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If $2 \bar{i} + 3 \bar{j} - 4 \bar{k}$ and $- \bar{i} + 2 \bar{j} + \bar{k}$ are the two diagonals of a parallelogram then the area of the parallelogram in square units is

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$\frac{1}{2} \sqrt{170}$

$\sqrt{174}$

$\sqrt{\frac{87}{2}}$

$\frac{1}{4} \sqrt{174}$

answer is C.

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Detailed Solution

$\vec{a} = 2 i + 3 j - 4 k$

$\vec{b} = - i + 2 j + k$

area of parallelogram $= \frac{1}{2} |\bar{a} \times \bar{b}| = \sqrt{\frac{87}{2}}$

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