If the vectors vec{BC} = 2hat{i} + hat{j} + h | vedclass

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(B) Let $\vec{a} = \vec{BC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \vec{CD} = \hat{i} + 2\hat{j} - 2\hat{k}$.In parallelogram $ABCD$, the side

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$\frac{-10\sqrt{2}}{3}$

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(B) Let $\vec{a} = \vec{BC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \vec{CD} = \hat{i} + 2\hat{j} - 2\hat{k}$.In parallelogram $ABCD$, the sides are $\vec{BC}$ and $\vec{CD}$. The diagonals are $\vec{AC} = \vec{BC} + \vec{CD} = (2\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} - 2\hat{k}) = 3\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{BD} = \vec{BC} - \vec{CD} = (2\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} - 2\hat{k}) = \hat{i} - \hat{j} + 3\hat{k}$.Let $\vec{d_1} = \vec{AC} = 3\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{d_2} = \vec{BD} = \hat{i} - \hat{j} + 3\hat{k}$.$|\vec{d_1}| = \sqrt{3^2 + 3^2 + (-1)^2} = \sqrt{9 + 9 + 1} = \sqrt{19}$.$|\vec{d_2}| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{1 + 1 + 9} = \sqrt{11}$.$\vec{d_1} \cdot \vec{d_2} = (3)(1) + (3)(-1) + (-1)(3) = 3 - 3 - 3 = -3$.$\cos 	heta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} = \frac{-3}{\sqrt{19} \cdot \sqrt{11}} = \frac{-3}{\sqrt{209}}$.Since $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{9}{209} = \frac{200}{209}$, we have $\sin 	heta = \frac{\sqrt{200}}{\sqrt{209}} = \frac{10\sqrt{2}}{\sqrt{209}}$.Therefore, $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{10\sqrt{2} / \sqrt{209}}{-3 / \sqrt{209}} = \frac{-10\sqrt{2}}{3}$.

If the vectors $\vec{BC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{CD} = \hat{i} + 2\hat{j} - 2\hat{k}$ represent two adjacent sides of a parallelogram $ABCD$ and $\theta$ is the angle between its diagonals $\vec{AC}$ and $\vec{BD}$, then $\tan \theta =$

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