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The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by: $$cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$$
Given $\vec{a}=3\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}-2\hat{j}+4\hat{k}$, the dot product is: $$\vec{a} \cdot \vec{b} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12$$
The magnitude of $\vec{a}$ is: $$|\vec{a}| = \sqrt{(3)^2 + (1)^2 + (2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$$
The magnitude of $\vec{b}$ is: $$|\vec{b}| = \sqrt{(2)^2 + (-2)^2 + (4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$$
$$cos(\theta) = \frac{12}{\sqrt{14} \cdot 2\sqrt{6}} = \frac{6}{\sqrt{14 \cdot 6}} = \frac{6}{\sqrt{84}} = \frac{6}{\sqrt{4 \cdot 21}} = \frac{6}{2\sqrt{21}} = \frac{3}{\sqrt{21}}$$ Rationalizing the denominator: $$cos(\theta) = \frac{3\sqrt{21}}{21} = \frac{\sqrt{21}}{7}$$
$$\theta = cos^{-1}(\frac{\sqrt{21}}{7})$$
Final Answer: $cos^{-1}(\frac{\sqrt{21}}{7})$
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