(i)

\begin{align*} \overrightarrow{CM} & = -\overrightarrow{MC} \\ & = -{1 \over 2} \overrightarrow{BC} \\ & = -{1 \over 2} \overrightarrow{AD} \\ & = -{1 \over 2} \textbf{q} \end{align*}

(ii)

\begin{align*} \overrightarrow{DB} & = \overrightarrow{DA} + \overrightarrow{AB} \\ & = -\overrightarrow{AD} + \overrightarrow{AB} \\ & = -\textbf{q} + \textbf{p} \end{align*}

(iii)

\begin{align*} \overrightarrow{AM} & = \overrightarrow{AB} + \overrightarrow{BM} \\ & = \overrightarrow{AB} + {1 \over 2} \overrightarrow{BC} \\ & = \overrightarrow{AB} + {1 \over 2} \overrightarrow{AD} \\ & = \textbf{p} + {1 \over 2} \textbf{q} \end{align*}

(iv)

\begin{align*} \overrightarrow{MD} & = \overrightarrow{MB} + \overrightarrow{BD} \\ & = -\overrightarrow{BM} - \overrightarrow{DB} \\ & = -{1 \over 2}\overrightarrow{BC} - \overrightarrow{DB} \\ & = -{1 \over 2}\textbf{q} - (\textbf{p} - \textbf{q}) \phantom{0000} [\text{from part ii}]\\ & = -{1 \over 2}\textbf{q} - \textbf{p} + \textbf{q} \\ & = {1 \over 2}\textbf{q} - \textbf{p} \end{align*}

(i)

\begin{align*} \overrightarrow{BC} & = \overrightarrow{BA} + \overrightarrow{AC} \\ & = -\overrightarrow{AB} + \overrightarrow{AC} \\ & = - \textbf{u} + \textbf{v} \end{align*}

(ii)

\begin{align*} \overrightarrow{AM} & = {1 \over 2} \overrightarrow{AB} \\ & = {1 \over 2} \textbf{u} \end{align*}

(iii)

\begin{align*} \overrightarrow{AN} & = {1 \over 2} \overrightarrow{AC} \\ & = {1 \over 2} \textbf{v} \end{align*}

(iv)

\begin{align*} \overrightarrow{MN} & = \overrightarrow{MA} + \overrightarrow{AN} \\ & = -\overrightarrow{AM} + \overrightarrow{AN} \\ & = -{1 \over 2} \textbf{u} + {1 \over 2} \textbf{v} \end{align*}

Last part

\begin{align*} \overrightarrow{MN} & = -{1 \over 2} \textbf{u} + {1 \over 2} \textbf{v} \\ & = {1 \over 2} (-\textbf{u} + \textbf{v}) \\ & = {1 \over 2} \overrightarrow{BC} \phantom{00000} [\text{from part i}] \\ \\ 1. & \phantom{0} \overrightarrow{BC} \text{ and } \overrightarrow{MN} \text{ are parallel} \\ 2. & \phantom{0} \overrightarrow{MN} = {1 \over 2} \overrightarrow{BC} \end{align*}

\begin{align*} \overrightarrow{BM} & = \overrightarrow{BO} + \overrightarrow{OM} \\ & = -\overrightarrow{OB} + {1 \over 2} \overrightarrow{OA} \\ & = -\textbf{b} + {1 \over 2} \textbf{a} \end{align*}

(i)

\begin{align*} \overrightarrow{AB} & = \overrightarrow{AO} + \overrightarrow{OB} \\ & = -\overrightarrow{OA} + \overrightarrow{OB} \\ & = - \textbf{a} + \textbf{b} \end{align*}

(ii)

\begin{align*} AC & : CB \\ {2 \over 3} & : 1 \\ 2 & : 3 \\ \\ \overrightarrow{AC} & = {2 \over 5} \overrightarrow{AB} \\ & = {2 \over 5} ( -\textbf{a} + \textbf{b} ) \phantom{00000} [\text{from part i}] \end{align*}

