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The area of the parallelogram whose diagonals are represented by the vectors $\bar{a}=3 \hat{i}-\hat{j}-2…
The area of the parallelogram whose diagonals are represented by the vectors $\bar{a}=3 \hat{i}-\hat{j}-2 \hat{k}$ and $\bar{b}=-\hat{i}+3 \hat{j}-3 \hat{k}$ is
$\sqrt{266}$ sq. units $\frac{1}{2} \sqrt{266}$ sq. units 266 sq. units 122 sq. units Solution
Let $\overrightarrow{\mathrm{d}_1}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{d}_2}=-\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$
$\therefore\left|\overrightarrow{\mathrm{d}_1}\right|=\sqrt{14}$ and $\left|\overrightarrow{\mathrm{d}_2}\right|=\sqrt{19}$
$\begin{aligned}
& \text { Also } \overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
3 & -1 & -2 \\
-1 & 3 & -3
\end{array}\right|=9 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+8 \hat{\mathrm{k}} \\
& \therefore\left|\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right|=\sqrt{81+121+64}=\sqrt{266} \\
& \left|\mathrm{~d}_1 \times \mathrm{d}_2\right|^2=\left|\mathrm{d}_1\right|^2\left|\mathrm{~d}_2\right|^2 \sin ^2 \theta \\
& \therefore 266=(14)(19) \sin ^2 \theta \\
& \therefore \sin ^2 \theta=1 \quad \Rightarrow \sin \theta=1 \quad \ldots\left[\because 0 < \theta \leq \frac{\pi}{2}\right]
\end{aligned}$
Area of parallelogram $=\frac{1}{2}\left|\mathrm{~d}_1\right|\left|\mathrm{d}_2\right| \sin \theta$
$=\frac{\sqrt{266}}{2}$
Asked in: MHT CET 2021 (22 Sep Shift 1)
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