Long Question 7

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