Long Question 5

Question 5:
Diagram below shows a triangle KLM.


$\begin{array}{l}\text{It is given that }KP:PL=1:2,LR:RM=2:1,\overrightarrow{KP}=2\underset{˜}{x},\overrightarrow{KM}=3\underset{˜}{y}.\\ \text{(a) Express in terms of }\underset{˜}{x}\text{ and }\underset{˜}{y},\\ \text{(i) }\overrightarrow{MP}\\ \text{(ii) }\overrightarrow{MR}\\ \\ \text{(b) Given }\underset{˜}{x}=2\underset{˜}{i}\text{ and }\underset{˜}{y}=-\underset{˜}{i}+4\underset{˜}{j},\text{ find }\overrightarrow{\left|MR\right|}.\\ \\ \text{(c) Given }\overrightarrow{MQ}=h\overrightarrow{MP}\text{ and }\overrightarrow{QR}=n\overrightarrow{KR},\text{ where }h\text{ and }n\text{ are constants, }\\ \text{ find the value of }h\text{ and of }n\text{.}\end{array}$


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

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