Let position vectors of point A,B and C of triangle ABC represents be `hati+hatj+2hatk, hati+2hatj+hatk` and `2hati+hatj+hatk`. Let `l_(1)+l_(2)` and `l_(3)` be the length of perpendicular drawn from the orthocenter 'O' on the sides AB, BC and CA, then `(l_(1)+l_(2)+l_(3))` equals

Options

A. `(2)/(sqrt(6))`
B. `(3)/(sqrt(6))`
C. `(sqrt(6))/(2)`
D. `(sqrt(6))/(3)`.

Answer

C

Explanation

Clearly, triangle formed by the given points `hati+hatj+2hatk`,
`hati+2hatj+hatk and 2hati+hatj+hatk` is equilateral as `AB=BC=AC=sqrt(2)`.
`therefore`Distance of orthocentre 'O' from the sides is equal to inradius of the triangle.
`thereforel_(1)=l_(2)=l_(3)=`inradius`=r=(Delta)/(s)=((sqrt(3))/(4)(sqrt(2))^(2))/((3)/(2)(sqrt(2)))=(1)/(sqrt(6))`
`implies(l_(1)+l_(2)+l_(3))=(3)/(sqrt(6))=(sqrt(6))/(2)`