Long Question 6

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