(iii)

\begin{align*} \overrightarrow{OC} & = \overrightarrow{OA} + \overrightarrow{AC} \\ & = \textbf{a} + {2 \over 5} (-\textbf{a} + \textbf{b}) \\ & = \textbf{a} - {2 \over 5} \textbf{a} + {2 \over 5} \textbf{b} \\ & = {3 \over 5} \textbf{a} + {2 \over 5} \textbf{b} \\ & = {1 \over 5} (3 \textbf{a} + 2 \textbf{b}) \end{align*}

(i)

\begin{align*} \overrightarrow{QR} & = \overrightarrow{QO} + \overrightarrow{OR} \\ & = -\overrightarrow{OQ} + \overrightarrow{OR} \\ & = -{4 \choose 2} + {5 \choose 4} \\ & = {-4 + 5 \choose -2 + 4} \\ & = {1 \choose 2} \\ \\ \overrightarrow{PS} & = \overrightarrow{QR} \\ & = {1 \choose 2} \\ \\ \overrightarrow{OS} & = \overrightarrow{OP} + \overrightarrow{PS} \\ & = {1 \choose 0} + {1 \choose 2} \\ & = {2 \choose 2} \\ \\ \therefore & \phantom{.} S(2, 2) \end{align*}

(ii)

\begin{align*} \overrightarrow{RQ} & = \overrightarrow{RO} + \overrightarrow{OQ} \\ & = -\overrightarrow{OR} + \overrightarrow{OQ} \\ & = -{5 \choose 4} + {4 \choose 2} \\ & = {-5 + 4 \choose -4 + 2} \\ & = {-1 \choose -2} \\ \\ \overrightarrow{PS} & = \overrightarrow{RQ} \\ & = {-1 \choose -2} \\ \\ \overrightarrow{OS} & = \overrightarrow{OP} + \overrightarrow{PS} \\ & = {1 \choose 0} + {-1 \choose -2} \\ & = {1 - 1 \choose 0 - 2} \\ & = {0 \choose -2} \\ \\ \therefore & \phantom{.} S (0, -2) \end{align*}

(i)

\begin{align*} BD & : DC \\ 3 & : 1 \\ \\ \overrightarrow{BC} & = {4 \over 3} \overrightarrow{BD} \\ & = {4 \over 3} \textbf{q} \end{align*}

(ii)

\begin{align*} \overrightarrow{AD} & = \overrightarrow{AB} + \overrightarrow{BD} \\ & = -\overrightarrow{BA} + \overrightarrow{BD} \\ & = -\textbf{p} + \textbf{q} \end{align*}

(iii)

\begin{align*} \overrightarrow{CA} & = \overrightarrow{CB} + \overrightarrow{BA} \\ & = -\overrightarrow{BC} + \overrightarrow{BA} \\ & = -{4 \over 3} \textbf{q} + \textbf{p} \phantom{00000} [\text{Use } \overrightarrow{BC} \text{ from part i}] \end{align*}

(i)

\begin{align*} \overrightarrow{MR} & = \overrightarrow{MQ} + \overrightarrow{QR} \\ & = \overrightarrow{PM} + \overrightarrow{PS} \\ & = 2 \textbf{b} + \textbf{a} \end{align*}

(ii)

\begin{align*} SR & : SN \\ 3 & : 1 \\ \\ \overrightarrow{SR} & = \overrightarrow{PQ} \\ & = 2 \overrightarrow{PM} \\ & = 2( 2 \textbf{b} ) \\ & = 4 \textbf{b} \\ \\ \overrightarrow{RN} & = - \overrightarrow{NR} \\ & = - {2 \over 3} \overrightarrow{SR} \\ & = - {2 \over 3} (4 \textbf{b}) \\ & = - {8 \over 3} \textbf{b} \end{align*}

(iii)

\begin{align*} \overrightarrow{NM} & = \overrightarrow{NR} + \overrightarrow{RM} \\ & = -\overrightarrow{RN} - \overrightarrow{MR} \\ & = - \left(-{8 \over 3} \textbf{b} \right) - (2 \textbf{b} + \textbf{a}) \\ & = {8 \over 3} \textbf{b} - 2 \textbf{b} - \textbf{a} \\ & = {2 \over 3} \textbf{b} - \textbf{a} \end{align*}

