The vectors vec{AB} = 3hat{i} - 2hat{j} + 2ha | vedclass

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(D) Let the adjacent sides of the parallelogram be $\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$.The diagonals of th

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(D) Let the adjacent sides of the parallelogram be $\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$.The diagonals of the parallelogram are $\vec{d_1} = \vec{AC} = \vec{AB} + \vec{BC} = (3\hat{i} - 2\hat{j} + 2\hat{k}) + (\hat{i} - 2\hat{k}) = 4\hat{i} - 2\hat{j}$.The other diagonal is $\vec{d_2} = \vec{BD} = \vec{BC} - \vec{AB} = (\hat{i} - 2\hat{k}) - (3\hat{i} - 2\hat{j} + 2\hat{k}) = -2\hat{i} + 2\hat{j} - 4\hat{k}$.Let $	heta$ be the angle between the diagonals $\vec{d_1}$ and $\vec{d_2}$.$\cos 	heta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} = \frac{(4\hat{i} - 2\hat{j} + 0\hat{k}) \cdot (-2\hat{i} + 2\hat{j} - 4\hat{k})}{\sqrt{4^2 + (-2)^2 + 0^2} \sqrt{(-2)^2 + 2^2 + (-4)^2}}$$\cos 	heta = \frac{(4)(-2) + (-2)(2) + (0)(-4)}{\sqrt{16 + 4} \sqrt{4 + 4 + 16}} = \frac{-8 - 4}{\sqrt{20} \sqrt{24}} = \frac{-12}{\sqrt{480}} = \frac{-12}{4\sqrt{30}} = -\frac{3}{\sqrt{30}} = -\sqrt{\frac{9}{30}} = -\sqrt{\frac{3}{10}}$.Thus,$	heta = \cos^{-1}\left(-\sqrt{\frac{3}{10}}ight)$.Since this value is not among the options,the correct answer is $D$.

The vectors $\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$ are the adjacent sides of a parallelogram. The angle between its diagonals is

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