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If the vectors $\overline{B C}=2 \bar{i}+\bar{j}+\bar{k}$ and $\overline{C D}=\bar{i}+2 \bar{j}-2 \bar{k}$ represent two adjacent sides of a parallelogram $\mathrm{ABCD}$ and $\theta$ is the angle between its diagonals $\overline{A C}$ and $\overline{B D}$ then $\tan \theta=$
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The correct answer is:
$\frac{-10 \sqrt{2}}{3}$
The diagonals of parallelogram are represented by $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$ where $\vec{a}, \vec{b}$ are adjacent sides of parallelogram
$\begin{aligned} & \therefore \quad \overrightarrow{d_1}=\overrightarrow{B C}+\overrightarrow{C D}=3 \hat{i}+3 \hat{j}-\hat{k} \\ & \therefore \quad \cos \theta=\frac{\overrightarrow{d_1} \cdot \overrightarrow{d_2}}{\left|\overrightarrow{d_1}\right| \cdot\left|\overrightarrow{d_2}\right|}=\frac{3-3-3}{\sqrt{19} \cdot \sqrt{11}}=\frac{-3}{\sqrt{209}} \\ & \therefore \quad \sin \theta=\frac{\sqrt{200}}{\sqrt{209}} \Rightarrow \tan \theta=\frac{-10 \sqrt{2}}{3} .\end{aligned}$
$\begin{aligned} & \therefore \quad \overrightarrow{d_1}=\overrightarrow{B C}+\overrightarrow{C D}=3 \hat{i}+3 \hat{j}-\hat{k} \\ & \therefore \quad \cos \theta=\frac{\overrightarrow{d_1} \cdot \overrightarrow{d_2}}{\left|\overrightarrow{d_1}\right| \cdot\left|\overrightarrow{d_2}\right|}=\frac{3-3-3}{\sqrt{19} \cdot \sqrt{11}}=\frac{-3}{\sqrt{209}} \\ & \therefore \quad \sin \theta=\frac{\sqrt{200}}{\sqrt{209}} \Rightarrow \tan \theta=\frac{-10 \sqrt{2}}{3} .\end{aligned}$
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