If |overline{a}|=3,|overline{b}|=5,overline{b | vedclass

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(B) Given: $|\overline{a}|=3$,$|\overline{b}|=5$,$\overline{b} \cdot \overline{c}=10$,and the angle $	heta$ between $\overline{b}$ and $\overline{c}$

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) Given: $|\overline{a}|=3$,$|\overline{b}|=5$,$\overline{b} \cdot \overline{c}=10$,and the angle $	heta$ between $\overline{b}$ and $\overline{c}$ is $\frac{\pi}{3}$.First,find $|\overline{c}|$: $\overline{b} \cdot \overline{c} = |\overline{b}| |\overline{c}| \cos(\frac{\pi}{3}) = 10$.$5 	imes |\overline{c}| 	imes \frac{1}{2} = 10 \implies |\overline{c}| = 4$.Next,find $|\overline{b} 	imes \overline{c}|$: $|\overline{b} 	imes \overline{c}| = |\overline{b}| |\overline{c}| \sin(\frac{\pi}{3}) = 5 	imes 4 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3}$.Since $\overline{a}$ is perpendicular to $\overline{b} 	imes \overline{c}$,the angle between $\overline{a}$ and $(\overline{b} 	imes \overline{c})$ is $\frac{\pi}{2}$.Therefore,$|\overline{a} 	imes (\overline{b} 	imes \overline{c})| = |\overline{a}| |\overline{b} 	imes \overline{c}| \sin(\frac{\pi}{2}) = 3 	imes 10\sqrt{3} 	imes 1 = 30\sqrt{3}$.Wait,re-evaluating the options provided,there might be a typo in the question or options. Given the standard interpretation,the result is $30\sqrt{3}$. If we assume the question implies $\overline{a}$ is in the plane of $\overline{b}$ and $\overline{c}$ or similar,the result changes. However,based on the provided perpendicular condition,the calculation holds.

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