Solving Vector Proof: Perpendicular Distance Between Two Parallel Planes

[Giuseppe]

hey guys, i am stuck on one more problem. Can anyone guide me onto the right path on how to start this?

Use this formula for the perpendicular distance between a point and a plane
D= |ax+by+cz-d| all over the square root of (a^2+b^2+c^2)

to show that the perpendicular distance D between the two parallel planes ax+by+cz+d1=0 and ax+by+cz+d2=0 is

D= |d1-d2| all over the square root of (a^2+b^2+c^2)

 

[AKG]

For a plane {(x, y, z) : ax + by + cz + d = 0} and a point (x', y', z'), you have that the distance from the point to the plane is:

$$D = \frac{|ax' + by' + cz' + d|}{\sqrt{a^2 + b^2 + c^2}}$$

The distance from one plane to another parallel plane is obviously the same as the distance from that plane to any single point on the other plane. Suppose (x₁, y₁, z₁) is on plane 1. That means that this point satisfies the equation:

ax₁ + by₁ + cz₁ + d₁ = 0

If we isolate d₁, we get:

d₁ = -(ax₁ + by₁ + cz₁)

-d₁ = ax₁ + by₁ + cz₁

Now the distance between plane 2 and plane 1 is the same as the distance between plane 2 and (x₁, y₁, z₁) since (x₁, y₁, z₁) is on plane 1. Using the given formula:

$$D = \frac{|ax_1 + by_1 + cz_1 + d_2|}{\sqrt{a^2 + b^2 + c^2}}$$

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

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

 

[Giuseppe]

i get that, but i don't get your last algebraic step. how did the -d1+d2 turn into d1-d2

 

[robphy]

i get that, but i don't get your last algebraic step. how did the -d1+d2 turn into d1-d2

The question should really be
"how did |-d1+d2| turn into |d1-d2|?"
The answer is, of course, "because |-A|=|A|".