Find the vector equation of the plane which is at a distance of 5 units from the origin and which is normal to the vector `2hati + hatj + 2hatk`.

Explanation

Let `vecn = 2hati+ hatj+2 hatk` be a normal vector to the plane.
`therefore` Unit vector along the normal to the plane
`hatn =(veca)/(|veca|)`
`=(2hati + hatj + 2hatk)/(sqrt(2^(2)+1^(2)+2^(2)))`
`=(2hati+hatj+2hatk)/(3)`
`therefore `Equation of the required plane is
`vecr . hatn = p`
`rArr" "vecr.((2hati + hatj + 2hatk)/(3))=5`
`rArr" "vecr. (2hati+hatj+2hatk)=15`