$\begin{array}{l}\overrightarrow{QT}=\overrightarrow{QA}+\overrightarrow{AT}\\ \text{ = }\overrightarrow{QA}+\frac{3}{2}\overrightarrow{AS}\leftarrow \boxed{\begin{array}{l}AS=2ST\\ AT=3ST=\frac{3}{2}AS\end{array}}\\ =-6\underset{˜}{a}+\frac{3}{2}\left(6\underset{˜}{a}+\underset{˜}{b}\right)\\ =3\underset{˜}{a}+\frac{3}{2}\underset{˜}{b}\\ =\frac{3}{2}\left(2\underset{˜}{a}+\underset{˜}{b}\right)\\ =\frac{3}{2}\overrightarrow{QC}\\ \text{Maka }Q,C\text{ dan }T\text{ adalah segaris}\text{.}\end{array}$

Soalan 10 (3 markah):
Rajah menunjukkan vektor-vektor $\overrightarrow{OP},\overrightarrow{OQ}\text{ dan }\overrightarrow{OM}$ dilukis pada grid segi empat sama.

Rajah

$\begin{array}{l}\left(\text{a}\right)\text{ Ungkapkan }\overrightarrow{OM}\text{ dalam bentuk }h\underset{˜}{p}+k\underset{˜}{q}\text{, }\\ \text{dengan keadaan }h\text{ dan }k\text{ ialah pemalar}.\\ \left(\text{b}\right)\text{ Pada Rajah 3, tanda dan label titik }N\\ \text{dengan keadaan }\overrightarrow{MN}+\overrightarrow{OQ}=2\overrightarrow{OP}.\end{array}$


Penyelesaian:
(a)
$\overrightarrow{OM}=\underset{˜}{p}+2\underset{˜}{q}$

(b)

$\begin{array}{l}\overrightarrow{MN}+\overrightarrow{OQ}=2\overrightarrow{OP}\\ \overrightarrow{MN}=2\overrightarrow{OP}-\overrightarrow{OQ}\\ =2\underset{˜}{p}-\underset{˜}{q}\end{array}$

Soalan 11 (4 markah):
$\begin{array}{l}A\left(2,3\right)\text{ dan }B\left(-2,\text{ 5}\right)\text{ terletak pada}\\ \text{suatu satah Cartes}\text{.}\\ \text{Diberi bahawa }3\overrightarrow{OA}=2\overrightarrow{OB}+\overrightarrow{OC}.\\ \text{Cari}\\ \left(a\right)\text{ koordinat }C,\\ \left(b\right)\left|\overrightarrow{AC}\right|\end{array}$


Penyelesaian:
$\begin{array}{l}\text{Diberi }A\left(2,3\right)\text{ dan }B\left(-2,5\right)\\ \text{Maka, }\overrightarrow{OA}=2\underset{˜}{i}+3\underset{˜}{j}\text{ dan }\overrightarrow{OB}=-2\underset{˜}{i}+5\underset{˜}{j}\end{array}$

(a)
$\begin{array}{l}3\overrightarrow{OA}=2\overrightarrow{OB}+\overrightarrow{OC}\\ \overrightarrow{OC}=3\overrightarrow{OA}-2\overrightarrow{OB}\\ =3\left(2\underset{˜}{i}+3\underset{˜}{j}\right)-2\left(-2\underset{˜}{i}+5\underset{˜}{j}\right)\\ =6\underset{˜}{i}+9\underset{˜}{j}+4\underset{˜}{i}-10\underset{˜}{j}\\ =10\underset{˜}{i}-\underset{˜}{j}\\ \\ \text{Maka, koordinat }C\text{ ialah }\left(10,-1\right)\end{array}$


(b)
$\begin{array}{l}\overrightarrow{AC}=\overrightarrow{AO}+\overrightarrow{OC}\\ =-\overrightarrow{OA}+\overrightarrow{OC}\\ =-\left(2\underset{˜}{i}+3\underset{˜}{j}\right)+10\underset{˜}{i}-\underset{˜}{j}\\ =-2\underset{˜}{i}+10\underset{˜}{i}-3\underset{˜}{j}-\underset{˜}{j}\\ =8\underset{˜}{i}-4\underset{˜}{j}\\ \\ \left|\overrightarrow{AC}\right|=\sqrt{8^2+{\left(-4\right)}^2}\\ =\sqrt{80}\text{ unit}\\ =\sqrt{16\times 5}\text{ unit}\\ =4\sqrt{5}\text{ unit}\end{array}$

Soalan 8 (3 markah):
Rajah menunjukkan trapezium ABCD.

