Question

Find the distance between the planes 3x+y-4z=2 and 3x+y-4z=24

Answer 1

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


you helped this method.

Answer 2

Answer:

The distance between two planes is $d=\frac{22}{\sqrt{26}}$

Step-by-step explanation:

Given : The planes 3x+y-4z=2 and 3x+y-4z=24

To find : The distance between the planes?

Solution :

The distance between two planes is

If planes are in form

$P_1=ax+by+cz+d_1=0$

$P_2=ax+by+cz+d_2=0$

Then, formula is

$d=\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$

On comparing with the given planes,

a=3 , b=1, c=-4, $d_1=-2,d_2=-24$

Substitute in the formula,

$d=\frac{|-2-(-24)|}{\sqrt{3^2+1^2+(-4)^2}}$

$d=\frac{|-22|}{\sqrt{9+1+16}}$

$d=\frac{22}{\sqrt{26}}$

Therefore, The distance between two planes is $d=\frac{22}{\sqrt{26}}$