Long Question 2

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