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The area of the parallelogram whose sides are represented by the vectors $\hat{j} + 3 \hat{k} and \hat{i} + 2 \hat{j} - \hat{k}$ is

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$\sqrt{61} sq . unit$

$\sqrt{59} sq . unit$

$\sqrt{49} sq . unit$

$\sqrt{52} sq . unit$

answer is B.

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Detailed Solution

$\vec{A} = \hat{j} + 3 \hat{k}, \vec{B} = \hat{i} + 2 \hat{j} - \hat{k}$

$\vec{C} = \vec{A} \times \vec{B} = (\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 3 \\ 1 & 2 & - 1 \end{matrix}) = - 7 \hat{i} + 3 \hat{j} - \hat{k}$

Hence area = $| \vec{C} | = \sqrt{49 + 9 + 1} = \sqrt{59} sq . unit$

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