Find the direction cosines of the normal to the plane is `(3x-6y+2z=7`.

Answer

`3/7,-6/7,2/7`

Explanation

Given equation in vector from is `vecr.(3hati-6hatj+2hatk)=7`.
`|3hati-6hatj+2hatk)|=sqrt(3^(2)+(-6)^(2)+2^(2))=sqrt(49)=7`.
`therefore vecr.(3/7hati-6/7hatj+2/7hatk)=1`.
`therefore` d.c.'s of the normal to the plane are `3/7,-6/7,2/7`.