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(B) Given $\overline{a} 	imes \overline{b} = 2 \overline{a} 	imes \overline{c}$,we can write $\overline{a} 	imes (\overline{b} - 2 \overline{c}) =

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(B) Given $\overline{a} 	imes \overline{b} = 2 \overline{a} 	imes \overline{c}$,we can write $\overline{a} 	imes (\overline{b} - 2 \overline{c}) = \overline{0}$.This implies that the vector $(\overline{b} - 2 \overline{c})$ is parallel to $\overline{a}$,which is consistent with $\overline{b} - 2 \overline{c} = \lambda \overline{a}$.Let $\alpha$ be the angle between $\overline{b}$ and $\overline{c}$.Given $|\overline{b} 	imes \overline{c}| = \sqrt{15}$,we have $|\overline{b}| |\overline{c}| \sin \alpha = \sqrt{15}$.Substituting the values,$(4)(1) \sin \alpha = \sqrt{15}$,so $\sin \alpha = \frac{\sqrt{15}}{4}$.Then $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \frac{15}{16}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.Now,consider $|\overline{b} - 2 \overline{c}|^2 = |\lambda \overline{a}|^2$.$|\overline{b}|^2 + 4|\overline{c}|^2 - 4(\overline{b} \cdot \overline{c}) = \lambda^2 |\overline{a}|^2$.$16 + 4(1) - 4(|\overline{b}| |\overline{c}| \cos \alpha) = \lambda^2 (1)^2$.$20 - 4(4 	imes 1 	imes \frac{1}{4}) = \lambda^2$.$20 - 4 = \lambda^2$,which gives $\lambda^2 = 16$.Thus,$\lambda = \pm 4$. Since the options provide $-4$,the correct choice is $B$.

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