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

If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors such that $|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=1$, $\overline{\mathrm{c}}=\lambda(\overline{\mathrm{a}} \times \overline{\mathrm{b}})$ and $|\overline{\mathrm{a}}|=\frac{1}{\sqrt{3}},|\overline{\mathrm{b}}|=\frac{1}{\sqrt{2}},|\overline{\mathrm{c}}|=\frac{1}{\sqrt{6}}$, then the angle between $\bar{a}$ and $\bar{b}$ is
  1. $\frac{\pi}{6}$
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{2}$

Solution

Let $\theta$ be the angle between $\bar{a}$ and $\bar{b}$. $\begin{aligned} & \text { Since } \overline{\mathrm{c}}=\lambda(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \\ & \Rightarrow \overline{\mathrm{c}} \perp \overline{\mathrm{a}}, \overline{\mathrm{c}} \perp \overline{\mathrm{b}} \\ & \Rightarrow \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}=0 \end{aligned}$ Now, $\begin{aligned} & |\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=1 \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=1 \\ & \Rightarrow|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}})=1 \\ & \Rightarrow \frac{1}{3}+\frac{1}{2}+\frac{1}{6}+2\{|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \cos \theta\}=1 \\ & \Rightarrow \cos \theta=0 \\ & \Rightarrow \theta=\frac{\pi}{2} \end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 1)

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