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

$\begin{array}{1 1} (A)\;\Large\frac{3 \overrightarrow{b}- \overrightarrow{a}}{2} \\(B)\;\Large\frac{3 \overrightarrow{a}- \overrightarrow{b}}{2} \\ (C)\;\Large\frac{9 \overrightarrow{a}-4 \overrightarrow{b}}{2} \\ (D)\;\Large\frac{3 \overrightarrow{a}+ \overrightarrow{b}}{2}\end{array}$

Toolbox:
• Three points $A,B\; and\; C$ are said to be collinear if $AB= \lambda \;BC$
Let $\overrightarrow{OA}=\overrightarrow{a}$ and $\overrightarrow{OB}=\overrightarrow{b}$ and $\overrightarrow{OC}=\overrightarrow{c}$
Given $\overrightarrow{BC}=1.5 \overrightarrow{BA}$
we know that $\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$
and $\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}$
$\overrightarrow{OC}-\overrightarrow{OB}=1.5 (\overrightarrow{OA}-\overrightarrow{OB})$
=>$\overrightarrow{OC}-\overrightarrow{b}=1.5 (\overrightarrow{a}-\overrightarrow{b})$
$\overrightarrow{OC}=1.5 \overrightarrow{a}-1.5 \overrightarrow{b}+\overrightarrow{b}$
On simplifying we get,
$\overrightarrow{OC}=1.5 \overrightarrow{a}-.5 \overrightarrow{b}$
$\qquad =\large\frac{3}{2}$$\overrightarrow{a}-\large\frac{1}{2}$$ \overrightarrow{b}$
$\Large\overrightarrow{c}=\frac{3 \overrightarrow{a}- \overrightarrow{b}}{2}$