(i)

\begin{align*} \overrightarrow{BC} & = \overrightarrow{BA} + \overrightarrow{AC} \\ & = -\overrightarrow{AB} + \overrightarrow{AC} \\ & = - \textbf{u} + \textbf{v} \end{align*}

(ii)

\begin{align*} \overrightarrow{BE} & = {2 \over 5} \overrightarrow{BC} \\ & = {2 \over 5} ( - \textbf{u} + \textbf{v} ) \phantom{00000} [\text{from part i}] \end{align*}

(iii)

\begin{align*} \overrightarrow{AD} & = \overrightarrow{AC} + \overrightarrow{CD} \\ & = \textbf{v} + {3 \over 2} \textbf{u} \end{align*}

(iv)

\begin{align*} \overrightarrow{AE} & = \overrightarrow{AB} + \overrightarrow{BE} \\ & = \textbf{u} + {2 \over 5} ( - \textbf{u} + \textbf{v} ) \phantom{00000} [\text{From part ii}] \\ & = \textbf{u} - {2 \over 5} \textbf{u} + {2 \over 5} \textbf{v} \\ & = {3 \over 5} \textbf{u} + {2 \over 5} \textbf{v} \\ & = {1 \over 5} (3 \textbf{u} + 2 \textbf{v}) \end{align*}

(v)

\begin{align*} \overrightarrow{BD} & = \overrightarrow{BA} + \overrightarrow{AD} \\ & = -\overrightarrow{AB} + \overrightarrow{AD} \\ & = - \textbf{u} + \textbf{v} + {3 \over 2} \textbf{u} \\ & = {1 \over 2} \textbf{u} + \textbf{v} \end{align*}

(i)

\begin{align*} \overrightarrow{PR} & = \overrightarrow{PO} + \overrightarrow{OR} \\ & = -\overrightarrow{OP} + \overrightarrow{OR} \\ & = - 15 \textbf{a} + 15\textbf{b} \end{align*}

(ii)

\begin{align*} AR & : PR \\ {3 \over 4} & : 1 \\ 3 & : 4 \\ \\ \overrightarrow{PA} & = {1 \over 4} \overrightarrow{PR} \\ & = {1 \over 4} (- 15 \textbf{a} + 15\textbf{b}) \\ & = {15 \over 4} ( - \textbf{a} + \textbf{b} ) \end{align*}

(iii)

\begin{align*} \overrightarrow{OA} & = \overrightarrow{OP} + \overrightarrow{PA} \\ & = 15 \textbf{a} + {15 \over 4} ( - \textbf{a} + \textbf{b} ) \\ & = 15 \textbf{a} - {15 \over 4} \textbf{a} + {15 \over 4} \textbf{b} \\ & = {45 \over 4} \textbf{a} + {15 \over 4} \textbf{b} \\ & = {15 \over 4} (3 \textbf{a} + \textbf{b} ) \end{align*}

(iv)

\begin{align*} \overrightarrow{PB} & = {1 \over 3} \overrightarrow{PQ} \\ & = {1 \over 3} \overrightarrow{OR} \\ & = {1 \over 3} (15 \textbf{b}) \\ & = 5 \textbf{b} \\ \\ \overrightarrow{OB} & = \overrightarrow{OP} + \overrightarrow{PB} \\ & = 15 \textbf{a} + 5 \textbf{b} \end{align*}

(i)

\begin{align*} OQ & : QC \\ {2 \over 3} & : 1 \\ 2 & : 3 \\ \\ \overrightarrow{PC} & = \overrightarrow{PO} + \overrightarrow{OC} \\ & = -\overrightarrow{OP} + {5 \over 2} \overrightarrow{OQ} \\ & = - 8 \textbf{p} + {5 \over 2} (8 \textbf{q}) \\ & = - 8 \textbf{p} + 20 \textbf{q} \end{align*}