Rajah

$\begin{array}{l}\text{Diberi }\underset{˜}{p}=\left(\begin{array}{l}3\\ 4\end{array}\right)\text{ dan }\underset{˜}{q}=\left(\begin{array}{l}k-1\\ 2\end{array}\right)\text{, dengan}\\ \text{keadaan }k\text{ ialah pemalar, cari nilai }k\text{.}\end{array}$


Penyelesaian:

$\begin{array}{l}\underset{˜}{p}=m\underset{˜}{q}\\ \left(\begin{array}{l}3\\ 4\end{array}\right)=m\left(\begin{array}{l}k-1\\ 2\end{array}\right)\\ \left(\begin{array}{l}3\\ 4\end{array}\right)=\left(\begin{array}{l}mk-m\\ 2m\end{array}\right)\\ \\ mk-m=3\mathrm{..........}\left(1\right)\\ 2m=4\mathrm{................}\left(2\right)\\ \\ \text{Dari}\left(2\right):\\ 2m=4\\ m=2\\ \\ \text{Gantikan }m=2\text{ ke dalam }\left(1\right):\\ 2k-2=3\\ 2k=3+2\\ 2k=5\\ k=\frac{5}{2}\end{array}$

Soalan 9 (3 markah):
Rajah menunjukkan vektor-vektor $\overrightarrow{AB},\overrightarrow{AC}\text{ dan }\overrightarrow{AD}$ yang dilukis pada grid segi empat sama bersisi 1 unit.

Rajah

$\begin{array}{l}\left(\text{a}\right)\text{ Cari }\left|-\overrightarrow{BA}\right|.\\ \left(\text{b}\right)\text{ Diberi }\overrightarrow{AB}=\underset{˜}{b}\text{ dan }\overrightarrow{AC}=\underset{˜}{c},\\ \text{ungkapkan dalam sebutan }\underset{˜}{b}\text{ dan }\underset{˜}{c}\\ \left(i\right)\overrightarrow{BC},\\ \left(ii\right)\overrightarrow{AD}\end{array}$

Penyelesaian:
(a)
$\left|-\overrightarrow{BA}\right|=\sqrt{3^2+4^2}=5\text{ unit}$

(b)(i)
$\begin{array}{l}\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}\\ =-\underset{˜}{b}+\underset{˜}{c}\\ =\underset{˜}{c}-\underset{˜}{b}\end{array}$

(b)(ii)
$\begin{array}{l}\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}\\ =\underset{˜}{b}+2\overrightarrow{BC}\\ =\underset{˜}{b}+2\left(\underset{˜}{c}-\underset{˜}{b}\right)\\ =2\underset{˜}{c}-\underset{˜}{b}\end{array}$

Soalan 10 (10 markah):
Rajah 5 menunjukkan segi tiga OAQ dan segi tiga OPB dengan keadaan titik P berada pada OA dan titik Q berada pada OB. Garis lurus AQ dan garis lurus PB bersilang pada titik R.

$\begin{array}{l}\text{Diberi bahawa }\overrightarrow{OA}=18\underset{˜}{x},\overrightarrow{OB}=16\underset{˜}{y},OP:PA=1:2,OQ:QB=3:1,\\ \overrightarrow{PR}=m\overrightarrow{PB}\text{ dan }\overrightarrow{QR}=n\overrightarrow{QA},\text{ dengan keadaan }m\text{ dan }n\text{ ialah pemalar}\text{.}\\ \left(a\right)\text{ Ungkapkan }\overrightarrow{OR}\text{ dalam sebutan}\\ \left(i\right)m,\underset{˜}{x}\text{ dan }\underset{˜}{y},\\ \left(ii\right)n,\underset{˜}{x}\text{ dan }\underset{˜}{y},\\ \left(b\right)\text{ Seterusnya, cari nilai }m\text{ dan nilai }n.\\ \left(c\right)\text{ Diberi }\left|\underset{˜}{x}\right|=2\text{ unit, }\left|\underset{˜}{y}\right|=1\text{ unit dan }OA\text{ berserenjang kepada }OB,\text{ hitung }\left|\overrightarrow{PR}\right|.\end{array}$


