$\begin{array}{l}=\left|\overrightarrow{PQ}\right|+\left|\overrightarrow{RS}\right|\\ =\left|\sqrt{6^2+8^2}\right|+\left|\sqrt{6^2+8^2}\right|\\ =10+10\\ =20\text{units}\end{array}$

The diagram above shows a parallelogram OABC. M is the midpoint of BC. Vector $\overrightarrow{OA}=\underset{˜}{a}\text{ and }\overrightarrow{OC}=\underset{˜}{c}.$  Find the following vectors in terms of $\underset{˜}{a}\text{ and }\underset{˜}{c}.$
$\begin{array}{l}\left(a\right)\overrightarrow{OB}\\ \left(b\right)\overrightarrow{MB}\\ \left(c\right)\overrightarrow{OM}\end{array}$

Question 10 (3 marks):
Diagram 3 shows vectors $\overrightarrow{OP},\overrightarrow{OQ}\text{ and }\overrightarrow{OM}$ drawn on a square grid.

Diagram

$\begin{array}{l}\left(\text{a}\right)\text{ Express }\overrightarrow{OM}\text{ in the form }h\underset{˜}{p}+k\underset{˜}{q}\text{, }\\ \text{where }h\text{ and }k\text{ are constants}.\\ \left(\text{b}\right)\text{ On the diagram 3, mark and label }\\ \text{the point }N\text{ such that }\overrightarrow{MN}+\overrightarrow{OQ}=2\overrightarrow{OP}.\end{array}$


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

(b)

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

Question 11 (4 marks):
$\begin{array}{l}A\left(2,3\right)\text{ and }B\left(-2,\text{ 5}\right)\text{ lie on a }\\ \text{Cartesian plane}\text{.}\\ \text{It is given that }3\overrightarrow{OA}=2\overrightarrow{OB}+\overrightarrow{OC}.\\ \text{Find}\\ \left(a\right)\text{ the coordinates of }C,\\ \left(b\right)\left|\overrightarrow{AC}\right|\end{array}$

Solution:
$\begin{array}{l}\text{Given }A\left(2,3\right)\text{ and }B\left(-2,5\right)\\ \text{Thus, }\overrightarrow{OA}=2\underset{˜}{i}+3\underset{˜}{j}\text{ and }\overrightarrow{OB}=-2\underset{˜}{i}+5\underset{˜}{j}\end{array}$

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


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

Question 8 (3 marks):
Diagram shows vectors $\overrightarrow{AB},\overrightarrow{AC}\text{ and }\overrightarrow{AD}$ drawn on a square grid with sides of 1 unit.

Diagram

$\begin{array}{l}\left(\text{a}\right)\text{ Find }\left|-\overrightarrow{BA}\right|.\\ \left(\text{b}\right)\text{ Given }\overrightarrow{AB}=\underset{˜}{b}\text{ and }\overrightarrow{AC}=\underset{˜}{c},\\ \text{express in terms of }\underset{˜}{b}\text{ and }\underset{˜}{c}\\ \left(i\right)\overrightarrow{BC},\\ \left(ii\right)\overrightarrow{AD}\end{array}$


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

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

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

Question 9 (3 marks):
Diagram shows a trapezium ABCD.

Diagram

$\begin{array}{l}\text{Given }\underset{˜}{p}=\left(\begin{array}{l}3\\ 4\end{array}\right)\text{ and }\underset{˜}{q}=\left(\begin{array}{l}k-1\\ 2\end{array}\right)\text{, }\\ \text{where }k\text{ is a constant, find value of }k\text{.}\end{array}$


Solution:

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

Question 10 (10 marks):
Diagram shows a triangle ABC. The straight line AE intersects with the straight line BC at point D. Point V lies on the straight line AE.

$\begin{array}{l}\text{It is given that }\overrightarrow{BD}=\frac{1}{3}\overrightarrow{BC},\overrightarrow{AC}=6\underset{˜}{x}\text{ and }\overrightarrow{AB}=9\underset{˜}{y}.\\ \left(a\right)\text{ Express in terms of }\underset{˜}{x}\text{ and / or }\underset{˜}{y}:\\ \left(i\right)\overrightarrow{BC},\\ \left(ii\right)\overrightarrow{AD}.\\ \left(b\right)\text{ It is given that }\overrightarrow{AV}=m\overrightarrow{AD}\text{ and }\overrightarrow{BV}=n\left(\underset{˜}{x}-9\underset{˜}{y}\right),\text{ where }m\text{ and }n\text{ are constants}\text{.}\\ \text{ Find the value of }m\text{ and of }n.\\ \left(c\right)\text{ Given }\overrightarrow{AE}=h\underset{˜}{x}+9\underset{˜}{y},\text{ where }h\text{ is a constant, find the value of }h.\end{array}$