(ii)

\begin{align*} \overrightarrow{PB} & = {1 \over 4} \overrightarrow{PC} \\ & = {1 \over 4} (- 8 \textbf{p} + 20 \textbf{q} ) \\ & = -2 \textbf{p} + 5 \textbf{q} \end{align*}

(iii)

\begin{align*} \overrightarrow{OB} & = \overrightarrow{OP} + \overrightarrow{PB} \\ & = 8 \textbf{p} + (- 2 \textbf{p} + 5 \textbf{q} ) \\ & = 8 \textbf{p} - 2 \textbf{p} + 5 \textbf{q} \\ & = 6 \textbf{p} + 5 \textbf{q} \end{align*}

(iv)

\begin{align*} \overrightarrow{QB} & = \overrightarrow{QO} + \overrightarrow{OB} \\ & = - \overrightarrow{OQ} + \overrightarrow{OB} \\ & = - (8 \textbf{q} ) + 6 \textbf{p} + 5 \textbf{q} \\ & = - 8 \textbf{q} + 6 \textbf{p} + 5 \textbf{q} \\ & = 6 \textbf{p} - 3 \textbf{q} \end{align*}

(a)(i)

\begin{align*} \overrightarrow{PQ} & = \overrightarrow{PO} + \overrightarrow{OQ} \\ & = - \overrightarrow{OP} + \overrightarrow{OQ} \\ & = - {1 \choose 11} + {2 \choose 8} \\ & = {-1 \choose -11} + {2 \choose 8} \\ & = {1 \choose -3} \end{align*}

(a)(ii)

\begin{align*} \overrightarrow{SR} & = \overrightarrow{SO} + \overrightarrow{OR} \\ & = - \overrightarrow{OS} + \overrightarrow{OR} \\ & = - {-2 \choose 8} + {-1 \choose 7} \\ & = {2 \choose -8} + {-1 \choose 7} \\ & = {1 \choose -1} \end{align*}

(a)(iii)

\begin{align*} \overrightarrow{RQ} & = \overrightarrow{RO} + \overrightarrow{OQ} \\ & = - \overrightarrow{OR} + \overrightarrow{OQ} \\ & = - {-1 \choose 7} + {2 \choose 8} \\ & = {1 \choose -7} + {2 \choose 8} \\ & = {3 \choose 1} \end{align*}

(a)(iv)

\begin{align*} \overrightarrow{TQ} & = \overrightarrow{TO} + \overrightarrow{OQ} \\ & = - \overrightarrow{OT} + \overrightarrow{OQ} \\ & = - {-4 \choose 6} + {2 \choose 8} \\ & = {4 \choose -6} + {2 \choose 8} \\ & = {6 \choose 2} \end{align*}

(b)

\begin{align*} \overrightarrow{TQ} & = {6 \choose 2} \\ & = 2 {3 \choose 1} \\ & = 2 \overrightarrow{RQ} \\ \\ TQ & : RQ \\ 2 & : 1 \\ \\ \therefore {RQ \over TQ} & = {1 \over 2} \end{align*}

(i)

\begin{align*} \overrightarrow{BC} & = \overrightarrow{BA} + \overrightarrow{AC} \\ & = - \overrightarrow{AB} + \overrightarrow{AC} \\ & = - {4 \choose -5} + {6 \choose 3} \\ & = {-4 \choose 5} + {6 \choose 3} \\ & = {2 \choose 8} \end{align*}

(ii)

\begin{align*} \overrightarrow{BM} & = {1 \over 2} \overrightarrow{BC} \\ & = {1 \over 2} {2 \choose 8} \\ & = {1 \choose 4} \\ \\ \overrightarrow{AM} & = \overrightarrow{AB} + \overrightarrow{BM} \\ & = {4 \choose -5} + {1 \choose 4} \\ & = {5 \choose -1} \end{align*}

