The area of the parallelogram whose diagonals are \(3\hat i+\hat j-2\hat k\) and \(\hat i-3\hat j+4\hat k\) is
A. \(20\sqrt3\)
B. \(5\sqrt3\)
C. \(10\sqrt3\)
D. \(5\sqrt2\)
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Answer: B. \(5\sqrt3\)
\(\vec d_1\times\vec d_2=-2\hat i-14\hat j-10\hat k\), magnitude \(\sqrt{300}=10\sqrt3\). Area \(=\frac12|\vec d_1\times\vec d_2|=5\sqrt3\).
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· 23 Oct 2026
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