If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar unit vectors such that $\overline{\mathrm{a}}…

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

Solution

$\begin{aligned} & \bar{a} \times(\bar{b} \times \bar{c})=\frac{\bar{b}+\bar{c}}{\sqrt{2}} \\ & \Rightarrow(\bar{a} \cdot \bar{c}) \bar{b}-(\bar{a} \cdot \bar{b}) \bar{c}=\frac{\bar{b}+\bar{c}}{\sqrt{2}} \\ & \Rightarrow\left(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}-\frac{1}{\sqrt{2}}\right) \bar{b}-\left(\bar{a} \cdot \bar{b}+\frac{1}{\sqrt{2}}\right) \bar{c}=0\end{aligned}$ Since $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non-coplanar unit vectors, $\begin{aligned} & \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}+\frac{1}{\sqrt{2}}=0 \\ & \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}=-\frac{1}{\sqrt{2}} \\ & \Rightarrow|\overline{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=-\frac{1}{\sqrt{2}} \\ & \Rightarrow \cos \theta=-\frac{1}{\sqrt{2}} \\ & \Rightarrow \theta=\frac{3 \pi}{4} \end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 2)

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