(iii)

\begin{align*} \overrightarrow{AD} & = \overrightarrow{BC} \\ & = {2 \choose 8} \\ \\ \overrightarrow{OD} & = \overrightarrow{OA} + \overrightarrow{AD} \\ & = {1 \choose 2} + {2 \choose 8} \\ & = {3 \choose 10} \\ \\ \therefore & \phantom{.} D(3, 10) \end{align*}

(a)(i)

\begin{align*} AR & : SR \\ {1 \over 3} & : 1 \\ 1 & : 3 \\ \\ \overrightarrow{SA} & = {2 \over 3} \overrightarrow{SR} \\ & = {2 \over 3} \overrightarrow{PQ} \\ & = {2 \over 3} \textbf{b} \end{align*}

(a)(ii)

\begin{align*} BQ & : RB \\ 2 & : 1 \\ \\ \overrightarrow{QB} & = {2 \over 3} \overrightarrow{QR} \\ & = {2 \over 3} \overrightarrow{PS} \\ & = {2 \over 3} \textbf{a} \end{align*}

(a)(iii)

\begin{align*} \overrightarrow{PB} & = \overrightarrow{PQ} + \overrightarrow{QB} \\ & = \textbf{b} + {2 \over 3} \textbf{a} \end{align*}

(a)(iv)

\begin{align*} \overrightarrow{QS} & = \overrightarrow{QP} + \overrightarrow{PS} \\ & = - \overrightarrow{PQ} + \overrightarrow{PS} \\ & = - \textbf{b} + \textbf{a} \end{align*}

(a)(v)

\begin{align*} \overrightarrow{BA} & = \overrightarrow{BR} + \overrightarrow{RA} \\ & = {1 \over 3} \overrightarrow{QR} - {1 \over 3} \overrightarrow{SR} \\ & = {1 \over 3} \overrightarrow{PS} - {1 \over 3} \overrightarrow{PQ} \\ & = {1 \over 3} \textbf{a} - {1 \over 3} \textbf{b} \\ & = {1 \over 3} (\textbf{a} - \textbf{b}) \end{align*}

(b)(i)

\begin{align*} \overrightarrow{BA} & = {1 \over 3} (\textbf{a} - \textbf{b}) \\ & = {1 \over 3} \overrightarrow{QS} \phantom{00000} [\text{from part aiv}] \\ \\ BA & : QS \\ {1 \over 3} & : 1 \\ 1 & : 3 \\ \\ \therefore {BA \over QS} & = {1 \over 3} \end{align*}

(b)(ii) Since BA and QS are parallel, triangles ABR and SQR are similar.

\begin{align*} \text{For similar figures, }{A_1 \over A_2} & = \left( {l_1 \over l_2} \right)^2 \\ \\ { \text{Area of } \triangle ABR \over \text{Area of } \triangle SQR } & = \left( BA \over QS \right)^2 \\ & = \left( 1 \over 3 \right)^2 \\ & = {1 \over 9} \end{align*}

(b)(iii) Notice PQRS is made up of triangles SQR and SQP. SQR and SQP are congruent triangles, which means they have the same area.

\begin{align*} { \text{Area of } \triangle ABR \over \text{Area of } PQRS} & = {1 \over 9+9} \\ & = {1 \over 18} \end{align*}

(a)(i)

\begin{align*} \overrightarrow{RS} & = \overrightarrow{RP} + \overrightarrow{PS} \\ & = -\overrightarrow{PR} + \overrightarrow{PS} \\ & = - (3 \textbf{a} + 12 \textbf{b} ) + 5 \textbf{b} \\ & = - 3 \textbf{a} - 12 \textbf{b} + 5 \textbf{b} \\ & = - 3 \textbf{a} - 7 \textbf{b} \end{align*}

(a)(ii)

\begin{align*} PR & : PT \\ 3 & : 1 \\ \\ \overrightarrow{RT} & = {2 \over 3} \overrightarrow{RP} \\ & = -{2 \over 3} \overrightarrow{PR} \\ & = -{2 \over 3} ( 3 \textbf{a} + 12 \textbf{b} ) \\ & = - 2 \textbf{a} - 8 \textbf{b} \end{align*}

