$\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are three vectors such that…

$\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are three vectors such that $\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}=\overline{0}$ and $|\overline{\mathrm{a}}|=3,|\overline{\mathrm{b}}|=5,|\overline{\mathrm{c}}|=7$, then the angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ is
  1. $\frac{\pi}{4}$
  2. $\frac{\pi}{2}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{6}$

Solution

$\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}=0 \Rightarrow \overline{\mathrm{c}}=-(\overline{\mathrm{a}}+\overline{\mathrm{b}})$ and let angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ be $\theta$ $\begin{aligned} & \therefore|\overline{\mathrm{c}}|^2=(\overline{\mathrm{a}}+\overline{\mathrm{b}})^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \\ & =|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+2|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}| \cdot \cos \theta \\ & \therefore(7)^2=(3)^2+(5)^2+2(3)(5) \cos \theta \\ & \therefore 49=9+25+30 \cos \theta \Rightarrow \cos \theta=\frac{15}{30}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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