If bar{a}, bar{b}, bar{c} are three coplanar | vedclass

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(A) Given that $\bar{b} \cdot \bar{c} = |\bar{b}| |\bar{c}| \cos(45^{\circ}) = 8$. Substituting the values,we get $2 	imes |\bar{c}| 	imes \frac{1}{

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$8$

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(A) Given that $\bar{b} \cdot \bar{c} = |\bar{b}| |\bar{c}| \cos(45^{\circ}) = 8$. Substituting the values,we get $2 	imes |\bar{c}| 	imes \frac{1}{\sqrt{2}} = 8$,which implies $|\bar{c}| = 4 \sqrt{2}$. Since $\bar{a}, \bar{b}, \bar{c}$ are coplanar,the vector $\bar{b} 	imes \bar{c}$ is perpendicular to the plane containing $\bar{a}, \bar{b}, \bar{c}$. Therefore,$\bar{a}$ is perpendicular to $\bar{b} 	imes \bar{c}$. Using the property $|\bar{a} 	imes (\bar{b} 	imes \bar{c})| = |\bar{a}| |\bar{b} 	imes \bar{c}| \sin(90^{\circ}) = |\bar{a}| |\bar{b} 	imes \bar{c}|$. We know $|\bar{b} 	imes \bar{c}| = |\bar{b}| |\bar{c}| \sin(45^{\circ}) = 2 	imes 4 \sqrt{2} 	imes \frac{1}{\sqrt{2}} = 8$. Thus,$|\bar{a} 	imes (\bar{b} 	imes \bar{c})| = 1 	imes 8 = 8$.

If $\bar{a}, \bar{b}, \bar{c}$ are three coplanar vectors such that $|\bar{a}|=1, |\bar{b}|=2$,$\bar{b} \cdot \bar{c}=8$,and the angle between $\bar{b}$ and $\bar{c}$ is $45^{\circ}$,then $|\bar{a} \times (\bar{b} \times \bar{c})|=$

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