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Find the area of the parallelogram whose diagonals are vectors 3hati+hatj-2hatk and hati-3hatj+4hatk. |
Answer» Solution :Given that the DIAGONAL of a PARALLELOGRAM are `veca = 3hati+hatj-2hatk, vecb = hati-3hatj+4hatk` `|vecaxxvecb| = SQRT(4+196+100) = sqrt(300) = 10sqrt3` therefore AREA of the parallelogram = `1/2 xx10sqrt3 = 5SQRT3 sq. units. |
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