# What is the distance between the parallel planes ##3x + y – 4z = 2## and ##3x + y – 4z = 24##?

##d = 22/sqrt(26)##

Any plane can be represented as

##Pi-><< p-p_0, vec n >> = 0##

where ##p = {x,y,x}## is a generic plane point, ##p_0 in Pi## is a plane fixed point and ##vec n = {a,b,c}## a normal to ##Pi##. Here ##vec n={3,1,-4}##

The proposed planes can be represented as

##Pi_1->3(x-x_0)+(y-y_0)-4(z-z_0)=0## the ##p_0## coordinates are obtained according to

##3x_0+y_0-4z_0 =2##.

Choosing ##x_0=z_0=0## we obtain

##p_0 = {0,2,0}##

Analog reasonement for ##Pi_2##

then for ##p’_0##

##3x_0+y_0-4z_0 =24##. Choosing ##x_0=z_0=0## we obtain

##p’_0={0,24,0}##

The distance between ##Pi_1## and ##Pi_2## is given by the projection of ##p’_0-p_0## over the normalized normal vector ##(vec n)/abs(vec n)##

so ##d = abs(<< p’_0-p_0,(vec n)/abs(vec n)>> )##

or

##d= abs(<< {0,22,0},({3,1,-4})/sqrt(3^2+1+4^2))) =22/sqrt(26)##