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(D) Given that $\bar{c} = (2 \bar{a} 	imes \bar{b}) - 3 \bar{b}$.To find the angle $	heta$ between $\bar{b}$ and $\bar{c}$,we use the dot product fo

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

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(D) Given that $\bar{c} = (2 \bar{a} 	imes \bar{b}) - 3 \bar{b}$.To find the angle $	heta$ between $\bar{b}$ and $\bar{c}$,we use the dot product formula: $\cos 	heta = \frac{\bar{b} \cdot \bar{c}}{|\bar{b}| |\bar{c}|}$.First,calculate $\bar{b} \cdot \bar{c} = \bar{b} \cdot (2 \bar{a} 	imes \bar{b} - 3 \bar{b}) = 2 \bar{b} \cdot (\bar{a} 	imes \bar{b}) - 3 |\bar{b}|^2$.Since $\bar{b} \cdot (\bar{a} 	imes \bar{b}) = 0$ (as the cross product is perpendicular to both vectors),we have $\bar{b} \cdot \bar{c} = 0 - 3(4)^2 = -48$.Next,calculate $|\bar{c}|^2 = |2(\bar{a} 	imes \bar{b}) - 3 \bar{b}|^2 = 4|\bar{a} 	imes \bar{b}|^2 + 9|\bar{b}|^2 - 12 \bar{b} \cdot (\bar{a} 	imes \bar{b}) = 4|\bar{a} 	imes \bar{b}|^2 + 9(16) - 0$.We know $|\bar{a} 	imes \bar{b}|^2 = |\bar{a}|^2 |\bar{b}|^2 - (\bar{a} \cdot \bar{b})^2 = (1)^2(4)^2 - (2)^2 = 16 - 4 = 12$.So,$|\bar{c}|^2 = 4(12) + 144 = 48 + 144 = 192$,which means $|\bar{c}| = \sqrt{192} = 8\sqrt{3}$.Now,$\cos 	heta = \frac{-48}{4 	imes 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = -\frac{3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$.Thus,$	heta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

Let $\bar{a}$ and $\bar{b}$ be two vectors 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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