(a)(iii)

\begin{align*} \overrightarrow{RQ} & = \overrightarrow{RP} + \overrightarrow{PQ} \\ & = - \overrightarrow{PR} + \overrightarrow{PQ} \\ & = - (3 \textbf{a} + 12 \textbf{b} ) + 4 \textbf{a} + \textbf{b} \\ & = - 3 \textbf{a} - 12 \textbf{b} + 4 \textbf{a} + \textbf{b} \\ & = \textbf{a} - 11 \textbf{b} \end{align*}

(b)

\begin{align*} \overrightarrow{QT} & = \overrightarrow{QR} + \overrightarrow{RT} \\ & = - \overrightarrow{RQ} + \overrightarrow{RT} \\ & = -(\textbf{a} - 11 \textbf{b}) + (- 2 \textbf{a} - 8 \textbf{b}) \\ & = -\textbf{a} + 11 \textbf{b} - 2 \textbf{a} - 8 \textbf{b} \\ & = -3 \textbf{a} + 3 \textbf{b} \\ & = 3 \textbf{b} - 3 \textbf{a} \\ & = 3 ( \textbf{b} - \textbf{a} ) \phantom{000} \text{ (Shown)} \end{align*}

(c)

\begin{align*} \overrightarrow{QS} & = \overrightarrow{QR} + \overrightarrow{RS} \\ & = -\overrightarrow{RQ} + \overrightarrow{RS} \\ & = -( \textbf{a} - 11 \textbf{b} ) + (- 3 \textbf{a} - 7 \textbf{b}) \\ & = - \textbf{a} + 11 \textbf{b} - 3 \textbf{a} - 7 \textbf{b} \\ & = -4 \textbf{a} + 4 \textbf{b} \\ & = 4 \textbf{b} - 4 \textbf{a} \\ & = 4 (\textbf{b} - \textbf{a}) \end{align*}

(d)(i)

\begin{align*} \overrightarrow{QS} & = 4 (\textbf{b} - \textbf{a}) \\ & = {4 \over 3} [ 3 (\textbf{b} - \textbf{a}) ] \\ & = {4 \over 3} \overrightarrow{QT} \\ \\ QT & : QS \\ 1 & : {4 \over 3} \\ 3 & : 4 \\ \\ \therefore {QT \over QS} & = {3 \over 4} \end{align*}

(d)(ii) Triangles PQT and PQS share the same height.

\begin{align*} {\text{Area of } \triangle PQT \over \text{Area of } \triangle PQS} & = { {1 \over 2} \times QT \times h \over {1 \over 2} \times QS \times h} \\ & = {QT \over QS} \\ & = {3 \over 4} \end{align*}

(d)(iii) Triangles PQT and RQT share the same height.

\begin{align*} {\text{Area of } \triangle PQT \over \text{Area of } \triangle RQT} & = { {1 \over 2} \times PT \times h \over {1 \over 2} \times RT \times h} \\ & = {PT \over RT} \\ & = {1 \over 2} \end{align*}

(a)(i)

\begin{align*} \overrightarrow{OT} & = \overrightarrow{OC} + \overrightarrow{CT} \\ & = \overrightarrow{OC} - \overrightarrow{TC} \\ & = \textbf{q} - 3( \textbf{p} - \textbf{q} ) \\ & = \textbf{q} - 3 \textbf{p} + 3 \textbf{q} \\ & = 4 \textbf{q} - 3 \textbf{p} \end{align*}

(a)(ii)

\begin{align*} \overrightarrow{AT} & = \overrightarrow{AO} + \overrightarrow{OT} \\ & = - \overrightarrow{OA} + \overrightarrow{OT} \\ & = - \textbf{p} + ( 4 \textbf{q} - 3 \textbf{p} ) \\ & = - \textbf{p} + 4 \textbf{q} - 3 \textbf{p} \\ & = 4 \textbf{q} - 4 \textbf{p} \end{align*}

