In the diagram above, vector $\overrightarrow{OP}=\underset{˜}{p}\text{, }\overrightarrow{OR}=\underset{˜}{r}$ and Q divides PR in the ratio of 2 : 3. Find the following vectors in terms of $\underset{˜}{p}\text{ and }\underset{˜}{r}$
$\begin{array}{l}\left(a\right)\overrightarrow{PR}\\ \left(b\right)\overrightarrow{OQ}\\ \left(c\right)\overrightarrow{QM}\text{ if }M\text{ is the midpoint of }OR.\end{array}$

(b)
$\begin{array}{l}\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{PQ}\\ =\overrightarrow{OP}+\frac{2}{5}\overrightarrow{PR}\\ =\underset{˜}{p}+\frac{2}{5}\left(-\underset{˜}{p}+\underset{˜}{r}\right)\\ =\underset{˜}{p}-\frac{2}{5}\underset{˜}{p}+\frac{2}{5}\underset{˜}{r}\\ =\frac{3}{5}\underset{˜}{p}+\frac{2}{5}\underset{˜}{r}\end{array}$

(c)
$\begin{array}{l}\overrightarrow{QM}=\overrightarrow{QO}+\overrightarrow{OM}\\ =-\overrightarrow{OQ}+\overrightarrow{OM}\\ =-\overrightarrow{OQ}+\frac{1}{2}\overrightarrow{OR}\\ =-\left(\frac{3}{5}\underset{˜}{p}+\frac{2}{5}\underset{˜}{r}\right)+\frac{1}{2}\underset{˜}{r}\\ =-\frac{3}{5}\underset{˜}{p}-\frac{2}{5}\underset{˜}{r}+\frac{1}{2}\underset{˜}{r}\\ =-\frac{3}{5}\underset{˜}{p}+\frac{1}{10}\underset{˜}{r}\end{array}$

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

Question 4:

$\begin{array}{l}\text{Given that }\overrightarrow{AB}=7\underset{˜}{p},\overrightarrow{AD}=5\underset{˜}{q}\text{ and }DE=3EB,\text{ express, in terms of }\underset{˜}{p}\text{ and }\underset{˜}{q}.\\ \text{(a) }\overrightarrow{BD}\\ \text{(b) }\overrightarrow{EC}\end{array}$

Solution:
(a)
$\begin{array}{l}\text{Note: for parallelogram,}\overrightarrow{AB}=\overrightarrow{DC}=7\underset{˜}{p},\overrightarrow{AD}=\overrightarrow{BC}=5\underset{˜}{q}.\\ \overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}\\ \overrightarrow{BD}=-7\underset{˜}{p}+5\underset{˜}{q}\end{array}$  


(b)
$\begin{array}{l}\overrightarrow{DE}\text{=3}\overrightarrow{EB}\\ \frac{\overrightarrow{EB}}{\overrightarrow{DE}}=\frac{1}{3}\to EB:DE=1:3\\ \therefore \overrightarrow{EB}=\frac{1}{4}\overrightarrow{DB}\\ =\frac{1}{4}\left(-\overrightarrow{BD}\right)\\ =\frac{1}{4}\left[-\left(-7\underset{˜}{p}+5\underset{˜}{q}\right)\right]\leftarrow \text{From (a)}\\ \text{=}\frac{7}{4}\underset{˜}{p}+\frac{5}{4}\underset{˜}{q}\\ \\ \overrightarrow{EC}=\overrightarrow{EB}+\overrightarrow{BC}\\ \overrightarrow{EC}=\frac{7}{4}\underset{˜}{p}+\frac{5}{4}\underset{˜}{q}+5\underset{˜}{q}\\ \overrightarrow{EC}=\frac{7}{4}\underset{˜}{p}+\frac{25}{4}\underset{˜}{q}\end{array}$

Question 5:


Use the above information to find the values of h and k when r = 2p – 3q.

Solution:
$\begin{array}{l}r=2p-3q\\ \left(h-1\right)\underset{˜}{a}+\left(h+k\right)\underset{˜}{b}=2\left(5\underset{˜}{a}-7\underset{˜}{b}\right)-3\left(-2\underset{˜}{a}+3\underset{˜}{b}\right)\\ \left(h-1\right)\underset{˜}{a}+\left(h+k\right)\underset{˜}{b}=10\underset{˜}{a}-14\underset{˜}{b}+6\underset{˜}{a}-9\underset{˜}{b}\\ \left(h-1\right)\underset{˜}{a}+\left(h+k\right)\underset{˜}{b}=16\underset{˜}{a}-23\underset{˜}{b}\\ \\ \text{Comparing vector:}\\ h-1=16\\ h=17\\ \\ h+k=-23\\ 17+k=-23\\ k=-40\end{array}$

Question 4:
Diagram below shows quadrilateral ABCD. The straight line AC intersects the straight line BD at point E.


