Short Question 1 – 3

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