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

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$\frac{\pi}{6}$

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(A) Given,$|\overline{a}|=1, |\overline{b}|=1$ and $|\overline{c}|=2$. Using the vector triple product formula $\overline{a} 	imes (\overline{b} 	imes \overline{c}) = (\overline{a} \cdot \overline{c})\overline{b} - (\overline{a} \cdot \overline{b})\overline{c}$,we have: $\overline{a} 	imes (\overline{a} 	imes \overline{c}) = (\overline{a} \cdot \overline{c})\overline{a} - (\overline{a} \cdot \overline{a})\overline{c}$. Since $|\overline{a}|=1$,we have $\overline{a} \cdot \overline{a} = 1$. Substituting this into the given equation $\overline{a} 	imes (\overline{a} 	imes \overline{c}) + \overline{b} = \overline{0}$: $(\overline{a} \cdot \overline{c})\overline{a} - \overline{c} + \overline{b} = \overline{0}$. Rearranging gives $(\overline{a} \cdot \overline{c})\overline{a} - \overline{c} = -\overline{b}$. Taking the squared magnitude on both sides: $|(\overline{a} \cdot \overline{c})\overline{a} - \overline{c}|^2 = |-\overline{b}|^2$. $(\overline{a} \cdot \overline{c})^2 |\overline{a}|^2 + |\overline{c}|^2 - 2(\overline{a} \cdot \overline{c})(\overline{a} \cdot \overline{c}) = |\overline{b}|^2$. $(\overline{a} \cdot \overline{c})^2(1) + 4 - 2(\overline{a} \cdot \overline{c})^2 = 1$. $-(\overline{a} \cdot \overline{c})^2 = -3 \Rightarrow (\overline{a} \cdot \overline{c})^2 = 3$. Thus,$\overline{a} \cdot \overline{c} = \sqrt{3}$ (since the angle is acute,$\cos 	heta > 0$). $|\overline{a}||\overline{c}| \cos 	heta = \sqrt{3} \Rightarrow (1)(2) \cos 	heta = \sqrt{3}$. $\cos 	heta = \frac{\sqrt{3}}{2} \Rightarrow 	heta = \frac{\pi}{6}$.

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

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