Vectors bar{a} and bar{b} are such that |bar{ | vedclass

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(B) Given: $|\bar{a}|=1, |\bar{b}|=4$ and $\bar{a} \cdot \bar{b}=2$. We know that $|\bar{a} 	imes \bar{b}|^2 = |\bar{a}|^2 |\bar{b}|^2 - (\bar{a} \cd

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$\frac{5 \pi}{6}$

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(B) Given: $|\bar{a}|=1, |\bar{b}|=4$ and $\bar{a} \cdot \bar{b}=2$. We know that $|\bar{a} 	imes \bar{b}|^2 = |\bar{a}|^2 |\bar{b}|^2 - (\bar{a} \cdot \bar{b})^2$. $|\bar{a} 	imes \bar{b}|^2 = (1)^2(4)^2 - (2)^2 = 16 - 4 = 12$. Given $\bar{c} = 2(\bar{a} 	imes \bar{b}) - 3\bar{b}$. Since $(\bar{a} 	imes \bar{b}) \perp \bar{b}$,the vectors $2(\bar{a} 	imes \bar{b})$ and $-3\bar{b}$ are orthogonal. $|\bar{c}|^2 = |2(\bar{a} 	imes \bar{b})|^2 + |-3\bar{b}|^2 = 4|\bar{a} 	imes \bar{b}|^2 + 9|\bar{b}|^2$. $|\bar{c}|^2 = 4(12) + 9(16) = 48 + 144 = 192$. $|\bar{c}| = \sqrt{192} = 8\sqrt{3}$. Now,$\bar{b} \cdot \bar{c} = \bar{b} \cdot (2(\bar{a} 	imes \bar{b}) - 3\bar{b}) = 2(\bar{b} \cdot (\bar{a} 	imes \bar{b})) - 3(\bar{b} \cdot \bar{b})$. Since $\bar{b} \cdot (\bar{a} 	imes \bar{b}) = 0$,we have $\bar{b} \cdot \bar{c} = -3|\bar{b}|^2 = -3(16) = -48$. Let $	heta$ be the angle between $\bar{b}$ and $\bar{c}$. $\cos 	heta = \frac{\bar{b} \cdot \bar{c}}{|\bar{b}| |\bar{c}|} = \frac{-48}{4 	imes 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$. $	heta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}ight) = \frac{5\pi}{6}$.

Vectors $\bar{a}$ and $\bar{b}$ are such that $|\bar{a}|=1$,$|\bar{b}|=4$ and $\bar{a} \cdot \bar{b}=2$. If $\bar{c}=2 \bar{a} \times \bar{b}-3 \bar{b}$,then the angle between $\bar{b}$ and $\bar{c}$ is

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