The equation of a plane which is perpendicular to `(2hati-3hatj+hatk)` and at a distance of 5 units from the origin is

Options

A. `2x-3y+z=5`
B. `2x-3y+z=5sqrt(14)`
C. `x/2-y/3+z/1=5`
D. `x/2-y/3+z/1=5/sqrt(14)`

Answer

b

Explanation

We have `|vecn|=sqrt(2^(2)+(-3)^(2)+1^(2))=sqrt(14)`.
`therefore vecr.hatn=5 rArr vecr.vecn/sqrt(14)=5`.
`rArr vecr.vecn=5sqrt(14)`
`therefore (xhati+yathj+zhatk).(2hati-3hatj+hatk)=5sqrt(14) rArr 2x-3y+z=5sqrt(14)`.