The area of the parallelogram having diagonals a = 3i+ j - 2k and b= i - 3j + 4k is

10 \sqrt{3}

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5 \sqrt{3}

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Solution

The correct option is D $5 \sqrt{3}$
Here $a = 3 i + j - 2 k, b = i - 3 j + 4 k$
Thus $a \times b = ∣ ∣ ∣ ∣ \begin{matrix} i & j & k 3 & 1 & - 2 1 & - 3 & 4 \end{matrix} ∣ ∣ ∣ ∣ = - 2 i - 14 j - 10 k = - 2 (i + 7 j + 5 k)$
Hence area of parallelogram $= \frac{1}{2} | a \times b | = 5 \sqrt{3}$

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