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

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(C) Given $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos(\frac{\pi}{3}) = 10$.Since $|\vec{b}| = 5$,we have $5 |\vec{c}| (\frac{1}{2}) = 10$,which

Vedclass · Accepted Answer

$30$

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(C) Given $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos(\frac{\pi}{3}) = 10$.Since $|\vec{b}| = 5$,we have $5 |\vec{c}| (\frac{1}{2}) = 10$,which implies $|\vec{c}| = 4$.Now,the magnitude of the cross product is $|\vec{b} 	imes \vec{c}| = |\vec{b}| |\vec{c}| \sin(\frac{\pi}{3}) = 5 	imes 4 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3}$.Since $\vec{a}$ is perpendicular to $\vec{b} 	imes \vec{c}$,the angle between $\vec{a}$ and $\vec{b} 	imes \vec{c}$ is $\frac{\pi}{2}$.Thus,$|\vec{a} 	imes (\vec{b} 	imes \vec{c})| = |\vec{a}| |\vec{b} 	imes \vec{c}| \sin(\frac{\pi}{2}) = \sqrt{3} 	imes 10\sqrt{3} 	imes 1 = 30$.

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