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Question
The two adjacent sides of a parallelogram are `2hati - 4hatj + 5hatk` and `hati - 2hatj - 3hatk`. Find the unit vector parallel to its diagonal. Also, find its area.
Solution
The adjacent sides of a parallelogram are given by `veca = 2hati - 4hatj + 5hatk, vecb = hati - 2hatj - 3hatk`
Then the diagonal of a parallelogram is given by `veca + vecb`.
`veca + vecb = (2 + 1)hati + (-4 - 2)hatj + (5 - 3)hatk = 3hati - 6hatj + 2hatk`
Thus, the unit vector parallel to the diagonal is,
`(veca + vecb)/|veca + vecb| = ((3hati - 6hatj + 2hatk))/sqrt(3^2 + (-6)^2 + 2^2)`
`= ((3hati - 6hatj + 2hatk))/sqrt(9 + 36 + 4)`
`= 3/7hati- 6/7hatj + 2/7hatk`
Area of parallelogram ABCD
⇒ `|veca xx vecb| = |(hati, hatj, hatk), (2, -4, 5), (1, -2, -3)|`
`= hati(12 + 10) - hatj(-6 - 5) + hatk(-4 + 4)`
`= 22hati + 11hatj = 11(2hati + hatj)`
`= |veca xx vecb| = 11sqrt(2^2 + 1^2) `
`= 11sqrt5`
Therefore the area of the parallelogram is `11sqrt5` square units.
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