Short Question 8 & 9

Question 8 (3 marks):
Diagram shows vectors $\overrightarrow{AB},\overrightarrow{AC}\text{ and }\overrightarrow{AD}$ drawn on a square grid with sides of 1 unit.

Diagram

$\begin{array}{l}\left(\text{a}\right)\text{ Find }\left|-\overrightarrow{BA}\right|.\\ \left(\text{b}\right)\text{ Given }\overrightarrow{AB}=\underset{˜}{b}\text{ and }\overrightarrow{AC}=\underset{˜}{c},\\ \text{express in terms of }\underset{˜}{b}\text{ and }\underset{˜}{c}\\ \left(i\right)\overrightarrow{BC},\\ \left(ii\right)\overrightarrow{AD}\end{array}$


Solution:
(a)
$\left|-\overrightarrow{BA}\right|=\sqrt{3^2+4^2}=5\text{ units}$

(b)(i)
$\begin{array}{l}\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}\\ =-\underset{˜}{b}+\underset{˜}{c}\\ =\underset{˜}{c}-\underset{˜}{b}\end{array}$

(b)(ii)
$\begin{array}{l}\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}\\ =\underset{˜}{b}+2\overrightarrow{BC}\\ =\underset{˜}{b}+2\left(\underset{˜}{c}-\underset{˜}{b}\right)\\ =2\underset{˜}{c}-\underset{˜}{b}\end{array}$

Question 9 (3 marks):
Diagram shows a trapezium ABCD.

Diagram

$\begin{array}{l}\text{Given }\underset{˜}{p}=\left(\begin{array}{l}3\\ 4\end{array}\right)\text{ and }\underset{˜}{q}=\left(\begin{array}{l}k-1\\ 2\end{array}\right)\text{, }\\ \text{where }k\text{ is a constant, find value of }k\text{.}\end{array}$


Solution:

$\begin{array}{l}\underset{˜}{p}=m\underset{˜}{q}\\ \left(\begin{array}{l}3\\ 4\end{array}\right)=m\left(\begin{array}{l}k-1\\ 2\end{array}\right)\\ \left(\begin{array}{l}3\\ 4\end{array}\right)=\left(\begin{array}{l}mk-m\\ 2m\end{array}\right)\\ \\ mk-m=3\mathrm{..........}\left(1\right)\\ 2m=4\mathrm{................}\left(2\right)\\ \\ \text{From}\left(2\right):\\ 2m=4\\ m=2\\ \\ \text{Substitute }m=2\text{ into }\left(1\right):\\ 2k-2=3\\ 2k=3+2\\ 2k=5\\ k=\frac{5}{2}\end{array}$