Vertices of a triangle are `A(3, 1, 2), B(1, -1, 2)` and `C(2, 1, 1)`. Area of the triangle will be :

Options

A. `sqrt(2)"units"`
B. `2sqrt(2)" units"`
C. ` sqrt(3)"units"`
D. `2sqrt(3)"units"`

Answer

C

Explanation

`vecA = 3hati +hatj+ 2hatk " "veca = vecAB = (hati - hatj +2hatk)- (3hati + hatj + 2hatk)`
`vecB = hati -hatj+ 2hatk " "veca = - 2hati - 2hatj`
`vecC = 2hati -hatj+ hatk " "vecb = vecBC = (2hati + hatj + hatk) - (hati- hatj + 2hatk)`
Area = `=(1)/(2) |veca xx vecb|= (1)/(2) |{:(hati, hatj, hatk),(-2, -2, 0),(1, 2, -1):}|`
`=(1)/(2)[(2-0)hati-(2-0)hatj + (-4+2)hatk]`
`=(1)/(2)|2hati -2hatj- 2hatk|=(1)/(2) xx2sqrt(3)= sqrt(3) "unit"`