If a vector and b vector are the position vectors of A and B ,respectively ,find the position vector of a point C in BA produced such that BA =1.5 BA.

if C is the point in BA produced such that BC = 1.5 BA
$BC=\frac{3}{2}.BA\Rightarrow \frac{BC}{BA}=\frac{3}{2}$
$$\begin{gathered} \frac{BA}{BC}=\frac{2}{3} \\ \frac{BA-BC}{BC}=\frac{2-3}{3} \\ \frac{-AC}{BC}=\frac{-1}{3}\Rightarrow \frac{AC}{BC}=\frac{1}{3} \end{gathered}$$
thus C divides AB externally in the ratio 3:1
therefore position vector of C is $\frac{3*a-1*b}{3-1}=\frac{3a-b}{2}$

hope this helps you