The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines `r=(hati+hatj)+lambda(hati+2hatj-hatk)" and "r=(hati+hatj)+mu(-hati+hatj-2hatk)` is

Options

A. 3
B. `(1)/(3)`
C. `sqrt(3)`
D. `(1)/(sqrt(3))`

Answer

C

Explanation

Length of the perpendicular drawn from point `(x_(1), y_(1), z_(1))` to the plane `ax+by+cz+d=0`is
`d_(1)=(|ax_(1)+by_(1)+cz_(1)+d|)/(sqrt(a^(2)+b^(2)+c ^(2)))`
Given line vectors
`r=(hati+hatj)+lambda(hati+2hatj-hatk)" and "" "...(i)`
`r=(hati+hatj)+mu(-hati+hatj-2hatk)" "...(ii)`
Now, a vector perpendicular to the given vectors (i) and (ii) is
`n=|{:(" "hati,hatj," "hatk),(" "1,2,-1),(-1,1,-2):}|`
`=hati(-4+1)-hatj(-2-1)+hatk(1+2)`
`=-3hati+3hatj+3hatk`
`:.`The equation of plane containing given vecotrs (i) and (ii) is
`-3(x-1)+3(y-1)+3(z-0)=0`
`implies" "-3x+3y+3z=0`
`implies" "x-y-z=0" "...(iii)`
Now, the length of perpendicular drawn from the point (2, 1, 4) to the plane `x-y-z=0`, is
`d_(1)=(|2-1-4|)/(sqrt(1+1+1))`
`=(3)/(sqrt(3))=sqrt(3)`