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If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors with magnitudes $\sqrt{3}$, 1,2 respectively, such that $\overline{\mathrm{a}} \times(\overline{\mathrm{a}} \times \overline{\mathrm{c}})+3 \overline{\mathrm{b}}=\overline{0}$, if $\theta$ is the angle between $\bar{a}$ and $\bar{c}$, then $\sec ^2 \theta$ is
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$\frac{4}{3}$
$\begin{aligned} & \text { Given }|\overline{\mathrm{a}}|=\sqrt{3},|\overline{\mathrm{b}}|=1,|\overline{\mathrm{c}}|=2 \\ & \overline{\mathrm{a}} \times(\overline{\mathrm{a}} \times \overline{\mathrm{c}})+3 \overline{\mathrm{b}}=\overline{0} \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{a}}) \overline{\mathrm{c}}+3 \overline{\mathrm{b}}=\overline{0} \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-3 \overline{\mathrm{c}}=-3 \overline{\mathrm{b}} \quad \ldots\left[\overline{\mathrm{a}} \cdot \overline{\mathrm{a}}=|\overline{\mathrm{a}}|^2=3\right] \\ & \Rightarrow|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-3 \overline{\mathrm{c}}|=|-3 \overline{\mathrm{b}}| \\ & \Rightarrow\left|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^{-} \overline{\mathrm{a}}-3 \overline{\mathrm{c}}\right|^2=|-3 \overline{\mathrm{b}}|^2 \\ & \Rightarrow\left|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^{-}\right|^2+9|\overline{\mathrm{c}}|^2-6\{(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})\}=9|\overline{\mathrm{b}}|^2 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2|\overline{\mathrm{a}}|^2+9|\overline{\mathrm{c}}|^2-6(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})=9|\overline{\mathrm{b}}|^2 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2\left(|\overline{\mathrm{a}}|^2-6\right)+9|\overline{\mathrm{c}}|^2=9|\overline{\mathrm{b}}|^2 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2(3-6)+9(2)^2=9(1)^2\end{aligned}$
$\begin{aligned} & \Rightarrow-3(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2=-27 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2=9 \\ & \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}= \pm 3 \\ & \Rightarrow|\overline{\mathrm{a}}||\overline{\mathrm{c}}| \cos \theta= \pm 3 \\ & \Rightarrow(\sqrt{3})(2) \cos \theta= \pm 3 \\ & \Rightarrow \cos \theta= \pm \frac{\sqrt{3}}{2} \\ & \Rightarrow \sec \theta= \pm \frac{2}{\sqrt{3}} \\ & \Rightarrow \sec ^2 \theta=\frac{4}{3}\end{aligned}$
$\begin{aligned} & \Rightarrow-3(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2=-27 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2=9 \\ & \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}= \pm 3 \\ & \Rightarrow|\overline{\mathrm{a}}||\overline{\mathrm{c}}| \cos \theta= \pm 3 \\ & \Rightarrow(\sqrt{3})(2) \cos \theta= \pm 3 \\ & \Rightarrow \cos \theta= \pm \frac{\sqrt{3}}{2} \\ & \Rightarrow \sec \theta= \pm \frac{2}{\sqrt{3}} \\ & \Rightarrow \sec ^2 \theta=\frac{4}{3}\end{aligned}$
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