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

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(B) Using the vector triple product formula,$\vec{a} = \vec{b} 	imes (\vec{b} 	imes \vec{c}) = (\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec

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$2$

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(B) Using the vector triple product formula,$\vec{a} = \vec{b} 	imes (\vec{b} 	imes \vec{c}) = (\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{c}$.Given $|\vec{a}| = \sqrt{2}$,$|\vec{b}| = 1$,$|\vec{c}| = 2$,and $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos 	heta = 1 \cdot 2 \cdot \cos 	heta = 2 \cos 	heta$.Substituting these values,$\vec{a} = (2 \cos 	heta) \vec{b} - (1)^2 \vec{c} = 2 \cos 	heta \vec{b} - \vec{c}$.Now,calculate $|\vec{a}|^2 = |2 \cos 	heta \vec{b} - \vec{c}|^2 = (2 \cos 	heta)^2 |\vec{b}|^2 + |\vec{c}|^2 - 2(2 \cos 	heta) (\vec{b} \cdot \vec{c})$.$|\vec{a}|^2 = 4 \cos^2 	heta (1) + 4 - 4 \cos 	heta (2 \cos 	heta) = 4 \cos^2 	heta + 4 - 8 \cos^2 	heta = 4 - 4 \cos^2 	heta$.Since $|\vec{a}|^2 = (\sqrt{2})^2 = 2$,we have $2 = 4 - 4 \cos^2 	heta$.$4 \cos^2 	heta = 2 \Rightarrow \cos^2 	heta = \frac{1}{2}$.Since $0 Therefore,$1 + 	an 	heta = 1 + 	an(\frac{\pi}{4}) = 1 + 1 = 2$.

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