Let vec{a}, vec{b}, and vec{c} be three unit | vedclass

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(A) Given that $\vec{a}, \vec{b},$ and $\vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1.$Using the vector triple product formula,$

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$30$

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(A) Given that $\vec{a}, \vec{b},$ and $\vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1.$Using the vector triple product formula,$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}.$Given $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{1}{2} \vec{b},$ we have $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{1}{2} \vec{b}.$Since $\vec{b}$ and $\vec{c}$ are non-parallel,they are linearly independent. Comparing the coefficients of $\vec{b}$ and $\vec{c},$ we get $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = 0.$We know $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \alpha = 1 \cdot 1 \cdot \cos \alpha = 0 \implies \alpha = 90^{\circ}.$We know $\vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \beta = 1 \cdot 1 \cdot \cos \beta = \frac{1}{2} \implies \beta = 60^{\circ}.$Thus,$|\alpha - \beta| = |90^{\circ} - 60^{\circ}| = 30^{\circ}.$

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