The area of the parallelogram whose diagonals are <span class="math-tex">\(\rm \vec{a}= 3\hat{i} + \hat{j} - 2\hat{k}\)</span> and <span class="math-tex">\(\rm \vec{b}=\hat{i}-3\hat{j}+4\hat{k}\)</span> is:
When the diagonals of a parallelogram are \(\vec{a}\) and \(\vec{b}\), its area = \(\frac{1}{2}|\vec{a} \times \vec{b}|\).
\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(12 + 2) + \hat{k}(-9 - 1) = -2\hat{i} - 14\hat{j} - 10\hat{k}\)
\(|\vec{a} \times \vec{b}| = \sqrt{4 + 196 + 100} = \sqrt{300} = 10\sqrt{3}\)
Area = \(\frac{1}{2} \times 10\sqrt{3} = 5\sqrt{3}\)
Answer: (B)
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