Find the vector and equation of a plane which is at a distance of 10 units from origin and normal vector from origin to this plane is `2hati-hatj+2hatk`.

Explanation

Here,` vecn=2hati-hatj+2hatk` and `p = 10`
`:. hatn = (vecn)/(|vecn|)= (2hati-hatj+2hatk)/(sqrt(2^(2)+(-1)^(2)+2^(2)))=(2hati-hatj+2hatk)/(3)`
Equation of plane
`vecr.hatn=p`
`rArr vecr.((2hati-hatj+2hatk))/(3) = 10`
`rArr vecr.(2hati-hatj+2hatk) = 30`.