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(D) Let $	heta$ be the angle between $\bar{a}$ and $\bar{b}$.Since $\overline{c}=\lambda(\overline{a} 	imes \overline{b})$,it implies $\overline{c}$

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) Let $	heta$ be the angle between $\bar{a}$ and $\bar{b}$.Since $\overline{c}=\lambda(\overline{a} 	imes \overline{b})$,it implies $\overline{c}$ is perpendicular to both $\overline{a}$ and $\overline{b}$.Thus,$\overline{c} \cdot \overline{a} = 0$ and $\overline{c} \cdot \overline{b} = 0$.Given $|\overline{a}+\overline{b}+\overline{c}|=1$,squaring both sides gives $|\overline{a}+\overline{b}+\overline{c}|^2 = 1$.Expanding this,we get $|\overline{a}|^2+|\overline{b}|^2+|\overline{c}|^2+2(\overline{a} \cdot \overline{b}+\overline{b} \cdot \overline{c}+\overline{c} \cdot \overline{a}) = 1$.Substituting the values $|\overline{a}|^2 = \frac{1}{3}$,$|\overline{b}|^2 = \frac{1}{2}$,$|\overline{c}|^2 = \frac{1}{6}$ and the dot products involving $\overline{c}$ as $0$:$\frac{1}{3} + \frac{1}{2} + \frac{1}{6} + 2(\overline{a} \cdot \overline{b}) + 0 + 0 = 1$.$\frac{2+3+1}{6} + 2|\overline{a}||\overline{b}| \cos 	heta = 1$.$1 + 2(\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{2}}) \cos 	heta = 1$.$2(\frac{1}{\sqrt{6}}) \cos 	heta = 0$.Therefore,$\cos 	heta = 0$,which means $	heta = \frac{\pi}{2}$.

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