Vector Algebra Question - CBSE

Question

Area of a parallelogram whose adjacent sides are given by $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is:
(a) $\sqrt{42}$ sq. units
(b) $\sqrt{21}$ sq. units
(c) $42$ sq. units
(d) $2\sqrt{42}$ sq. units

Accepted Answer

$\vec{a} imes \vec{b} = 5\hat{i} + \hat{j} - 4\hat{k} \Rightarrow |\vec{a} imes \vec{b}| = \sqrt{42}$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Evaluating area $= \sqrt{42}$ sq. units: 1.0 Mark

Question:

Step-by-Step Solution

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