Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ be three vectors having…

Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ be three vectors having magnitude 1,1 and 2 respectively. If $\overline{\mathrm{a}} \times(\overline{\mathrm{a}} \times \overline{\mathrm{c}})+\overline{\mathrm{b}}=\overline{0}$, then the acute angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{c}}$ is
  1. $\frac{\pi}{6}$
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{2}$

Solution

Given, $|\bar{a}|=1,|\bar{b}|=1$ and $|\bar{c}|=2$ $\begin{aligned} & \text { Also, } \overline{\mathrm{a}} \times(\overline{\mathrm{a}} \times \overline{\mathrm{c}})+\overline{\mathrm{b}}=\overline{0} \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{a}}) \overline{\mathrm{c}}+\overline{\mathrm{b}}=\overline{\mathrm{c}} \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}+\overline{\mathrm{b}}=\overline{0} \quad \quad \ldots\left[\because \overline{\mathrm{a}} \cdot \overline{\mathrm{a}}=|\overline{\mathrm{a}}|^2=1\right] \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}=-\overline{\mathrm{b}} \\ & \Rightarrow|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}|=|-\overline{\mathrm{b}}| \\ & \Rightarrow\left|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^{\overline{\mathrm{a}}-\overline{\mathrm{c}}}\right|^2=|\overline{\mathrm{b}}|^2 \\ & \Rightarrow|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}|^2+|\overline{\mathrm{c}}|^2-2\{(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}\}=|\overline{\mathrm{b}}|^2 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2|\overline{\mathrm{a}}|^2+|\overline{\mathrm{c}}|^2-2(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})=|\overline{\mathrm{b}}|^2 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2\left\{|\overline{\mathrm{a}}|^2-2\right\}+|\overline{\mathrm{c}}|^2=|\overline{\mathrm{b}}|^2 \end{aligned}$ $\begin{aligned} & \Rightarrow-(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2+4=1 \quad \ldots\left[\because|\overline{\mathrm{~b}}|^2=1,|\overline{\mathrm{c}}|^2=4\right] \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2=3 \\ & \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}= \pm \sqrt{3} \\ & \Rightarrow|\overline{\mathrm{a}}||\overline{\mathrm{c}}| \cos \theta=\sqrt{3} \end{aligned}$ where $\theta$ is an acute angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{c}}$ $\Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6}$

Asked in: MHT CET 2024 (04 May Shift 1)

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