(a)(iii)

\begin{align*} \overrightarrow{AB} & = \overrightarrow{OC} \phantom{00000} [\text{Opposite sides of parallelogram}] \\ & = \textbf{q} \\ \\ \overrightarrow{OB} & = \overrightarrow{OA} + \overrightarrow{AB} \\ & = \textbf{p} + \textbf{q} \end{align*}

(a)(iv)

\begin{align*} \overrightarrow{BT} & = \overrightarrow{BO} + \overrightarrow{OT} \\ & = - \overrightarrow{OB} + \overrightarrow{OT} \\ & = - (\textbf{p} + \textbf{q}) + ( 4 \textbf{q} - 3 \textbf{p} ) \\ & = - \textbf{p} - \textbf{q} + 4 \textbf{q} - 3 \textbf{p} \\ & = 3 \textbf{q} - 4 \textbf{p} \end{align*}

(a)(v)

\begin{align*} BT & : BR \\ 4 & : 1 \\ \\ \overrightarrow{TR} & = {3 \over 4} \overrightarrow{TB} \\ & = - {3 \over 4} \overrightarrow{BT} \\ & = - {3 \over 4} (3 \textbf{q} - 4 \textbf{p} ) \\ & = {3 \over 4} ( 4 \textbf{p} - 3 \textbf{q} ) \end{align*}

(b)

\begin{align*} \overrightarrow{CR} & = \overrightarrow{CT} + \overrightarrow{TR} \\ & = -\overrightarrow{TC} + \overrightarrow{TR} \\ & = - 3 ( \textbf{p} - \textbf{q} ) + {3 \over 4} ( 4 \textbf{p} - 3 \textbf{q} ) \\ & = -3 \textbf{p} + 3 \textbf{q} + 3 \textbf{p} - {9 \over 4} \textbf{q} \\ & = {3 \over 4} \textbf{q} \phantom{000} (\text{Shown)} \end{align*}

(c)(i)

\begin{align*} \overrightarrow{CR} & = {3 \over 4} \textbf{q} \\ & = {3 \over 4} \overrightarrow{OC} \\ \\ CR & : OC \\ {3 \over 4} & : 1 \\ 3 & : 4 \\ \\ \therefore {CR \over OC} & = {3 \over 4} \end{align*}

(c)(ii) Since $\overrightarrow{CR} = {3 \over 4} \overrightarrow{OC}$, the points $O$, $C$ and $R$ lie in a straight line. Thus, $CR$ is parallel to $AB$ and triangles $TCR$ and $TAB$ are similar.

\begin{align*} CR & : OC : AB \\ 3 & : \phantom{.}4 \phantom{-} : 4 \\ \\ \text{For similar figures, } {A_1 \over A_2} & = \left(l_1 \over l_2\right)^2 \\ \\ { \text{Area of } \triangle TCR \over \text{Area of } \triangle TAB} & = \left( {CR \over AB} \right)^2 \\ & = \left(3 \over 4\right)^2 \\ & = {9 \over 16} \end{align*}

(a)(i)

\begin{align*} \overrightarrow{QP} & = \overrightarrow{QO} + \overrightarrow{OP} \\ & = - \overrightarrow{OQ} + \overrightarrow{OP} \\ & = - \textbf{q} + \textbf{p} \end{align*}

(a)(ii)

\begin{align*} \overrightarrow{QS} & = {2 \over 5} \overrightarrow{QP} \\ & = {2 \over 5} ( - \textbf{q} + \textbf{p} ) \end{align*}

(a)(iii)

\begin{align*} \overrightarrow{OS} & = \overrightarrow{OQ} + \overrightarrow{QS} \\ & = \textbf{q} + {2 \over 5} ( - \textbf{q} + \textbf{p} ) \\ & = \textbf{q} - {2 \over 5} \textbf{q} + {2 \over 5} \textbf{p} \\ & = {3 \over 5} \textbf{q} + {2 \over 5} \textbf{p} \\ & = {1 \over 5} ( 3 \textbf{q} + 2 \textbf{p} ) \end{align*}

