For each of the following planes, find the direction cosines of the normal to the plane and the distance of the plane from the origin:
i) `2x+3y-z=5`, ii) z=3 , iii) `3y+5=0`.

Answer

i) `(2/sqrt(14),3/sqrt(14),-1/sqrt(14)), 5/sqrt(14)`, ii) (0,0,1),3 , iii) `(0,-1,0),5/3`

Explanation

i) Given equation in vector form is `vecr.(2hati+3hatj-hatk)=5`.
`|2hati+3hatj-hatk|=sqrt(2^(2)+3^(2)+(-1)^(2))=sqrt(14)`.
`therefore vecr.(2/sqrt(14)hati+3/sqrt(14)hatj-1/sqrt(14)hatk)=5/sqrt(14)`.
d.c's of `hatn` are `2/sqrt(14),3/sqrt(14),-1/sqrt(14)` and `p=5/sqrt(14)`.
ii) Given equation in vector from is `vecr.(0hati+0hatj+1hatk)=3`.
This is `vecr.hatn=p`.
D.c.'s of `hatn` are 0,0,1 and p=3.
iii) Given equation in vector form is `vecr.(0hati-1.hatj+0.hatk)=5/3 [therefore -y=5/3]`.
Clearly, `|0hati-1.hatj+0.hatk|=sqrt(0^(2)+(-1)^(2)+0^(2))=1`.
`therefore vecr.hatn=p`, where `hatn=(0hati-1.hatj+0hatk)` and `p=5/3`.
D.c.'s of `hatn` are `0,-1,0` and `p=5/3`.