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(D) Given $\vec{c} = 2(\vec{a} 	imes \vec{b}) - 3\vec{b}$.Taking the dot product with $\vec{b}$:$\vec{b} \cdot \vec{c} = \vec{b} \cdot (2(\vec{a} 	i

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$48$

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(D) Given $\vec{c} = 2(\vec{a} 	imes \vec{b}) - 3\vec{b}$.Taking the dot product with $\vec{b}$:$\vec{b} \cdot \vec{c} = \vec{b} \cdot (2(\vec{a} 	imes \vec{b}) - 3\vec{b}) = 2(\vec{b} \cdot (\vec{a} 	imes \vec{b})) - 3|\vec{b}|^2$.Since $\vec{b} \cdot (\vec{a} 	imes \vec{b}) = 0$,we have $\vec{b} \cdot \vec{c} = -3|\vec{b}|^2 = -3(4)^2 = -48$.Also,$\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}| \cos \alpha = 4|\vec{c}| \cos \alpha$.Thus,$4|\vec{c}| \cos \alpha = -48 \Rightarrow |\vec{c}| \cos \alpha = -12$.Now,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 $(\vec{a} 	imes \vec{b}) \cdot \vec{b} = 0$,$|\vec{c}|^2 = 4|\vec{a} 	imes \vec{b}|^2 + 9(16) = 4|\vec{a}|^2|\vec{b}|^2 \sin^2 	heta + 144$.Given $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta = 1 \cdot 4 \cdot \cos 	heta = 2 \Rightarrow \cos 	heta = 1/2 \Rightarrow \sin^2 	heta = 3/4$.$|\vec{c}|^2 = 4(1)(16)(3/4) + 144 = 48 + 144 = 192$.We have $|\vec{c}|^2 \cos^2 \alpha = (-12)^2 = 144$.$192 \cos^2 \alpha = 144 \Rightarrow \cos^2 \alpha = 144/192 = 3/4$.Then $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - 3/4 = 1/4$.Therefore,$192 \sin^2 \alpha = 192 	imes (1/4) = 48$.

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=1$,$|\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\alpha$,then $192 \sin^2 \alpha$ is equal to

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