Penyelesaian: 
(a)(i)
$\begin{array}{l}\overrightarrow{OR}=\overrightarrow{OP}+\overrightarrow{PR}\\ =\frac{1}{3}\overrightarrow{OA}+m\overrightarrow{PB}\\ =\frac{1}{3}\left(18\underset{˜}{x}\right)+m\left(\overrightarrow{PO}+\overrightarrow{OB}\right)\\ =6\underset{˜}{x}+m\left(-6\underset{˜}{x}+16\underset{˜}{y}\right)\end{array}$

(a)(ii)
$\begin{array}{l}\overrightarrow{OR}=\overrightarrow{OQ}+\overrightarrow{QR}\\ =\frac{3}{4}\overrightarrow{OB}+n\overrightarrow{QA}\\ =\frac{3}{4}\left(16\underset{˜}{y}\right)+n\left(\overrightarrow{QO}+\overrightarrow{OA}\right)\\ =12\underset{˜}{y}+n\left(-12\underset{˜}{y}+18\underset{˜}{x}\right)\\ =\left(12-12n\right)\underset{˜}{y}+18n\underset{˜}{x}\end{array}$


(b)
$\begin{array}{l}6\underset{˜}{x}+m\left(-6\underset{˜}{x}+16\underset{˜}{y}\right)=\left(12-12n\right)\underset{˜}{y}+18n\underset{˜}{x}\\ 6\underset{˜}{x}-6m\underset{˜}{x}+16m\underset{˜}{y}=18n\underset{˜}{x}+12\underset{˜}{y}-12n\underset{˜}{y}\\ \text{Dengan cara perbandingan;}\\ 6-6m=18n\\ 1-m=3n\\ m=1-3n\mathrm{..............}\left(1\right)\\ \\ 16m=12-12n\\ 4m=3-3n\mathrm{..............}\left(2\right)\\ \\ \text{Gantikan (1) ke dalam (2),}\\ 4\left(1-3n\right)=3-3n\\ 4-12n=3-3n\\ 9n=1\\ n=\frac{1}{9}\\ \\ \text{Gantikan }n=\frac{1}{9}\text{ ke dalam (1),}\\ m=1-3\left(\frac{1}{9}\right)\\ m=\frac{2}{3}\end{array}$


(c)
$\begin{array}{l}\left|\underset{˜}{x}\right|=2\text{, }\left|\underset{˜}{y}\right|=1\\ \overrightarrow{PR}=\frac{2}{3}\overrightarrow{PB}\\ =\frac{2}{3}\left(-6\underset{˜}{x}+16\underset{˜}{y}\right)\\ =-4\underset{˜}{x}+\frac{32}{3}\underset{˜}{y}\\ \\ \left|\overrightarrow{PR}\right|=\sqrt{{\left[-4\left(2\right)\right]}^2+{\left[\frac{32}{3}\left(1\right)\right]}^2}\\ =\sqrt{\frac{1600}{9}}\\ =\frac{40}{3}\text{ units}\end{array}$

Soalan 9 (10 markah):
Rajah 5 menunjukkan segi tiga ABC. Garis lurus AE bersilang dengan garis lurus BC di titik D. Titik V terletak pada garis lurus AE.

$\begin{array}{l}\text{Diberi bahawa }\overrightarrow{BD}=\frac{1}{3}\overrightarrow{BC},\overrightarrow{AC}=6\underset{˜}{x}\text{ dan }\overrightarrow{AB}=9\underset{˜}{y}.\\ \left(a\right)\text{ Ungkapkan dalam sebutan }\underset{˜}{x}\text{ dan / atau }\underset{˜}{y}:\\ \left(i\right)\overrightarrow{BC},\\ \left(ii\right)\overrightarrow{AD}.\\ \left(b\right)\text{ Diberi bahawa }\overrightarrow{AV}=m\overrightarrow{AD}\text{ dan }\overrightarrow{BV}=n\left(\underset{˜}{x}-9\underset{˜}{y}\right),\\ \text{ dengan keadaan }m\text{ dan }n\text{ ialah pemalar}\text{.}\\ \text{ Cari nilai }m\text{ dan nilai }n.\\ \left(c\right)\text{ Diberi }\overrightarrow{AE}=h\underset{˜}{x}+9\underset{˜}{y},\text{ dengan keadaan }h\text{ ialah pemalar, }\\ \text{ cari nilai }h.\end{array}$


