Let vec{a}=a_1 hat{i}+a_2 hat{j}+a_3 hat{k} a | vedclass

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(C) Given $|\vec{a}|=1$,$|\vec{b}|=4$,and $\vec{a} \cdot \vec{b}=2$. We are given $\vec{c}=2(\vec{a} 	imes \vec{b})-3 \vec{b}$. To find the angle $

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$\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

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(C) Given $|\vec{a}|=1$,$|\vec{b}|=4$,and $\vec{a} \cdot \vec{b}=2$. We are given $\vec{c}=2(\vec{a} 	imes \vec{b})-3 \vec{b}$. To find the angle $	heta$ between $\vec{b}$ and $\vec{c}$,we use the formula $\cos 	heta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|}$. First,calculate $\vec{b} \cdot \vec{c}$: $\vec{b} \cdot \vec{c} = \vec{b} \cdot (2(\vec{a} 	imes \vec{b}) - 3 \vec{b}) = 2(\vec{b} \cdot (\vec{a} 	imes \vec{b})) - 3(\vec{b} \cdot \vec{b})$. Since $\vec{b} \cdot (\vec{a} 	imes \vec{b}) = 0$ (as the cross product is perpendicular to both vectors),we have $\vec{b} \cdot \vec{c} = 0 - 3|\vec{b}|^2 = -3(4^2) = -3(16) = -48$. Next,calculate $|\vec{c}|^2$: $|\vec{c}|^2 = |2(\vec{a} 	imes \vec{b}) - 3 \vec{b}|^2 = 4|\vec{a} 	imes \vec{b}|^2 + 9|\vec{b}|^2 - 12(\vec{a} 	imes \vec{b}) \cdot \vec{b}$. Since $(\vec{a} 	imes \vec{b}) \cdot \vec{b} = 0$,we have $|\vec{c}|^2 = 4|\vec{a} 	imes \vec{b}|^2 + 9|\vec{b}|^2$. Using $|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = (1)^2(4)^2 - (2)^2 = 16 - 4 = 12$. Thus,$|\vec{c}|^2 = 4(12) + 9(16) = 48 + 144 = 192$. So,$|\vec{c}| = \sqrt{192} = \sqrt{64 	imes 3} = 8\sqrt{3}$. Finally,$\cos 	heta = \frac{-48}{4 	imes 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$. Therefore,$	heta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}ight)$.

Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ and $\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$ be two vectors such that $|\vec{a}|=1$,$\vec{a} \cdot \vec{b}=2$,and $|\vec{b}|=4$. If $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$,then the angle between $\vec{b}$ and $\vec{c}$ is equal to:

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