Short Question 10 & 11

Question 10 (3 marks):
Diagram 3 shows vectors $\overrightarrow{OP},\overrightarrow{OQ}\text{ and }\overrightarrow{OM}$ drawn on a square grid.

Diagram

$\begin{array}{l}\left(\text{a}\right)\text{ Express }\overrightarrow{OM}\text{ in the form }h\underset{˜}{p}+k\underset{˜}{q}\text{, }\\ \text{where }h\text{ and }k\text{ are constants}.\\ \left(\text{b}\right)\text{ On the diagram 3, mark and label }\\ \text{the point }N\text{ such that }\overrightarrow{MN}+\overrightarrow{OQ}=2\overrightarrow{OP}.\end{array}$


Solution:
(a)
$\overrightarrow{OM}=\underset{˜}{p}+2\underset{˜}{q}$

(b)

$\begin{array}{l}\overrightarrow{MN}+\overrightarrow{OQ}=2\overrightarrow{OP}\\ \overrightarrow{MN}=2\overrightarrow{OP}-\overrightarrow{OQ}\\ =2\underset{˜}{p}-\underset{˜}{q}\end{array}$

Question 11 (4 marks):
$\begin{array}{l}A\left(2,3\right)\text{ and }B\left(-2,\text{ 5}\right)\text{ lie on a }\\ \text{Cartesian plane}\text{.}\\ \text{It is given that }3\overrightarrow{OA}=2\overrightarrow{OB}+\overrightarrow{OC}.\\ \text{Find}\\ \left(a\right)\text{ the coordinates of }C,\\ \left(b\right)\left|\overrightarrow{AC}\right|\end{array}$

Solution:
$\begin{array}{l}\text{Given }A\left(2,3\right)\text{ and }B\left(-2,5\right)\\ \text{Thus, }\overrightarrow{OA}=2\underset{˜}{i}+3\underset{˜}{j}\text{ and }\overrightarrow{OB}=-2\underset{˜}{i}+5\underset{˜}{j}\end{array}$

(a)
$\begin{array}{l}3\overrightarrow{OA}=2\overrightarrow{OB}+\overrightarrow{OC}\\ \overrightarrow{OC}=3\overrightarrow{OA}-2\overrightarrow{OB}\\ =3\left(2\underset{˜}{i}+3\underset{˜}{j}\right)-2\left(-2\underset{˜}{i}+5\underset{˜}{j}\right)\\ =6\underset{˜}{i}+9\underset{˜}{j}+4\underset{˜}{i}-10\underset{˜}{j}\\ =10\underset{˜}{i}-\underset{˜}{j}\\ \\ \text{Thus, coordinate of }C\text{ is }\left(10,-1\right)\end{array}$


(b)
$\begin{array}{l}\overrightarrow{AC}=\overrightarrow{AO}+\overrightarrow{OC}\\ =-\overrightarrow{OA}+\overrightarrow{OC}\\ =-\left(2\underset{˜}{i}+3\underset{˜}{j}\right)+10\underset{˜}{i}-\underset{˜}{j}\\ =-2\underset{˜}{i}+10\underset{˜}{i}-3\underset{˜}{j}-\underset{˜}{j}\\ =8\underset{˜}{i}-4\underset{˜}{j}\\ \\ \left|\overrightarrow{AC}\right|=\sqrt{8^2+{\left(-4\right)}^2}\\ =\sqrt{80}\text{ units}\\ =\sqrt{16\times 5}\text{ units}\\ =4\sqrt{5}\text{ units}\end{array}$