Penyelesaian: 
(a)(i)
$\begin{array}{l}\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}\\ =-9\underset{˜}{y}+6\underset{˜}{x}\\ =6\underset{˜}{x}-9\underset{˜}{y}\end{array}$

(a)(ii)
$\begin{array}{l}\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}\\ =9\underset{˜}{y}+\frac{1}{3}\overrightarrow{BC}\\ =9\underset{˜}{y}+\frac{1}{3}\left(6\underset{˜}{x}-9\underset{˜}{y}\right)\\ =9\underset{˜}{y}+2\underset{˜}{x}-3\underset{˜}{y}\\ =2\underset{˜}{x}+6\underset{˜}{y}\end{array}$


(b)
$\begin{array}{l}\text{Diberi }\overrightarrow{AV}=m\overrightarrow{AD}\\ =m\left(2\underset{˜}{x}+6\underset{˜}{y}\right)\\ =2m\underset{˜}{x}+6m\underset{˜}{y}\\ \\ \overrightarrow{AV}=\overrightarrow{AB}+\overrightarrow{BV}\\ \text{ = }9\underset{˜}{y}+n\left(\underset{˜}{x}-9\underset{˜}{y}\right)\\ =9\underset{˜}{y}+n\underset{˜}{x}-9n\underset{˜}{y}\\ =n\underset{˜}{x}+\left(9-9n\right)\underset{˜}{y}\\ \\ \text{Dengan menyamakan pekali bagi }\underset{˜}{x}\text{ dan }\underset{˜}{y}\text{, }\\ 2m\underset{˜}{x}+6m\underset{˜}{y}=n\underset{˜}{x}+\left(9-9n\right)\underset{˜}{y}\\ 2m=n\\ n=2m\mathrm{.............}\left(1\right)\\ 6m=9-9n\mathrm{.............}\left(2\right)\\ \text{Gantikan (1) ke dalam (2),}\\ 6m=9-9\left(2m\right)\\ 6m=9-18m\\ 24m=9\\ m=\frac{9}{24}=\frac{3}{8}\\ \\ \text{Daripada }\left(1\right):\\ n=2\left(\frac{3}{8}\right)=\frac{3}{4}\end{array}$


(c)
$\begin{array}{l}A,D\text{ dan }E\text{ adalah segaris}.\\ \overrightarrow{AD}=k\left(\overrightarrow{AE}\right)\\ \overrightarrow{AD}=k\left(h\underset{˜}{x}+9\underset{˜}{y}\right)\\ 2\underset{˜}{x}+6\underset{˜}{y}=kh\underset{˜}{x}+9k\underset{˜}{y}\\ \\ \text{Dengan menyamakan pekali bagi }\underset{˜}{y}:\\ 9k=6\\ k=\frac{6}{9}\\ k=\frac{2}{3}\\ \\ \text{Dengan menyamakan pekali bagi }\underset{˜}{x}:\\ kh=2\\ \left(\frac{2}{3}\right)h=2\\ h=2\times \frac{3}{2}\\ h=3\end{array}$

Soalan 8:
Rajah di bawah menunjukkan sisi empat OPQR. Garis lurus PR bersilang dengan garis lurus OQ di titik S.

