Question

Show that area of the parallelogram whose diagonals are given by `vec"a"` and `vec"b"` is `(|vec"a" xx vec"b"|)/2`. Also find the area of the parallelogram whose diagonals are `2hat"i" - hat"j" + hat"k"` and `hat"i" + 3hat"j" - hat"k"`.

Answer 1

Let ABCD be a parallelogram such that, 

`vec"AB" = vec"p"`

`vec"AD" = vec"q" = vec"BC"`

∴ By law of triangle, we get

`vec"AC" = vec"a" = vec"p" + vec"q"`  .....(i)

And `vec"BD" = vec"b" = -vec"p" + vec"q"` .....(ii)

Adding equation (i) and (ii) we get,

`vec"a" + vec"b" = 2vec"q"`

⇒ `vec"q" = ((vec"a" + vec"b")/2)`

Subtracting equation (ii) from equation (i) we get

`vec"a" - vec"b" = 2vec"p"`

⇒ `vec"p" = ((vec"a" - vec"b")/2)`

∴ `vec"p" xx vec"q" = 1/4(vec"a" + vec"b") xx (vec"a" - vec"b")`

= `1/4 (vec"a" xx vec"a" - vec"a" xx vec"b" + vec"b" xx vec"a" - vec"b" xx vec"b")`

= `1/4(-vec"a" xx vec"b" xx vec"b" xx vec"a")`  ......`[(because vec"a" xx vec"a" = 0),(vec"b" xx vec"b" = 0)]`

= `1/4(vec"a" xx vec"b" + vec"a" xx vec"b")`

= `1/4 * 2(vec"a" xx vec"b")`

= `|vec"a" xx vec"b"|/2`

So, the area of the parallelogram ABCD = `|vec"p" xx vec"q"| = 1/2|vec"a" xx vec"b"|`

Now area of parallelogram whose diagonals are `2hat"i" - hat"j" + hat"k"` and `hat"i" + 3hat"j" - hat"k"`

= `1/2|(2hat"i" - hat"j" + hat"k") xx (hat"i" + 3hat"j" - hat"k")|`

= `-|(hat"i", hat"j", hat"k"),(2, 1, 1),(1, 3, 1)|`

= `1/2 |hat"i"(1 - 3) - hat"j"(-2 - 1) + hat"k"(6 + 1)|`

= `1/2 - 2hat"i" + 3hat"j" + 7hat"k"|`

= `1/2 sqrt((-2)^2 + (3)^2 + (7)^2)`

= `1/2 sqrt(4 + 9 + 49)`

= `1/2 sqrt(62)` sq.units

Hence, the required area is `1/2 sqrt(62)` sq.units