Question:
The area of a parallelogram with diagonals $\mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and $\mathbf{b} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ is
The area of a parallelogram with diagonals $\mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and $\mathbf{b} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ is
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Area of parallelogram with diagonals $\mathbf{d_1}$ and $\mathbf{d_2}$: Area $= \dfrac{1}{2}|\mathbf{d_1}\times\mathbf{d_2}|$.
Updated On: May 2, 2026
- $10\sqrt{3}$
- $5\sqrt{3}$
- $5\sqrt{3}$
- $\dfrac{5}{\sqrt{3}}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Area of parallelogram when diagonals are given $= \dfrac{1}{2}|\mathbf{d_1} \times \mathbf{d_2}|$.
Step 2: Detailed Explanation:
$\mathbf{a}\times\mathbf{b} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\3&1&-2\\1&-3&4\end{vmatrix} = \mathbf{i}(4-6)-\mathbf{j}(12+2)+\mathbf{k}(-9-1) = -2\mathbf{i}-14\mathbf{j}-10\mathbf{k}$.
$|\mathbf{a}\times\mathbf{b}| = \sqrt{4+196+100} = \sqrt{300} = 10\sqrt{3}$.
Area $= \dfrac{1}{2}\times10\sqrt{3} = 5\sqrt{3}$.
Step 3: Final Answer:
Area $= 5\sqrt{3}$.
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