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(D) Given $|\overline{b}|=4, |\overline{c}|=1$ and $|\overline{b} 	imes \overline{c}|=\sqrt{15}$.Let $\alpha$ be the angle between $\overline{b}$ and

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$4$

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(D) Given $|\overline{b}|=4, |\overline{c}|=1$ and $|\overline{b} 	imes \overline{c}|=\sqrt{15}$.Let $\alpha$ be the angle between $\overline{b}$ and $\overline{c}$.$|\overline{b} 	imes \overline{c}| = |\overline{b}||\overline{c}| \sin \alpha = \sqrt{15}$.$4 	imes 1 	imes \sin \alpha = \sqrt{15} \implies \sin \alpha = \frac{\sqrt{15}}{4}$.Then,$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{15}{16} = \frac{1}{16} \implies \cos \alpha = \frac{1}{4}$ (assuming $\alpha$ is acute).Given $\overline{b} = 2\overline{c} + \lambda \overline{a}$,we have $\overline{b} - 2\overline{c} = \lambda \overline{a}$.Squaring both sides:$|\overline{b} - 2\overline{c}|^2 = |\lambda \overline{a}|^2$.$|\overline{b}|^2 + 4|\overline{c}|^2 - 4(\overline{b} \cdot \overline{c}) = \lambda^2 |\overline{a}|^2$.$|\overline{b}|^2 + 4|\overline{c}|^2 - 4(|\overline{b}||\overline{c}| \cos \alpha) = \lambda^2 |\overline{a}|^2$.$16 + 4(1) - 4(4 	imes 1 	imes \frac{1}{4}) = \lambda^2 (2)^2$.$16 + 4 - 4 = 4\lambda^2$.$16 = 4\lambda^2 \implies \lambda^2 = 4 \implies \lambda = \pm 2$.Since $4$ is an option,the value is $4$.

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors,$|\overline{a}|=2, |\overline{b}|=4, |\overline{c}|=1$,$|\overline{b} \times \overline{c}|=\sqrt{15}$ and $\overline{b}=2 \overline{c}+\lambda \overline{a}$,then the value of $\lambda$ is

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