2009 H2 Mathematics Paper 2 Question 2

Vectors I: Basics, Dot and Cross Products

Answers

(i)

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

(iii)

c=17(6−23){\displaystyle \mathbf{c}=\frac{1}{7} \begin{pmatrix} 6 \\ - 2 \\ 3 \end{pmatrix}}
∣a⋅c∣{\left|\mathbf{a}\cdot\mathbf{c}\right|} represents the length of projection of OA→{\overrightarrow{OA}} on OP→{\overrightarrow{OP}}

(iv)

a×p=28(53−8){\mathbf{a}\times\mathbf{p}=28 \begin{pmatrix} 5 \\ 3 \\ - 8 \end{pmatrix}}
∣a×p∣{\left|\mathbf{a}\times\mathbf{p}\right|} represents the area of the parallelogram formed by OA→{\overrightarrow{OA}} and OP→{\overrightarrow{OP}}
Area of △OAP=982 units2{\triangle OAP} \allowbreak = {98 \sqrt{2} \textrm{ units}^2}

Full solutions

(i)

By Ratio Theorem,
OP→=2OB→+OA→3=2(11−132)+(141414)3=(12−46)\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(12,−4,6)  ■{P \left( 12, - 4, 6 \right) \; \blacksquare}

(ii)

AB→=OB→−OA→=(−3−27−12)=−3(194)\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*}
AB→⋅OP→=−3(194)⋅(12−46)=−3(12−36+24)=0\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{AB} and OP{OP} are perpendicular   ■{\; \blacksquare}

(iii)

c=OP→∣OP→∣=1144+16+36(12−46)=114(12−46)=17(6−23)  ■\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 ∣a⋅c∣{\left|\mathbf{a}\cdot\mathbf{c}\right|} is the length of projection of OA→{\overrightarrow{OA}} on OP→  ■{\overrightarrow{OP} \; \blacksquare}

(iv)

a×p=(141414)×(12−46)=14(2)(111)×(6−23)=28(53−8)  ■\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 ∣a×p∣{\left|\mathbf{a}\times\mathbf{p}\right|} is the area of the parallelogram formed by OA→{\overrightarrow{OA}} and OP→  ■{\overrightarrow{OP} \; \blacksquare}
Area of △OAP=12∣a×p∣=12∣28(53−8)∣=1425+9+64=1498=982 units2  ■\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*}