Let bar{a}, bar{b} and bar{c} be three vector | vedclass

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(B) Given,$|\bar{a}|=1, |\bar{b}|=1$ and $|\bar{c}|=2$. Using the vector triple product formula $\bar{a} 	imes (\bar{a} 	imes \bar{c}) = (\bar{a} \c

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$\frac{\pi}{6}$

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(B) Given,$|\bar{a}|=1, |\bar{b}|=1$ and $|\bar{c}|=2$. Using the vector triple product formula $\bar{a} 	imes (\bar{a} 	imes \bar{c}) = (\bar{a} \cdot \bar{c})\bar{a} - (\bar{a} \cdot \bar{a})\bar{c}$. Given equation: $(\bar{a} \cdot \bar{c})\bar{a} - |\bar{a}|^2\bar{c} + \bar{b} = \bar{0}$. Since $|\bar{a}|=1$,we have $(\bar{a} \cdot \bar{c})\bar{a} - \bar{c} = -\bar{b}$. Taking the magnitude squared on both sides: $|(\bar{a} \cdot \bar{c})\bar{a} - \bar{c}|^2 = |-\bar{b}|^2$. $(\bar{a} \cdot \bar{c})^2 |\bar{a}|^2 + |\bar{c}|^2 - 2(\bar{a} \cdot \bar{c})(\bar{a} \cdot \bar{c}) = |\bar{b}|^2$. $(\bar{a} \cdot \bar{c})^2(1) + 4 - 2(\bar{a} \cdot \bar{c})^2 = 1$. $-(\bar{a} \cdot \bar{c})^2 = -3 \Rightarrow (\bar{a} \cdot \bar{c})^2 = 3$. Thus,$\bar{a} \cdot \bar{c} = \sqrt{3}$ (for acute angle). $|\bar{a}||\bar{c}| \cos 	heta = \sqrt{3} \Rightarrow (1)(2) \cos 	heta = \sqrt{3}$. $\cos 	heta = \frac{\sqrt{3}}{2} \Rightarrow 	heta = \frac{\pi}{6}$.

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