$\begin{array}{l}\text{Diberi bahawa }\overrightarrow{OP}=7\underset{˜}{x},\overrightarrow{OR}=5\underset{˜}{y},PS:SR=3:1\text{ dan }\overrightarrow{OR}\text{ selari dengan }\overrightarrow{PQ}.\\ \text{(a) Ungkapkan dalam sebutan }\underset{˜}{x}\text{ dan }\underset{˜}{y},\\ \text{(i) }\overrightarrow{PR}\\ \text{(ii) }\overrightarrow{OS}\\ \\ \text{(b) Guna }\overrightarrow{PQ}=m\overrightarrow{OR}\text{ dan }\overrightarrow{SQ}=n\overrightarrow{OS},\text{ dengan keadaan }m\text{ dan }n\text{ adalah pemalar,}\\ \text{ Cari nilai }m\text{ dan nilai }n.\\ \\ \text{(c) Diberi bahawa }\left|\underset{˜}{y}\right|=4\text{ unit dan luas }ORS{\text{ ialah 50 cm}}^2\text{, cari jarak tegak}\\ \text{ dari titik }S\text{ ke }OR\text{.}\end{array}$


Penyelesaian:
(a)(i)
$\begin{array}{l}\overrightarrow{PR}=\overrightarrow{PO}+\overrightarrow{OR}\\ =-7\underset{˜}{x}+5\underset{˜}{y}\end{array}$


(a)(ii)
$\begin{array}{l}\overrightarrow{OS}=\overrightarrow{OP}+\overrightarrow{PS}\\ =7\underset{˜}{x}+\frac{3}{4}\overrightarrow{PR}\\ =7\underset{˜}{x}+\frac{3}{4}\left(-7\underset{˜}{x}+5\underset{˜}{y}\right)\\ =7\underset{˜}{x}-\frac{21}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}\\ =\frac{7}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}\end{array}$


(b)
$\begin{array}{l}\overrightarrow{PS}=\overrightarrow{PQ}-\overrightarrow{SQ}\\ \frac{3}{4}\overrightarrow{PR}=m\overrightarrow{OR}-n\overrightarrow{OS}\\ \frac{3}{4}\left(-7\underset{˜}{x}+5\underset{˜}{y}\right)=m\left(5\underset{˜}{y}\right)-n\left(\frac{7}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}\right)\\ -\frac{21}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}=5m\underset{˜}{y}-\frac{7}{4}n\underset{˜}{x}-\frac{15}{4}m\underset{˜}{y}\\ -\frac{21}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}=-\frac{7}{4}n\underset{˜}{x}+\frac{5}{4}m\underset{˜}{y}\\ -21\underset{˜}{x}+15\underset{˜}{y}=-7n\underset{˜}{x}+5m\underset{˜}{y}\\ 7n=21\\ n=3\\ 5m=15\\ m=3\end{array}$


(c)
$\begin{array}{l}\text{Luas }\Delta ORS=50\\ \frac{1}{2}\times \left(5\underset{˜}{y}\right)\times t=50\\ \frac{1}{2}\times 5\left(4\right)\times t=50\\ 10t=50\\ t=5\\ \therefore \text{ Jarak tegak dari titik }S\text{ ke }OR=5\text{ unit}\text{.}\end{array}$

Soalan 7:
Rajah di bawah menunjukkan sisi empat OPBC. Garis lurus AC bersilang dengan garis lurus PQ di titik B.

$\begin{array}{l}\text{Diberi bahawa }\overrightarrow{OP}=\underset{˜}{a},\overrightarrow{OQ}=\underset{˜}{b},\overrightarrow{OA}=4\overrightarrow{AP},\overrightarrow{OC}=3\overrightarrow{OQ},\overrightarrow{PB}=h\overrightarrow{PQ}\text{ dan}\\ \overrightarrow{AB}=k\overrightarrow{AC}.\\ \text{(a) Ungkapkan }\overrightarrow{OB}\text{ dalam sebutan }h\text{, }\underset{˜}{a}\text{ dan }\underset{˜}{b}.\\ \text{(b) Ungkapkan }\overrightarrow{OB}\text{ dalam sebutan }k\text{, }\underset{˜}{a}\text{ dan }\underset{˜}{b}.\\ \\ \text{(c)(i) Cari nilai }h\text{ dan nilai }k.\\ \text{ (ii) Seterusnya, nyatakan }\overrightarrow{OB}\text{ dalam sebutan }\underset{˜}{a}\text{ dan }\underset{˜}{b}.\end{array}$


