Distance between the two planes `2x+3y +4z =4 and 4x+6y +8z=12` is

Options

A. 2 units
B. 4 units
C. 8 units
D. `2/(sqrt(29))` units

Answer

D

Explanation

Given planes are
`2x+3y+4z=4"……."(1)`
and `4x+6y+8z=12 rArr 2x+3y+4z=6"……."(2)`
Clearly two planes are parallel .
We know that the distance between the planes `ax+by+cz=d_(1)` and `ax+by+cz=d_(2)` is
`d=|(d_(2)-d_(1))/(sqrt(a^(2)+b^(2)+c^(2)))|rArrd=|(6-4)/(sqrt(2^(2)+3^(2)+4^(2)))|`
`rArr d = (2)/(sqrt(4+9+16P)) = 2/(sqrt(29))` unit