$\begin{array}{l}\text{It is given that }BE:ED=2:3,\overrightarrow{AB}=10\underset{˜}{x},\overrightarrow{AD}=25\underset{˜}{y}\text{ and }\overrightarrow{BC}=-\underset{˜}{x}+15\underset{˜}{y}.\\ \text{(a) Express in terms of }\underset{˜}{x}\text{ and }\underset{˜}{y},\\ \text{(i) }\overrightarrow{BD}\\ \text{(ii) }\overrightarrow{AE}\\ \\ \text{(b) Find the ratio }AE:EC.\end{array}$


Solution:
(a)(i)
$\begin{array}{l}\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}\\ =\overrightarrow{AD}-\overrightarrow{AB}\\ =25\underset{˜}{y}-10\underset{˜}{x}\end{array}$

(a)(ii)
$\begin{array}{l}\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}\\ =\overrightarrow{AB}+\frac{2}{5}\overrightarrow{BD}\\ =10\underset{˜}{x}+\frac{2}{5}\left(25\underset{˜}{y}-10\underset{˜}{x}\right)\\ =10\underset{˜}{x}+\frac{2}{5}\left(25\underset{˜}{y}-10\underset{˜}{x}\right)\\ =10\underset{˜}{x}+10\underset{˜}{y}-4\underset{˜}{x}\\ =6\underset{˜}{x}+10\underset{˜}{y}\\ =2\left(3\underset{˜}{x}+5\underset{˜}{y}\right)\end{array}$

(b)
$\begin{array}{l}\overrightarrow{EC}=\overrightarrow{EB}+\overrightarrow{BC}\\ =\overrightarrow{BC}-\overrightarrow{BE}\\ =\overrightarrow{BC}-\frac{2}{3}\overrightarrow{ED}\\ =\overrightarrow{BC}-\frac{2}{3}\left(\overrightarrow{EA}+\overrightarrow{AD}\right)\\ =-\underset{˜}{x}+15\underset{˜}{y}-\frac{2}{3}\left(-6\underset{˜}{x}-10\underset{˜}{y}+25\underset{˜}{y}\right)\\ =-\underset{˜}{x}+15\underset{˜}{y}-\frac{2}{3}\left(-6\underset{˜}{x}+15\underset{˜}{y}\right)\\ =-\underset{˜}{x}+15\underset{˜}{y}+4\underset{˜}{x}-10\underset{˜}{y}\\ =3\underset{˜}{x}+5\underset{˜}{y}\\ \frac{AE}{EC}=\frac{2\left(3\underset{˜}{x}+5\underset{˜}{y}\right)}{1\left(3\underset{˜}{x}+5\underset{˜}{y}\right)}\\ AE:EC=2:1\end{array}$

Question 2:
Given that   $\overrightarrow{AB}=\left(\begin{array}{c}10\\ 14\end{array}\right),\overrightarrow{OB}=\left(\begin{array}{c}4\\ 6\end{array}\right)$ and $\overrightarrow{CD}=\left(\begin{array}{c}m\\ 7\end{array}\right)$ , find
(a) the coordinates of A,
(b) the unit vector in the direction of $\overrightarrow{OA}$ .
(c) the value of m if CD is parallel to AB .

Solution:
(a)
$\begin{array}{l}\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}\\ \left(\begin{array}{c}10\\ 14\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}4\\ 6\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}10\\ 14\end{array}\right)-\left(\begin{array}{c}4\\ 6\end{array}\right)\\ \overrightarrow{AO}=\left(\begin{array}{c}6\\ 8\end{array}\right)\\ \overrightarrow{OA}=\left(\begin{array}{c}-6\\ -8\end{array}\right)\\ A=\left(-6,-8\right)\end{array}$


(b)
$\begin{array}{l}\left|\overrightarrow{OA}\right|=\sqrt{{\left(-6\right)}^2+{\left(-8\right)}^2}\\ \left|\overrightarrow{OA}\right|=\sqrt{100}=10\\ \\ \text{the unit vector in the direction of }\overrightarrow{OA}\\ =\frac{\overrightarrow{OA}}{\left|\overrightarrow{OA}\right|}\\ =\frac{\left(\begin{array}{c}-6\\ -8\end{array}\right)}{10}=\frac{1}{10}\left(\begin{array}{c}-6\\ -8\end{array}\right)\\ =\left(\begin{array}{c}-\frac{3}{5}\\ -\frac{4}{5}\end{array}\right)\end{array}$