Solution: 
(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)
$\begin{array}{l}\text{Given }\overrightarrow{AV}=m\overrightarrow{AD}\\ =m\left(2\underset{˜}{x}+6\underset{˜}{y}\right)\\ =2m\underset{˜}{x}+6m\underset{˜}{y}\\ \\ \overrightarrow{AV}=\overrightarrow{AB}+\overrightarrow{BV}\\ \text{ = }9\underset{˜}{y}+n\left(\underset{˜}{x}-9\underset{˜}{y}\right)\\ =9\underset{˜}{y}+n\underset{˜}{x}-9n\underset{˜}{y}\\ =n\underset{˜}{x}+\left(9-9n\right)\underset{˜}{y}\\ \\ \text{By equating the coefficients of }\underset{˜}{x}\text{ and }\underset{˜}{y}\text{, }\\ 2m\underset{˜}{x}+6m\underset{˜}{y}=n\underset{˜}{x}+\left(9-9n\right)\underset{˜}{y}\\ 2m=n\\ n=2m\mathrm{.............}\left(1\right)\\ 6m=9-9n\mathrm{.............}\left(2\right)\\ \text{Substitute (1) into (2),}\\ 6m=9-9\left(2m\right)\\ 6m=9-18m\\ 24m=9\\ m=\frac{9}{24}=\frac{3}{8}\\ \\ \text{From }\left(1\right):\\ n=2\left(\frac{3}{8}\right)=\frac{3}{4}\end{array}$


(c)
$\begin{array}{l}A,D\text{ and }E\text{ are collinear}.\\ \overrightarrow{AD}=k\left(\overrightarrow{AE}\right)\\ \overrightarrow{AD}=k\left(h\underset{˜}{x}+9\underset{˜}{y}\right)\\ 2\underset{˜}{x}+6\underset{˜}{y}=kh\underset{˜}{x}+9k\underset{˜}{y}\\ \\ \text{Equating the coefficients of }\underset{˜}{y}:\\ 9k=6\\ k=\frac{6}{9}\\ k=\frac{2}{3}\\ \\ \text{Equating the coefficients of }\underset{˜}{x}:\\ kh=2\\ \left(\frac{2}{3}\right)h=2\\ h=2\times \frac{3}{2}\\ h=3\end{array}$

Question 9 (10 marks):
Diagram 5 shows triangles OAQ and OPB where point P lies on OA and point Q lies on OB. The straight lines AQ and PB intersect at point R.
$\begin{array}{l}\text{It is given that }\overrightarrow{OA}=18\underset{˜}{x},\overrightarrow{OB}=16\underset{˜}{y},OP:PA=1:2,OQ:QB=3:1,\\ \overrightarrow{PR}=m\overrightarrow{PB}\text{ and }\overrightarrow{QR}=n\overrightarrow{QA},\text{ where }m\text{ and }n\text{ are constants}\text{.}\\ \left(a\right)\text{ Express }\overrightarrow{OR}\text{ in terms of}\\ \left(i\right)m,\underset{˜}{x}\text{ and }\underset{˜}{y},\\ \left(ii\right)n,\underset{˜}{x}\text{ and }\underset{˜}{y},\\ \left(b\right)\text{ Hence, find the value of }m\text{ and of }n.\\ \left(c\right)\text{ Given }\left|\underset{˜}{x}\right|=2\text{ units, }\left|\underset{˜}{y}\right|=1\text{ unit and }OA\text{ is perpendicular to }OB\text{ calculate }\left|\overrightarrow{PR}\right|.\end{array}$

Solution: 
(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)
$\begin{array}{l}6\underset{˜}{x}+m\left(-6\underset{˜}{x}+16\underset{˜}{y}\right)=\left(12-12n\right)\underset{˜}{y}+18n\underset{˜}{x}\\ 6\underset{˜}{x}-6m\underset{˜}{x}+16m\underset{˜}{y}=18n\underset{˜}{x}+12\underset{˜}{y}-12n\underset{˜}{y}\\ \text{by comparison;}\\ 6-6m=18n\\ 1-m=3n\\ m=1-3n\mathrm{..............}\left(1\right)\\ \\ 16m=12-12n\\ 4m=3-3n\mathrm{..............}\left(2\right)\\ \\ \text{Substitute (1) into (2),}\\ 4\left(1-3n\right)=3-3n\\ 4-12n=3-3n\\ 9n=1\\ n=\frac{1}{9}\\ \\ \text{Substitute }n=\frac{1}{9}\text{ into (1),}\\ m=1-3\left(\frac{1}{9}\right)\\ m=\frac{2}{3}\end{array}$

[adinserter block="3"]

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

Question 8:
Diagram below shows quadrilateral OPQR. The straight line PR intersects the straight line OQ at point S.


