NIMCET 2017 — Mathematics PYQ

Concept:
$(\vec{a}+\vec{b}+\vec{c}{)}^2=|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2+2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})$

Calculation:

Here, $|\vec{a}|=2$, $|\vec{b}|=3$, $|\vec{c}|=5$

$\vec{a}+\vec{b}+\vec{c}=0$

$|\vec{a}+\vec{b}+\vec{c}{|}^2=0$

$\Rightarrow {\vec{a}}^2+{\vec{b}}^2+{\vec{c}}^2+2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=0$

$\Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=-\frac{({\vec{a}}^2+{\vec{b}}^2+{\vec{c}}^2)}{2}$

$=-\frac{4+9+25}{2}$

$=-19$

Hence, option (4) is correct.