Let overline{a}, overline{b}, overline{c} be | vedclass

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(D) Given $\overline{b} \cdot \overline{c} = 10$. Since $\overline{b} \cdot \overline{c} = |\overline{b}| |\overline{c}| \cos(\frac{\pi}{3}) = 10$,we

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(D) Given $\overline{b} \cdot \overline{c} = 10$. Since $\overline{b} \cdot \overline{c} = |\overline{b}| |\overline{c}| \cos(\frac{\pi}{3}) = 10$,we have $(5) |\overline{c}| (\frac{1}{2}) = 10$,which implies $|\overline{c}| = 4$. We are given that $\overline{a}$ is perpendicular to $\overline{b} 	imes \overline{c}$,so the angle between $\overline{a}$ and $\overline{b} 	imes \overline{c}$ is $\frac{\pi}{2}$. Therefore,$|\overline{a} 	imes (\overline{b} 	imes \overline{c})| = |\overline{a}| |\overline{b} 	imes \overline{c}| \sin(\frac{\pi}{2}) = |\overline{a}| |\overline{b} 	imes \overline{c}|$. Now,$|\overline{b} 	imes \overline{c}| = |\overline{b}| |\overline{c}| \sin(\frac{\pi}{3}) = (5)(4)(\frac{\sqrt{3}}{2}) = 10\sqrt{3}$. Thus,$|\overline{a} 	imes (\overline{b} 	imes \overline{c})| = (\sqrt{3})(10\sqrt{3}) = 10 	imes 3 = 30$.

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