CBSE 2025 · Region 1 · Set 1 · Q23 · 2 marks
The diagonals of a parallelogram are given by $\displaystyle \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\displaystyle \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$. Find the area of the parallelogram.
Marking-scheme solution
\[\begin{aligned}
& \vec{\mathrm{a}} \times \vec{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & -1 & 1 \\
1 & 3 & -1
\end{array}\right|=-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\
& \text { Area of parallelogram }=\frac{1}{2}|\vec{\mathrm{a}} \times \vec{\mathrm{b}}| \\
& =\frac{1}{2} \sqrt{(-2)^{2}+3^{2}+7^{2}}=\frac{\sqrt{62}}{2}
\end{aligned}
\]
Vector AlgebraProduct of Two VectorsApplyvery_short_answermedium
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