Answer 1

The vertices of the triangle ABC are A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1).

`vec(AB) = vec(OB) - vec(OA) = hati- 3hatj + hatk`

`vec(AC) = vec(OC) - vec(OA) = 3hati + 3hatj - 4hatk`

Area of the ∆ABC

`=1/2 |vec(AB) xx vec(AC)|`

`= 1/2|(hati,hatj,hatk),(1,-3,1),(3,3,-4)|`

`= 1/2|9hati + 7hatj +12hatk|`

`=1/2 sqrt(81+49+144)`

= `sqrt(274)/2`square units