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Show that the area of the parallelogram having diagonals (3i + j - 2k) and (i -3j +4k ) is 5(3)^1/2.

$L e t \vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k} a n d \vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k} . T h e n, \Rightarrow \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & - 2 \\ 1 & - 3 & 4 \end{matrix} | = \hat{i} (4 - 6) - \hat{j} (12 + 2) + \hat{k} (- 9 - 1) = - 2 \hat{i} - 14 \hat{j} - 10 \hat{k} \Rightarrow | \vec{a} \times \vec{b} | = \sqrt{(- 2)^{2} + (- 14)^{2} + (- 10)^{2}} = \sqrt{300} ∴ A r e a o f p a r a l l e \log r a m = \frac{1}{2} | \vec{a} \times \vec{b} | = \frac{1}{2} \sqrt{300} {(u n i t)}^{2}$

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