If $\vec{a}$ and $\vec{b}$ are the position vectors of $A$ and $B$ respectively, then find the position vector of a point $C$ in $BA$ produced such that $BC = 1.5 BA$.

$\frac{3\vec{a} - \vec{b}}{2}$

The correct answer is Option (2) → $\frac{3\vec{a} - \vec{b}}{2}$ ##

Since, $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$

$∴\vec{BA} = \vec{OA} - \vec{OB} = \vec{a} - \vec{b}$

And since, $1.5 \vec{BA} = 1.5(\vec{a} - \vec{b})$ [multiply by 1.5 on both sides of equation]

$\vec{BC} = 1.5 \vec{BA} = 1.5(\vec{a} - \vec{b})$

$\vec{OC} - \vec{OB} = 1.5\vec{a} - 1.5\vec{b}$

$\vec{OC} = 1.5\vec{a} - 1.5\vec{b} + \vec{b}$ $[∵\vec{OB} = \vec{b}]$

$= 1.5\vec{a} - 0.5\vec{b}$

$= \frac{3\vec{a} - \vec{b}}{2}$