Vector Algebra Question - CBSE

Question

Find the area of a parallelogram whose diagonals are determined by $\vec{d_1} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d_2} = \hat{i} - 3\hat{j} + 4\hat{k}$.

Accepted Answer

$\vec{d_1} imes \vec{d_2} = -2\hat{i} - 14\hat{j} - 10\hat{k} \Rightarrow |\vec{d_1} imes \vec{d_2}| = 10\sqrt{3}$. [1.0 Mark]
$ ext{Area} = \dfrac{1}{2}(10\sqrt{3}) = 5\sqrt{3}$ sq. units. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Calculating cross product of diagonals: 1.0 Mark
Evaluating area $= 5\sqrt{3}$ sq. units: 1.0 Mark

Question:

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