Question

If $\overrightarrow{a,}\overrightarrow{b}$are the position vectors of A, B respectively, find the position vector of a point C in AB produced such that AC = 3 AB and that a point D in BA produced such that BD = 2BA.

Answer 1

Let the position vectors of C and D are $\overrightarrow{c}$ and $\overrightarrow{d}$ respectively. We have,
$$\begin{gathered} AC=3AB. \\ \Rightarrow AC=3(AC-BC). \\ \Rightarrow 2AC=3BC. \end{gathered}$$
$\Rightarrow \frac{AC}{BC}=\frac{3}{2}.$
So C divides AB in the ratio of $3:2$ externally.
$\overrightarrow{c}=\frac{2\overrightarrow{a}-3\overrightarrow{b}}{2-3}=3\overrightarrow{b}-2\overrightarrow{a}.$

Position vector of point C is $3\overrightarrow{b}-2\overrightarrow{a}$
Moreover,
$$\begin{gathered} BD=2BA. \\ \Rightarrow BD=2(BD-AD). \\ \Rightarrow BD=2AD. \end{gathered}$$
$\Rightarrow \frac{BD}{AD}=\frac{2}{1}$.
∴ $\overrightarrow{d}=\frac{\overrightarrow{b}-2\overrightarrow{a}}{1-2}=2\overrightarrow{a}-\overrightarrow{b}.$
Position vector of point D is $2\overrightarrow{a}-\overrightarrow{b}$