Penyelesaian:
(a)
$\begin{array}{l}\overrightarrow{OB}=\overrightarrow{OP}+\overrightarrow{PB}\\ =\underset{˜}{a}+h\overrightarrow{PQ}\\ =\underset{˜}{a}+h\left(\overrightarrow{PO}+\overrightarrow{OQ}\right)\\ =\underset{˜}{a}+h\left(-\underset{˜}{a}+\underset{˜}{b}\right)\\ =\underset{˜}{a}-h\underset{˜}{a}+h\underset{˜}{b}\\ \overrightarrow{OB}=\left(1-h\right)\underset{˜}{a}+h\underset{˜}{b}\end{array}$


(b)
$\begin{array}{l}\overrightarrow{OB}=\overrightarrow{OP}+\overrightarrow{PB}\\ =\underset{˜}{a}+\overrightarrow{PA}+\overrightarrow{AB}\\ =\underset{˜}{a}+\left(-\frac{1}{5}\overrightarrow{OP}\right)+k\overrightarrow{AC}\\ =\underset{˜}{a}+\left(-\frac{1}{5}\underset{˜}{a}\right)+k\left(\overrightarrow{AO}+\overrightarrow{OC}\right)\\ =\frac{4}{5}\underset{˜}{a}+k\left(-\frac{4}{5}\overrightarrow{OP}+3\overrightarrow{OQ}\right)\\ =\frac{4}{5}\underset{˜}{a}+k\left(-\frac{4}{5}\underset{˜}{a}+3\underset{˜}{b}\right)\\ =\frac{4}{5}\underset{˜}{a}-\frac{4}{5}k\underset{˜}{a}+3k\underset{˜}{b}\\ \overrightarrow{OB}=\frac{4}{5}\left(1-k\right)\underset{˜}{a}+3k\underset{˜}{b}\end{array}$


(c)(i)
$\begin{array}{l}\left(1-h\right)\underset{˜}{a}+h\underset{˜}{b}=\frac{4}{5}\left(1-k\right)\underset{˜}{a}+3k\underset{˜}{b}\\ 1-h=\frac{4}{5}-\frac{4}{5}k\mathrm{..........}\left(1\right)\\ h=3k\mathrm{..........}\left(2\right)\\ \text{Gantikan }\left(2\right)\text{ ke dalam }\left(1\right)\\ 1-3k=\frac{4}{5}-\frac{4}{5}k\\ 5-15k=4-4k\\ 11k=1\\ k=\frac{1}{11}\\ \\ \text{Gantikan }k=\frac{1}{11}\text{ ke dalam }\left(2\right)\\ h=3\left(\frac{1}{11}\right)\\ =\frac{3}{11}\end{array}$


(c)(ii)
$\begin{array}{l}\overrightarrow{OB}=\left(1-h\right)\underset{˜}{a}+h\underset{˜}{b}\\ \text{apabila }h=\frac{3}{11}\\ =\left(1-\frac{3}{11}\right)\underset{˜}{a}+\left(\frac{3}{11}\right)\underset{˜}{b}\\ =\frac{8}{11}\underset{˜}{a}+\frac{3}{11}\underset{˜}{b}\end{array}$

Soalan 6:
Diagram below shows a trapezium OABC and point D lies on AC.


$\begin{array}{l}\text{Diberi bahawa }\overrightarrow{OC}=18\underset{˜}{b},\overrightarrow{OA}=6\underset{˜}{a}\text{ dan }\overrightarrow{OC}=2\overrightarrow{AB}.\\ \text{(a) Ungkapkan dalam sebutan }\underset{˜}{a}\text{ dan }\underset{˜}{b},\\ \text{(i) }\overrightarrow{AC}\\ \text{(ii) }\overrightarrow{OB}\\ \\ \text{(b) Diberi bahawa }\overrightarrow{AD}=k\overrightarrow{AC},\text{ dengan keadaan }k\text{ ialah pemalar}\text{.}\\ \text{Cari nilai }k\text{ jika titik-titik }O\text{, }D\text{ dan }B\text{ adalah segaris}\text{.}\end{array}$


Penyelesaian:
(a)(i)
$\begin{array}{l}\overrightarrow{AC}=\overrightarrow{AO}+\overrightarrow{OC}\\ =-6\underset{˜}{a}+18\underset{˜}{b}\\ =18\underset{˜}{b}-6\underset{˜}{a}\end{array}$


