Question

Find the area of the parallelogram whose diagonals are `hati - 3hatj + hatk` and `hati + hatj + hatk`.

Answer 1

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