1}{7} \begin{pmatrix} 6 \\ - 2 \\ 3 \end{pmatrix}}

{\left|\mathbf{a}\cdot\mathbf{c}\right|}

represents the length of projection of

{\overrightarrow{OA}}

{\overrightarrow{OP}}

{\mathbf{a}\times\mathbf{p}=28 \begin{pmatrix} 5 \\ 3 \\ - 8 \end{pmatrix}}

{\left|\mathbf{a}\times\mathbf{p}\right|}

represents the area of the parallelogram formed by

{\overrightarrow{OA}}

and

{\overrightarrow{OP}}

Area of

{\triangle OAP} \allowbreak = {98 \sqrt{2} \textrm{ units}^2}

By Ratio Theorem,

\begin{align*} \overrightarrow{OP} &= \frac{2\overrightarrow{OB}+\overrightarrow{OA}}{3} \\ &= \frac{2\begin{pmatrix} 11 \\ - 13 \\ 2 \end{pmatrix}+\begin{pmatrix} 14 \\ 14 \\ 14 \end{pmatrix}}{3} \\ &= \begin{pmatrix} 12 \\ - 4 \\ 6 \end{pmatrix} \end{align*}

Coordinates of

{P \left( 12, - 4, 6 \right) \; \blacksquare}

\begin{align*} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix} - 3 \\ - 27 \\ - 12 \end{pmatrix} \\ &= - 3 \begin{pmatrix} 1 \\ 9 \\ 4 \end{pmatrix} \end{align*}

\begin{align*} & \overrightarrow{AB} \cdot \overrightarrow{OP} \\ &= - 3 \begin{pmatrix} 1 \\ 9 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 12 \\ - 4 \\ 6 \end{pmatrix} \\ &= -3(12 - 36 + 24) \\ &= 0 \end{align*}

Hence

{AB}

and

{OP}

are perpendicular

{\; \blacksquare}

\begin{align*} \mathbf{c} &= \frac{\overrightarrow{OP}}{\left|\overrightarrow{OP}\right|} \\ &= \frac{1}{\sqrt{144 + 16 + 36}} \begin{pmatrix} 12 \\ - 4 \\ 6 \end{pmatrix} \\ &= \frac{1}{14} \begin{pmatrix} 12 \\ - 4 \\ 6 \end{pmatrix} \\ &= \frac{1}{7} \begin{pmatrix} 6 \\ - 2 \\ 3 \end{pmatrix} \; \blacksquare \end{align*}

The geometric meaning of

{\left|\mathbf{a}\cdot\mathbf{c}\right|}

is the length of projection of

{\overrightarrow{OA}}

{\overrightarrow{OP} \; \blacksquare}

\begin{align*} & \mathbf{a} \times \mathbf{p} \\ &= \begin{pmatrix} 14 \\ 14 \\ 14 \end{pmatrix} \times \begin{pmatrix} 12 \\ - 4 \\ 6 \end{pmatrix} \\ &= 14(2) \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 6 \\ - 2 \\ 3 \end{pmatrix} \\ &= 28 \begin{pmatrix} 5 \\ 3 \\ - 8 \end{pmatrix} \; \blacksquare \end{align*}

The geometric meaning of

{\left|\mathbf{a}\times\mathbf{p}\right|}

is the area of the parallelogram formed by

{\overrightarrow{OA}}

and

{\overrightarrow{OP} \; \blacksquare}

\begin{align*} & \textrm{Area of } \triangle OAP \\ &= \frac{1}{2} \left|\mathbf{a}\times\mathbf{p}\right| \\ &= \frac{1}{2} \left| 28 \begin{pmatrix} 5 \\ 3 \\ - 8 \end{pmatrix} \right| \\ &= 14 \sqrt{25 + 9 + 64} \\ &= 14 \sqrt{98} \\ &= 98 \sqrt{2} \textrm{ units}^2 \; \blacksquare \end{align*}