(a)(iv)

\begin{align*} \overrightarrow{ST} & = \overrightarrow{SQ} + \overrightarrow{QT} \\ & = - \overrightarrow{QS} + {1 \over 2} \overrightarrow{OQ} \\ & = - {2 \over 5} ( - \textbf{q} + \textbf{p} ) + {1 \over 2} \textbf{q} \\ & = {2 \over 5} \textbf{q} - {2 \over 5} \textbf{p} + {1 \over 2} \textbf{q} \\ & = {9 \over 10} \textbf{q} - {2 \over 5} \textbf{p} \\ & = {1 \over 10} (9 \textbf{q} - 4 \textbf{p} ) \end{align*}

(b)(i)

\begin{align*} \overrightarrow{RS} & = \overrightarrow{RP} + \overrightarrow{PS} \\ & = {1 \over 3} \overrightarrow{OP} - \overrightarrow{SP} \\ & = {1 \over 3} \overrightarrow{OP} - {3 \over 5} \overrightarrow{QP} \\ & = {1 \over 3} \textbf{p} - {3 \over 5} (- \textbf{q} + \textbf{p} ) \\ & = {1 \over 3} \textbf{p} + {3 \over 5} \textbf{q} - {3 \over 5} \textbf{p} \\ & = {3 \over 5} \textbf{q} - {4 \over 15} \textbf{p} \\ & = {1 \over 15} ( 9 \textbf{q} - 4 \textbf{p} ) \\ & = {10 \over 15} \left[ {1 \over 10} ( 9 \textbf{q} - 4 \textbf{p} ) \right] \\ & = {2 \over 3} \overrightarrow{ST} \phantom{00} \text{ (Shown)} \end{align*}

(b)(ii)

\begin{align*} RS & : ST \\ {2 \over 3} & : 1 \\ 2 & : 3 \\ \\ 1. & \text{ Points } R, S \text{ and } T \text{ are collinear (i.e. lie in a straight line)} \\ 2. & \phantom{0} 3RS = 2ST \end{align*}

(i)

\begin{align*} \overrightarrow{AD} & = \overrightarrow{AB} + \overrightarrow{BD} \\ \\ \overrightarrow{PQ} & = \overrightarrow{PB} + \overrightarrow{BQ} \\ & = {1 \over 2} \overrightarrow{AB} + {1 \over 2} \overrightarrow{BD} \\ & = {1 \over 2} ( \overrightarrow{AB} + \overrightarrow{BD} ) \\ & = {1 \over 2} \overrightarrow{AD} \\ \\ \therefore PQ & \phantom{.} // \phantom{.} AD \text{ and } PQ = {1 \over 2} AD \end{align*}

(ii)

\begin{align*} \overrightarrow{SR} & = \overrightarrow{SC} + \overrightarrow{CR} \\ & = {1 \over 2} \overrightarrow{AC} + {1 \over 2} \overrightarrow{CD} \\ & = {1 \over 2} ( \overrightarrow{AC} + \overrightarrow{CD} ) \\ & = {1 \over 2} \overrightarrow{AD} \\ & = \overrightarrow{PQ} \\ \\ \overrightarrow{PS} & = \overrightarrow{PA} + \overrightarrow{AS} \\ & = {1 \over 2} \overrightarrow{BA} + {1 \over 2} \overrightarrow{AC} \\ & = {1 \over 2} ( \overrightarrow{BA} + \overrightarrow{AC} ) \\ & = {1 \over 2} \overrightarrow{BC} \\ \\ \overrightarrow{QR} & = \overrightarrow{QD} + \overrightarrow{DR} \\ & = {1 \over 2} \overrightarrow{BD} + {1 \over 2} \overrightarrow{DC} \\ & = {1 \over 2} ( \overrightarrow{BD} + \overrightarrow{DC} ) \\ & = {1 \over 2} \overrightarrow{BC} \\ & = \overrightarrow{PS} \\ \\ \\ \text{Since } \overrightarrow{PQ} = \overrightarrow{SR} & \text{ and } \overrightarrow{PS} = \overrightarrow{QR}, \phantom{.} PQRS \text{ is a parallelogram} \end{align*}