Find the vector equation of the plane whose Cartesian equation is `5y+8=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
`5y=-8 rArr -5y=8 rArr 0.x-5y+0z=8` ………..(i)
Let `vecr` be the position vector of any point P(x,y,z) on plane (i). Then, the above eqution may be written as
`vecr.(0hati-5hatj+0hatk)=8 rArr vecr.vecn=8`, where `vecn=0hati-5hatj+0hat`k
`rArr vecr.(0hati-5hatj+0hatk) rArr vecr.vecn=8`, where `vecn=0hati-5hatj+0hatk`
`rArr vecr.(vecn)/(|vecn|)=8|(vecn)|`, where `|vecn|=sqrt(0^(2)+(-5)^(2)+0^(2))=5`
`rArr vecr.(0/5hati-5/5hat+0/5hat)=8/5 rArr vecr.(0hati-hatj+0hatk)=8/5`.
`therefore` d.c.'s of the normal to the plane are `0,-1,0` and its distance from the origin is `8/5` units.