The distance of the point `(1, -2, 4)` from the plane passing through the point `(1,2,2) and ` perpendicular to the planes `x-y + 2z =3 and 2x - 2y+ z+ 12 =0,` is :

Options

A. `1/(sqrt(2))`
B. `2`
C. `sqrt(2)`
D. `2sqrt(2)`

Answer

D

Explanation

Let `vecn` be a vector normal to the required plane.
Then, its equation is
`(vecr-(hati+2hatj+2hatk)).vecn=0`…………i
This plane is perpendicular to the plane `x-y+2z=3` and `2x-2y+z+12=0` whose normals are `vecn_(1)=hati-hatj+2hatk` and `vecn_(2)=2hati-2hatj+hatk`
`:.vecn=vecn_(1)xxvecn_(2)=|(hati,hatj,hatk),(1,-1,2),(2,-2,1)|=3hati+3hatj+0hatk`
Substituting `vecn=3hati+3hatj` in i we obtain
`vecr.(3hati+3hatj)=9` or `vecr.(hati+hatj)=3`............ii
The distance of this plane from `veca=hati-2hatj+4hatk` is given by
`d=|((hati-2hatj+4hatk).(hati+hatj)-3)/(sqrt(1+1))|=2sqrt(2)`