Find the equation of a plane which is at a distance of `3\ sqrt(3)` units from origin and the normal to which is equally inclined to the coordinate axes.

Options

A. `x+y+z=9`
B. `x+y-z=9`
C. `x-y-z=9`
D. `x-y+z=9`

Answer

A

Explanation

Since, normal to the plane is equally inclined to the coordinate axis.
Therefore `cos alpha=cos beta=cos gamma=1/(sqrt(3))`
So, the normal is `vecN=1/(sqrt(3))hati+1/(sqrt(3))hatj+1/(sqrt(3))hatk` and plane is at a distance of `3sqrt(3)` units from origin.
The equation of plane is `vecr.vecN=3sqrt(3) [ :' vecN=(vecN)/(|N|)]`
[since vector equation of the plane at a distance `p` from the origin is `vecr.vecN=p`]
`implies(xhati+yhatj+zhatk).((1/(sqrt(3))hati+1/(sqrt(3)hatj+1/(sqrt93))hatk))/1=3sqrt(3)`
`implies x/(sqrt(3))+y/(sqrt(3))+z/(sqrt(3))=3sqrt(3)`
`:. x+y+z=3sqrt(3).sqrt(3)=9`
So, the required equation fo plane is `x+y+z=9`.