Question

find the distance between the planes r.(2i-3j+6k)-4=0 and r.(6i-9j+18k)+30=0

Answer 1

first we convert this

equation in Cartesian
form

(xi+yj+zk).(2i-3j+6k)-4=0

2x-3y+6z=4 -------------(1)

(xi+yj+zk)(6i-9j+18k)+30=0

6x-9y+18z=-30

2x-3y+6z=-10 -------------(2)

now you look (1) and (2) equation .both

are parellel planes equation in Cartesian form

now use formula distance between two

parellel plane .

distance=modulus (-10-4)/root(2^2+3^2+6^2)

=14/root(4+9+36)=14/root (49)=14/7

=2 unit is answer