NIMCET 2017 — Mathematics PYQ

Concept:
$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta$

Calculation:
Here, $\vec{a}+\vec{b}+\vec{c}=0$
$\Rightarrow \vec{b}+\vec{c}=-\vec{a}$

Taking magnitude and squaring both sides,
$\Rightarrow |\vec{b}+\vec{c}{|}^2=|-\vec{a}{|}^2$
$\Rightarrow |\vec{b}{|}^2+|\vec{c}{|}^2+2\vec{b}.\vec{c}=49$
$\Rightarrow 2|\vec{b}|.|\vec{c}|\cos \theta =49-(25+9)$
$\Rightarrow \cos \theta =\frac{15}{2\times 5\times 3}$
$\Rightarrow \theta ={\cos }^{-1}(\frac{1}{2})$
$\theta =60^{\circ }$
Hence, option (1) is correct.