(c)
$\begin{array}{l}\text{Given }\overrightarrow{CD}\text{ parallel }\overrightarrow{AB}\\ \therefore \overrightarrow{CD}=k\overrightarrow{AB}\\ \left(\begin{array}{c}m\\ 7\end{array}\right)=k\left(\begin{array}{c}10\\ 14\end{array}\right)\\ \left(\begin{array}{c}m\\ 7\end{array}\right)=\left(\begin{array}{c}10k\\ 14k\end{array}\right)\\ \\ 7=14k\\ k=\frac{1}{2}\\ \\ m=10k=10\left(\frac{1}{2}\right)=5\end{array}$

(A) Vectors in Cartesian Coordinates

$$\boxed{\left|\overrightarrow{OA}\right|=\sqrt{x^2+y^2}}$$

(B) Unit Vector in the Direction of a Vector
$$\boxed{\text{ Unit vector of }\underset{˜}{a},\underset{˜}{\widehat{a}}=\frac{x\underset{˜}{i}+y\underset{˜}{j}}{\sqrt{x^2+y^2}}}$$  

Solution:
$\begin{array}{l}\underset{˜}{r}=k\underset{˜}{i}-8\underset{˜}{j}\\ \\ \text{Given }\left|\underset{˜}{r}\right|=10\\ \sqrt{x^2+y^2}=10\\ \sqrt{k^2+{\left(-8\right)}^2}=10\\ k^2+64=100\\ k=\pm 6\\ \\ \text{Unit vector of }\widehat{\underset{˜}{r}}=\frac{x\underset{˜}{i}+y\underset{˜}{j}}{\sqrt{x^2+y^2}}\\ \\ \text{When }k=6,\text{ When }k=-6\\ \\ \widehat{\underset{˜}{r}}=\frac{6\underset{˜}{i}-8\underset{˜}{j}}{10}=\frac{3\underset{˜}{i}-4\underset{˜}{j}}{5}\text{ , }\widehat{\underset{˜}{r}}=\frac{-6\underset{˜}{i}-8\underset{˜}{j}}{10}=\frac{-3\underset{˜}{i}-4\underset{˜}{j}}{5}\\ \\ \widehat{\underset{˜}{r}}=\frac{1}{5}\left(3\underset{˜}{i}-4\underset{˜}{j}\right)\text{ , }\widehat{\underset{˜}{r}}=\frac{1}{5}\left(-3\underset{˜}{i}-4\underset{˜}{j}\right)\end{array}$

$\begin{array}{l}\underset{˜}{b}-\underset{˜}{a}=\left(\begin{array}{l}3\\ 7\end{array}\right)-\left(\begin{array}{l}6\\ 3\end{array}\right)=\left(\begin{array}{l}3-6\\ 7-3\end{array}\right)=\left(\begin{array}{l}-3\\ 4\end{array}\right)\\ \\ \left|\underset{˜}{b}-\underset{˜}{a}\right|=\sqrt{{\left(-3\right)}^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\end{array}$


(b)
$\begin{array}{l}\text{The unit vector in the direction of }\underset{˜}{b}-\underset{˜}{a}\\ =\frac{1}{5}\left(\begin{array}{l}-3\\ 4\end{array}\right)\\ =\left(\begin{array}{l}-\frac{3}{5}\\ \frac{4}{5}\end{array}\right)\end{array}$

Question 1:
Given that O (0, 0), A (–3, 4) and B (–9, 12), find in terms of the unit vectors,   $\underset{˜}{i}$ and   $\underset{˜.}{j}$
(b) the unit vector in the direction of  $\overrightarrow{AB}$

Solution:
(a) 
$\begin{array}{l}A=(-3,4),\text{ thus }\overrightarrow{OA}=-3\underset{˜}{i}+4\underset{˜}{j}\\ B=(-9,12),\text{ thus }\overrightarrow{OB}=-9\underset{˜}{i}+12\underset{˜}{j}\\ \overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}\\ \overrightarrow{AB}=-\left(-3\underset{˜}{i}+4\underset{˜}{j}\right)+\left(-9\underset{˜}{i}+12\underset{˜}{j}\right)\\ \overrightarrow{AB}=3\underset{˜}{i}-4\underset{˜}{j}-9\underset{˜}{i}+12\underset{˜}{j}\\ \overrightarrow{AB}=-6\underset{˜}{i}+8\underset{˜}{j}\end{array}$

(b)
$\begin{array}{l}\text{The magnitude of }\left|\overrightarrow{AB}\right|,\\ \left|\overrightarrow{AB}\right|=\sqrt{{\left(-6\right)}^2+{\left(8\right)}^2}=10\\ \\ \therefore \text{The unit vector in the direction of }\overrightarrow{AB},\\ \frac{\overrightarrow{AB}}{\left|\overrightarrow{AB}\right|}=\frac{1}{10}\left(-6\underset{˜}{i}+8\underset{˜}{j}\right)=-\frac{3}{5}\underset{˜}{i}+\frac{4}{5}\underset{˜}{j}\end{array}$

