If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are unit vectors and $ heta$ is angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{c}}$ and $\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+2 \overline{\mathrm{c}}=\overline{0}$, then $|\overline{\mathrm{a}} imes…

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If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are unit vectors and $	heta$ is angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{c}}$ and $\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+2 \overline{\mathrm{c}}=\overline{0}$, then $|\overline{\mathrm{a}} 	imes \overline{\mathrm{c}}|=$

Aicharya · Accepted Answer

$\begin{aligned} & \overline{\mathrm{a}}+2 \overline{\mathrm{b}}+2 \overline{\mathrm{c}}=\overline{0} \ & \Rightarrow \mathrm{a}+2 \overline{\mathrm{c}}=-2 \overline{\mathrm{b}} \end{aligned}$ Squaring on both sides, we get $\begin{aligned} & |\overline{\mathrm{a}}|^2+4 \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}+4|\overline{\mathrm{c}}|^2=4|\overline{\mathrm{b}}|^2 \ & \Rightarrow 1+4|\overline{\mathrm{a}}||\overrightarrow{\mathrm{c}}| \cos 	heta+4=4 \ & \Rightarrow \cos 	heta=-\frac{1}{4} \ & \Rightarrow \sin 	heta=\frac{\sqrt{15}}{4} \ & |\overline{\mathrm{a}} 	imes \overline{\mathrm{c}}|=|\overline{\mathrm{a}}||\overline{\mathrm{c}}| \sin 	heta \ & \quad=(1)(1)\left(\frac{\sqrt{15}}{4}ight)=\frac{\sqrt{15}}{4} \end{aligned}$

If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are unit vectors and $\theta$ is…

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