If the vectors of a trianlge are `A(hati+hatj+2hatk), B(3hati-hatj+2hati) and C(2hati-hatj+hatk)` the area of triangle is

Options

A. `2sqrt3` sq units
B. `sqrt3` sq units
C. `2sqrt3` sq units
D. 3 sq units

Answer

D

Explanation

Now , `AB=(3hat(i)-hat(j)+2hat(k))-(hat(i)+hat(j)+2hat(k))=2hat(i)-2hat(j)`
and `AC=(2hat(i)-hat(j)+hat(k))-(hat(i)+hat(j)+2hat(k))=hat(i)-2hat(j)-hat(k)`
`(ABxxAC)=|{:(hat(i),hat(j),hat(k)),(2,-2,0),(1,-2,-1):}|`
`=hat(i)(2-0)-hat(j)(-2-0)+hat(k)(-4+2)`
`=2hat(i)+2hat(j)+2hat(k)`
`therefore` Area of triangle `AB=(1)/(2)|ABxxAC|`
`=(1)/(2)sqrt((2)^2+(2)^2+(-2)^2)`
`=(2)/(2)sqrt(1+1+1)=sqrt(3)` sq units.