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 \cdot |\bar{c}| \cdot \frac{1}{\s

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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 $2 \cdot |\bar{c}| \cdot \frac{1}{\sqrt{2}} = 8$,which implies $|\bar{c}| = 4\sqrt{2}$. Since $\bar{a}, \bar{b}, \bar{c}$ are coplanar,the vector $\bar{v} = (\bar{b} 	imes \bar{c})$ is perpendicular to the plane containing $\bar{a}, \bar{b}, \bar{c}$. Thus,$\bar{a}$ is perpendicular to $(\bar{b} 	imes \bar{c})$. Therefore,$|\bar{a} 	imes (\bar{b} 	imes \bar{c})| = |\bar{a}| |\bar{b} 	imes \bar{c}| \sin(90^{\circ}) = |\bar{a}| |\bar{b} 	imes \bar{c}|$. We know $|\bar{b} 	imes \bar{c}| = |\bar{b}| |\bar{c}| \sin(45^{\circ}) = 2 \cdot 4\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 8$. Since $|\bar{a}| = 1$,the value is $1 \cdot 8 = 8$.

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 the value of $|\bar{a} \times(\bar{b} \times \bar{c})|$ is

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