Find the vectors equation of a plane whose Cartesian equation is `2x-3y+4z+6=0`. Find the direction cosines of the normal to the plane and its distance from the origin.

Explanation

The given equation of the plane is
`2x-3y+4z=6 rArr -2x+3y-4z=6`
If r is the position vector of any point P(x,y,z) on plane (i), then we may write it as
`(xhati+yhatj+zhatk).(-2hati+3hatj)-4hatk)=6`
`rArr hatr.(-2hati+3hatj-4hatk)=6`
`rArr vecr.vecn=6`, where vecn=(-2hati+3hatj-4hatk)`
`rArr vecr.vecn/(|vecn|)=6/(|vecn|)`, where `|vecn|=sqrt((-2)^(2)+3^(2)+(-4)^(2))=sqrt(29)`
`rArr vecr.(-2/sqrt(29)hati+3/sqrt(29)hati-4/sqrt(29)hatk)=6/sqrt(29)`, and its distance from the origin is `6/sqrt(29)`.