Question 2:
Given that A (–3, 2), B (4, 6) and C (m, n), find the value of m and of n such that    $2\overrightarrow{AB}+\overrightarrow{BC}=\left(\begin{array}{c}12\\ -3\end{array}\right)$

Solution:
$\begin{array}{l}A=\left(\begin{array}{l}-3\\ 2\end{array}\right),B=\left(\begin{array}{l}4\\ 6\end{array}\right)\text{ and }C=\left(\begin{array}{l}m\\ n\end{array}\right)\\ \\ \overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}\\ \overrightarrow{AB}=-\left(\begin{array}{l}-3\\ 2\end{array}\right)+\left(\begin{array}{l}4\\ 6\end{array}\right)=\left(\begin{array}{l}7\\ 4\end{array}\right)\\ \\ \overrightarrow{BC}=\overrightarrow{BO}+\overrightarrow{OC}\\ \overrightarrow{BC}=-\left(\begin{array}{l}4\\ 6\end{array}\right)+\left(\begin{array}{l}m\\ n\end{array}\right)=\left(\begin{array}{l}-4+m\\ -6+n\end{array}\right)\\ \\ \text{Given }2\overrightarrow{AB}+\overrightarrow{BC}=\left(\begin{array}{c}12\\ -3\end{array}\right)\\ 2\left(\begin{array}{l}7\\ 4\end{array}\right)+\left(\begin{array}{l}-4+m\\ -6+n\end{array}\right)=\left(\begin{array}{c}12\\ -3\end{array}\right)\\ \left(\begin{array}{l}14-4+m\\ 8-6+n\end{array}\right)=\left(\begin{array}{c}12\\ -3\end{array}\right)\\ \\ 10+m=12\\ m=2\\ \\ 2+n=-3\\ n=-5\end{array}$

Question 3:
$\text{Express }\overrightarrow{OD}\text{ in terms of }\underset{˜}{x}\text{ and }\underset{˜}{y}.$

Solution:
$\begin{array}{l}\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=3\underset{˜}{x}+12\underset{˜}{y}\\ \\ \overrightarrow{OD}=3\overrightarrow{DB}\\ \frac{\overrightarrow{OD}}{\overrightarrow{DB}}=\frac{3}{1}\\ \overrightarrow{OD}:\overrightarrow{DB}=3:1\\ \therefore \overrightarrow{OD}=\frac{3}{4}\overrightarrow{OB}\\ =\frac{3}{4}\left(3\underset{˜}{x}+12\underset{˜}{y}\right)\\ =\frac{9}{4}\underset{˜}{x}+9\underset{˜}{y}\end{array}$

$\begin{array}{l}\text{(a)(i) }\overrightarrow{AS}=\overrightarrow{AD}+\overrightarrow{DS}\\ =\overrightarrow{AD}+\overrightarrow{AQ}\leftarrow \boxed{\begin{array}{l}AQ:QB=3:1\text{ and }\\ DS:SC=3:1\\ \therefore \overrightarrow{AQ}=\overrightarrow{DS}\end{array}}\\ =\underset{˜}{b}+6\underset{˜}{a}\\ =6\underset{˜}{a}+\underset{˜}{b}\end{array}$


$\begin{array}{l}\text{(a)(ii) }\overrightarrow{QC}=\overrightarrow{QB}+\overrightarrow{BC}\\ =\frac{1}{3}\overrightarrow{AQ}+\overrightarrow{AD}\leftarrow \boxed{\begin{array}{l}AQ:QB=3:1\\ \frac{AQ}{QB}\text{=}\frac{3}{1}\Rightarrow QB=\frac{1}{3}AQ\\ \text{and for parallelogram, }\\ BC//AD,BC=AD\end{array}}\\ =\frac{1}{3}\left(6\underset{˜}{a}\right)+\underset{˜}{b}\\ =2\underset{˜}{a}+\underset{˜}{b}\end{array}$


$\begin{array}{l}\text{(b) }\overrightarrow{QT}=\overrightarrow{QA}+\overrightarrow{AT}\\ =\overrightarrow{QA}+\frac{3}{2}\overrightarrow{AS}\leftarrow \boxed{\begin{array}{l}AS=2ST\\ AT=3ST=\frac{3}{2}AS\end{array}}\\ =-6\underset{˜}{a}+\frac{3}{2}\left(6\underset{˜}{a}+\underset{˜}{b}\right)\\ =3\underset{˜}{a}+\frac{3}{2}\underset{˜}{b}\\ =\frac{3}{2}\left(2\underset{˜}{a}+\underset{˜}{b}\right)\\ =\frac{3}{2}\overrightarrow{QC}\\ \therefore \text{Points }Q,C\text{ and }T\text{ are collinear}\text{.}\end{array}$