The vector equation of a plane is `vecr.(3hati+2hatj-6hatk) = 56`. Convert it into normal form. Also find the length of perpendicular from origin and direction cosines of normal to the plane.

Explanation

`vecr.(3hati+2hatj-6hatk) = 56`
Here `vecn=3hati+2hatj-6hatk`
`rArr |vecn |=sqrt(9+4+36) = 7`
`:. vecr.((3hati+2hatj-6hatk))/(7) = (56)/(7)`
`rArr vecr.(3/7hati+2/7hatj-6/7hatk) = 8`
Which is the normal form of the plane.
Direction cosines of the perpendicular drawn from
` =3/7,2/7,(-6)/(7)`
origin and length of perpendicular from origin `= 8` units.