$\begin{array}{l}\text{It is given that }\overrightarrow{OP}=7\underset{˜}{x},\overrightarrow{OR}=5\underset{˜}{y},PS:SR=3:1\text{ and }\overrightarrow{OR}\text{ is parallel to }\overrightarrow{PQ}.\\ \text{(a) Express in terms of }\underset{˜}{x}\text{ and }\underset{˜}{y},\\ \text{(i) }\overrightarrow{PR}\\ \text{(ii) }\overrightarrow{OS}\\ \\ \text{(b) Using }\overrightarrow{PQ}=m\overrightarrow{OR}\text{ and }\overrightarrow{SQ}=n\overrightarrow{OS},\text{ where }m\text{ and }n\text{ are constants,}\\ \text{ Find the value of }m\text{ and of }n.\\ \\ \text{(c) Given that }\left|\underset{˜}{y}\right|=4\text{ units and the area of }ORS{\text{ is 50 cm}}^2\text{, find the}\\ \text{ perpendicular distance from point }S\text{ to }OR\text{.}\end{array}$


Solution:
(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)
$\begin{array}{l}\text{Area of }\Delta ORS=50\\ \frac{1}{2}\times \left(5\underset{˜}{y}\right)\times t=50\\ \frac{1}{2}\times 5\left(4\right)\times t=50\\ 10t=50\\ t=5\\ \therefore \text{ Perpendicular distance from point }S\text{ to }OR=5\text{ units}\text{.}\end{array}$

Question 7:
Diagram below shows quadrilateral OPBC. The straight line AC intersects the straight line PQ at point B.


$\begin{array}{l}\text{It is given that }\overrightarrow{OP}=\underset{˜}{a},\overrightarrow{OQ}=\underset{˜}{b},\overrightarrow{OA}=4\overrightarrow{AP},\overrightarrow{OC}=3\overrightarrow{OQ},\overrightarrow{PB}=h\overrightarrow{PQ}\text{ and}\\ \overrightarrow{AB}=k\overrightarrow{AC}.\\ \text{(a) Express }\overrightarrow{OB}\text{ in terms of }h\text{, }\underset{˜}{a}\text{ and }\underset{˜}{b}.\\ \text{(b) Express }\overrightarrow{OB}\text{ in terms of }k\text{, }\underset{˜}{a}\text{ and }\underset{˜}{b}.\\ \\ \text{(c)(i) Find the value of }h\text{ and of }k.\\ \text{(ii) Hence, state }\overrightarrow{OB}\text{ in terms of }\underset{˜}{a}\text{ and }\underset{˜}{b}.\end{array}$


Solution:
(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)
$\begin{array}{l}\left(1-h\right)\underset{˜}{a}+h\underset{˜}{b}=\frac{4}{5}\left(1-k\right)\underset{˜}{a}+3k\underset{˜}{b}\\ 1-h=\frac{4}{5}-\frac{4}{5}k\mathrm{..........}\left(1\right)\\ h=3k\mathrm{..........}\left(2\right)\\ \text{Substitute }\left(2\right)\text{ into the }\left(1\right)\\ 1-3k=\frac{4}{5}-\frac{4}{5}k\\ 5-15k=4-4k\\ 11k=1\\ k=\frac{1}{11}\\ \\ \text{Substitute }k=\frac{1}{11}\text{ into }\left(2\right)\\ h=3\left(\frac{1}{11}\right)\\ =\frac{3}{11}\end{array}$


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

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


$\begin{array}{l}\text{It is given that }\overrightarrow{OC}=18\underset{˜}{b},\overrightarrow{OA}=6\underset{˜}{a}\text{ and }\overrightarrow{OC}=2\overrightarrow{AB}.\\ \text{(a) Express in terms of }\underset{˜}{a}\text{ and }\underset{˜}{b},\\ \text{(i) }\overrightarrow{AC}\\ \text{(ii) }\overrightarrow{OB}\\ \\ \text{(b) It is given that }\overrightarrow{AD}=k\overrightarrow{AC},\text{ where }k\text{ is a constant}\text{.}\\ \text{Find the value of }k\text{ if the points }O\text{, }D\text{ and }B\text{ are collinear}\text{.}\end{array}$


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