(a)(ii)
$\begin{array}{l}\overrightarrow{OC}=2\overrightarrow{AB}\\ 18\underset{˜}{b}=2\left(\overrightarrow{AO}+\overrightarrow{OB}\right)\\ 18\underset{˜}{b}=2\left(-6\underset{˜}{a}+\overrightarrow{OB}\right)\\ 18\underset{˜}{b}=-12\underset{˜}{a}+2\overrightarrow{OB}\\ \overrightarrow{OB}=6\underset{˜}{a}+9\underset{˜}{b}\end{array}$


(b)
$\begin{array}{l}\overrightarrow{OD}=h\overrightarrow{OB}\\ =h\left(6\underset{˜}{a}+9\underset{˜}{b}\right)\\ =6h\underset{˜}{a}+9h\underset{˜}{b}\\ \\ \overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}\\ =6h\underset{˜}{a}+9h\underset{˜}{b}-6\underset{˜}{a}\\ =\underset{˜}{a}\left(6h-6\right)+9h\underset{˜}{b}\\ \\ \overrightarrow{AD}=k\overrightarrow{AC}\\ \underset{˜}{a}\left(6h-6\right)+9h\underset{˜}{b}=k\left(18\underset{˜}{b}-6\underset{˜}{a}\right)\\ \underset{˜}{a}\left(6h-6\right)+9h\underset{˜}{b}=-6k\underset{˜}{a}+18k\underset{˜}{b}\\ 6h-6=-6k\\ h-1=-k\\ h=1-k\mathrm{..........}\left(1\right)\\ \\ 9h=18k\\ h=2k\\ \text{Dari }\left(1\right),\\ 1-k=2k\\ 3k=1\\ k=\frac{1}{3}\end{array}$

Soalan 5:
Rajah di bawah menunjukkan sebuah segi tiga KLM.


$\begin{array}{l}\text{Diberi }KP:PL=1:2,LR:RM=2:1,\overrightarrow{KP}=2\underset{˜}{x},\overrightarrow{KM}=3\underset{˜}{y}.\\ \text{(a) Ungkapkan dalam sebutan }\underset{˜}{x}\text{ dan }\underset{˜}{y},\\ \text{(i) }\overrightarrow{MP}\\ \text{(ii) }\overrightarrow{MR}\\ \\ \text{(b) Diberi }\underset{˜}{x}=2\underset{˜}{i}\text{ dan }\underset{˜}{y}=-\underset{˜}{i}+4\underset{˜}{j},\text{ cari }\overrightarrow{\left|MR\right|}.\\ \\ \text{(c) Diberi }\overrightarrow{MQ}=h\overrightarrow{MP}\text{ dan }\overrightarrow{QR}=n\overrightarrow{KR},\text{ dengan keadaan }h\text{ dan }n\text{ ialah pemalar, }\\ \text{ cari nilai }h\text{ dan nilai }n\text{.}\end{array}$


Penyelesaian:
(a)(i)
$\begin{array}{l}\overrightarrow{MP}=\overrightarrow{MK}+\overrightarrow{KP}\\ =-3\underset{˜}{y}+2\underset{˜}{x}\\ =2\underset{˜}{x}-3\underset{˜}{y}\end{array}$

(a)(ii)
$\begin{array}{l}\overrightarrow{MR}=\frac{1}{3}\overrightarrow{ML}\\ =\frac{1}{3}\left(\overrightarrow{MK}+\overrightarrow{KL}\right)\\ =\frac{1}{3}\left(-3\underset{˜}{y}+6\underset{˜}{x}\right)\\ =2\underset{˜}{x}-\underset{˜}{y}\end{array}$

(b)
$\begin{array}{l}\overrightarrow{MR}=2\left(2\underset{˜}{i}\right)-\left(-\underset{˜}{i}+4\underset{˜}{j}\right)\\ =4\underset{˜}{i}+\underset{˜}{i}-4\underset{˜}{j}\\ =5\underset{˜}{i}-4\underset{˜}{j}\\ \left|\overrightarrow{MR}\right|=\sqrt{5^2+{\left(-4\right)}^2}\\ =\sqrt{41}\text{ unit}\end{array}$

