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(D) The area of a parallelogram with diagonals $\vec{a}$ and $\vec{b}$ is given by $	ext{Area} = \frac{1}{2} |\vec{a} 	imes \vec{b}| = 2 \sqrt{2}$.T

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$\frac{3 \pi}{4}$

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(D) The area of a parallelogram with diagonals $\vec{a}$ and $\vec{b}$ is given by $	ext{Area} = \frac{1}{2} |\vec{a} 	imes \vec{b}| = 2 \sqrt{2}$.Thus,$|\vec{a} 	imes \vec{b}| = 4 \sqrt{2}$.Given $|\vec{a}|=1$ and $|\vec{a} \cdot \vec{b}| = |\vec{a} 	imes \vec{b}|$,we have $|\vec{a}| |\vec{b}| \cos 	heta = |\vec{a}| |\vec{b}| \sin 	heta$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.Since $	heta$ is acute,$\cos 	heta = \sin 	heta \Rightarrow 	heta = \frac{\pi}{4}$.Now,$|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin \frac{\pi}{4} = 1 \cdot |\vec{b}| \cdot \frac{1}{\sqrt{2}} = 4 \sqrt{2} \Rightarrow |\vec{b}| = 8$.Given $\vec{c} = 2 \sqrt{2}(\vec{a} 	imes \vec{b}) - 2 \vec{b}$.Since $(\vec{a} 	imes \vec{b})$ is perpendicular to both $\vec{a}$ and $\vec{b}$,the vectors $(2 \sqrt{2}(\vec{a} 	imes \vec{b}))$ and $(-2 \vec{b})$ are orthogonal.Thus,$|\vec{c}|^2 = |2 \sqrt{2}(\vec{a} 	imes \vec{b})|^2 + |-2 \vec{b}|^2 = 8(4 \sqrt{2})^2 + 4(8)^2 = 8(32) + 4(64) = 256 + 256 = 512$.$|\vec{c}| = \sqrt{512} = 16 \sqrt{2}$.Now,$\vec{b} \cdot \vec{c} = \vec{b} \cdot (2 \sqrt{2}(\vec{a} 	imes \vec{b}) - 2 \vec{b}) = 2 \sqrt{2} (\vec{b} \cdot (\vec{a} 	imes \vec{b})) - 2 |\vec{b}|^2 = 0 - 2(8)^2 = -128$.Let $\alpha$ be the angle between $\vec{b}$ and $\vec{c}$. Then $\cos \alpha = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|} = \frac{-128}{8 \cdot 16 \sqrt{2}} = \frac{-128}{128 \sqrt{2}} = -\frac{1}{\sqrt{2}}$.Therefore,$\alpha = \frac{3 \pi}{4}$.

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