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Question
The diagonals of a parallelogram are given by `veca = 2hati - hatj + hatk and vecb = hati + 3hatj - hatk`. Find the area of the parallelogram.
Sum
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Solution
Diagonals of a parallelogram are
`veca = 2hati - hatj + hatk`
`vecb = hati + 3hatj - hatk`
`|veca xx vecb| = |(2hati - hatj + hatk) xx (hati + 3hatj - hatk)|`
= `|(hati, hatj, hatk), (2, -1, 1), (1, 3, -1)|`
= `|hati(1 - 3) - hatj(-2-1) + hatk(6 + 1)|`
=`|-2hati + 3hatj + 7hatk|`
Since the given vectors are diagonals of the parallelogram, the area is given by the formula:
Now, the area of the parallelogram = `1/2 xx |veca xx vecb|`
= `1/2 sqrt((-2)^2 + (3)^2 + (7)^2)`
= `1/2 sqrt(4 + 9 + 49)`
= `1/2 sqrt(62)` sq. units
= `sqrt(62)/2` sq. units
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