(c)
$\begin{array}{l}\overrightarrow{MQ}+\overrightarrow{QR}=\overrightarrow{MR}\\ h\overrightarrow{MP}+n\overrightarrow{KR}=\overrightarrow{MR}\\ h\left(2\underset{˜}{x}-3\underset{˜}{y}\right)+n\left(\overrightarrow{KM}+\overrightarrow{MR}\right)=2\underset{˜}{x}-\underset{˜}{y}\\ h\left(2\underset{˜}{x}-3\underset{˜}{y}\right)+n\left(3\underset{˜}{y}+2\underset{˜}{x}-\underset{˜}{y}\right)=2\underset{˜}{x}-\underset{˜}{y}\\ 2h\underset{˜}{x}-3h\underset{˜}{y}+2n\underset{˜}{x}+2n\underset{˜}{y}=2\underset{˜}{x}-\underset{˜}{y}\\ \left(2h+2n\right)\underset{˜}{x}+\left(-3h+2n\right)\underset{˜}{y}=2\underset{˜}{x}-\underset{˜}{y}\\ 2h+2n=\mathrm{2..........}(1)\\ -3h+2n=-\mathrm{1..........}(2)\\ \left(1\right)-\left(2\right):5h=3\\ h=\frac{3}{5}\\ \text{Dari }\left(1\right):h+n=1\\ \frac{3}{5}+n=1\\ n=1-\frac{3}{5}\\ n=\frac{2}{5}\end{array}$

Soalan 4:
Rajah di bawah menunjukkan sisi empat ABCD. Garis lurus AC bersilang dengan garis lurus BD di titik E.


$\begin{array}{l}\text{Diberi bahawa }BE:ED=2:3,\overrightarrow{AB}=10\underset{˜}{x},\overrightarrow{AD}=25\underset{˜}{y}\text{ dan }\overrightarrow{BC}=-\underset{˜}{x}+15\underset{˜}{y}.\\ \text{(a) Ungkapkan dalam sebutan }\underset{˜}{x}\text{ dan }\underset{˜}{y},\\ \text{(i) }\overrightarrow{BD}\\ \text{(ii) }\overrightarrow{AE}\\ \\ \text{(b) Cari nisbah }AE:EC.\end{array}$


Penyelesaian:
(a)(i)
$\begin{array}{l}\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}\\ =\overrightarrow{AD}-\overrightarrow{AB}\\ =25\underset{˜}{y}-10\underset{˜}{x}\end{array}$

(a)(ii)
$\begin{array}{l}\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}\\ =\overrightarrow{AB}+\frac{2}{5}\overrightarrow{BD}\\ =10\underset{˜}{x}+\frac{2}{5}\left(25\underset{˜}{y}-10\underset{˜}{x}\right)\\ =10\underset{˜}{x}+\frac{2}{5}\left(25\underset{˜}{y}-10\underset{˜}{x}\right)\\ =10\underset{˜}{x}+10\underset{˜}{y}-4\underset{˜}{x}\\ =6\underset{˜}{x}+10\underset{˜}{y}\\ =2\left(3\underset{˜}{x}+5\underset{˜}{y}\right)\end{array}$

(b)
$\begin{array}{l}\overrightarrow{EC}=\overrightarrow{EB}+\overrightarrow{BC}\\ =\overrightarrow{BC}-\overrightarrow{BE}\\ =\overrightarrow{BC}-\frac{2}{3}\overrightarrow{ED}\\ =\overrightarrow{BC}-\frac{2}{3}\left(\overrightarrow{EA}+\overrightarrow{AD}\right)\\ =-\underset{˜}{x}+15\underset{˜}{y}-\frac{2}{3}\left(-6\underset{˜}{x}-10\underset{˜}{y}+25\underset{˜}{y}\right)\\ =-\underset{˜}{x}+15\underset{˜}{y}-\frac{2}{3}\left(-6\underset{˜}{x}+15\underset{˜}{y}\right)\\ =-\underset{˜}{x}+15\underset{˜}{y}+4\underset{˜}{x}-10\underset{˜}{y}\\ =3\underset{˜}{x}+5\underset{˜}{y}\\ \frac{AE}{EC}=\frac{2\left(3\underset{˜}{x}+5\underset{˜}{y}\right)}{1\left(3\underset{˜}{x}+5\underset{˜}{y}\right)}\\ AE:EC=2:1\end{array}$