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(B) Given $\vec{a} 	imes (\vec{a} 	imes \vec{c}) + \vec{b} = \vec{0}$, we have $\vec{b} = -\vec{a} 	imes (\vec{a} 	imes \vec{c})$.Using the vector

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$\frac{\pi}{6}$

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

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

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