Find the vector equation of a plane which is at a distance of 8 units from the origin and which is normal to the vector `2 hat i+ hat j+2 hat kdot`

Options

A. `vecr.(2hati+hatj+2hatk)=8`
B. `vecr.(2hati+hatj+2hatk)=24`
C. `vecr.(2hati+hatj+2hatk)=4`
D. none of these

Answer

B

Explanation

Here `d=8` and `vecn=2hati+hatj+2hatk`
`:. hatn=(vecn)/(|vecn|)=(2hati+hatj+2hatk)/(sqrt(4+1+4))=2/3hati+1/3hatj+2/3hatk`
Hence the required equation of the plane is
`vecr.(2/3hati+1/3hatj+2/3hatk)=8impliesvecr.(2hati+hatj+2hatk)=24`