Mathematics - Vector Question with Solution | TestHub

Let $a+b+c=0\Rightarrow (a+b)=-c$ $\Rightarrow (a+b{)}^2=c^2$ $$\begin{array}{rl}& \Rightarrow a^2+b^2+2a\cdot b=c^2\\ & \Rightarrow 9+25+\mathrm{2.3.5}\cos \theta =49\\ & (\because |\vec{a}|=3,|\vec{b}|=5\text{ and }|\vec{c}|=7)\\ & \therefore \cos \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}\end{array}$$