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

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Answer 1 · 2016-07-14T18:51:34

#22/sqrt26#

The distance #d# btwn. two parallel planes, say

#Pi_1 : ax+by+cz+d_1=0#, &

#Pi_2 :ax+by+cz+d_2=0# is given by the formula :

#d=|d_1-d_2|/sqrt(a^2+b^2+c^2)#

In our case, #d=|-2-(-24)|/sqrt(3^2+1^2+(-4)^2)= 22/sqrt26#

Answer 2 · 2016-07-14T20:35:53

#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)#