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JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 84

The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8i^−6j^8\hat i-6\hat j and 3i^+4j^−12k^3\hat i+4\hat j-12\hat k, is :

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Answer and worked solution

Official answer: B

Area =12∣d⃗1×d⃗2∣=\frac12|\vec d_1\times\vec d_2| with (8,−6,0)×(3,4,−12)=(72,96,50)(8,-6,0)\times(3,4,-12)=(72,96,50), of length 130. The area is 65.

Topics:Cross product
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