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प्रश्न
Find the area of the parallelogram whose diagonals are `hati - 3hatj + hatk` and `hati + hatj + hatk`.
उत्तर
Diagonals of a parallelogram
`vecd_1 = hati - 3hatj + hatk`,
`vecd_2 = hati + hatj + hatk`
Now `vecd_1 xx vecd_2 = |(hati, hatj, hatk),(1, -3, 1),(1, 1, 1)|`
= `(-3 - 1)hati - (1 - 1)hatj + (1 + 3)hatk`
= `-4hati - 0hatj + 4hatk`
∵ Area of a parallelogram
= `1/2 |vecd_1 xx vecd_2|`
= `1/2 sqrt((-4)^2 + 0 + 4^2)`
= `1/2 xx sqrt(16 + 16)`
= `1/2 xx sqrt(32)`
= `1/2 xx 4sqrt(2)`
= `2sqrt(2)` sq.units
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