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.

Answer

`x+y+z=9`

Explanation

Let the required equation of the plane be `vecr.hatn=p,` where `p=3sqrt(3)`. Let `hatn=(cosalpha)hati+(cosalpha)hatk`, where `alpha` is acute.
Then, `cos^(2)alpha+cos^(2)alpha+cos^(2)alpha=1 rArr 3cos^(2)alpha =1 rArr cos^(2)alpha=1/3 rArr cosalpha=1/sqrt(3)`.
`therefore` the required equation is
`vecr.(1/sqrt(3)hati+1/sqrt(3)hatj+1/sqrt(3)hatk)=3sqrt(3)`.
Hence, `vecr.(hati+hatj+hatk)=9 rArr (xhati+yhatj+zhatk).(hati+hatj+hatk)=9 rArr x+y+z=9`.