Long Question 9

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

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