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(D) The angle $	heta$ between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by the dot product formula: $\cos 	heta = \frac{\ov

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$90$

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(D) The angle $	heta$ between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by the dot product formula: $\cos 	heta = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|}$.First,calculate the dot product: $\overrightarrow{A} \cdot \overrightarrow{B} = (2)(4) + (4)(2) + (4)(-4) = 8 + 8 - 16 = 0$.Since the dot product is $0$,we have $\cos 	heta = 0$.Therefore,$	heta = \cos^{-1}(0) = 90^o$.

Find the angle in $^o$ between two vectors $\overrightarrow{A} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and $\overrightarrow{B} = 4\hat{i} + 2\hat{j} - 4\hat{k}$.

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