Question

Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Answer 1

Here,

`vec (BC) = (hati + 5hatj + 5hatk) - (2hati + 3hatj + 5hatk)`

`= -hati + 2hatj`

`vec(BA) = (hati + hatj + 2hatk) - (2hati + 3hatj + 5hatk)`

`= -hati - 2hatj - 3hatk`

∴ `vec(BC) xxvec (BA) = abs((hati, hatj, hatk), (-1, 2, 0), (-1, -2, -3))`

`= (-6 + 0)hati - (3 + 0)hatj + (2 + 2)hatk`

`= -6hati - 3hatj + 4hatk`

So, `|vec(BC) xx vec( BA)| = sqrt(36 + 9 + 16)`

`= sqrt61`

∴ Area of `triangle ABC = 1/2 |vec(BC) xx vec(BA)|`

`= 1/2 (sqrt61)` sq. units.