Solving Vector Proof: Perpendicular Distance Between Two Parallel Planes

In summary, to find the perpendicular distance between two parallel planes, use the formula D= |d1-d2| all over the square root of (a^2+b^2+c^2) where d1 and d2 are the constants in the equations of the planes and a, b, and c are the coefficients of the variables. This formula can be derived by finding the distance between a point on one plane and the other plane, and then using the general formula for distance between a point and a plane.
  • #1
Giuseppe
42
0
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)
 
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  • #2
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:

[tex]D = \frac{|ax' + by' + cz' + d|}{\sqrt{a^2 + b^2 + c^2}}[/tex]

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:

[tex]D = \frac{|ax_1 + by_1 + cz_1 + d_2|}{\sqrt{a^2 + b^2 + c^2}}[/tex]

[tex]D = \frac{|-d_1 + d_2|}{\sqrt{a^2 + b^2 + c^2}}[/tex]

[tex]D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}[/tex]
 
  • #3
i get that, but i don't get your last algebraic step. how did the -d1+d2 turn into d1-d2
 
  • #4
Giuseppe said:
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|".
 

1. What is the formula for finding the perpendicular distance between two parallel planes?

The formula for finding the perpendicular distance between two parallel planes is given by d = |Ax + By + Cz + D| / √(A^2 + B^2 + C^2), where A, B, and C are the coefficients of the normal vector to the planes and D is the distance between the planes.

2. How do you determine if two planes are parallel?

Two planes are parallel if their normal vectors are parallel, meaning they have the same direction. This can be determined by comparing the coefficients of the normal vectors. If the coefficients are proportional, the planes are parallel.

3. Can the perpendicular distance between two parallel planes be negative?

No, the perpendicular distance between two parallel planes is always positive. This distance represents the shortest distance between the planes, so it cannot be negative.

4. Is it necessary for the planes to be parallel in order to find the perpendicular distance between them?

Yes, the planes must be parallel in order to find the perpendicular distance between them. If the planes are not parallel, there is no unique shortest distance between them.

5. Can the perpendicular distance between two parallel planes be 0?

Yes, the perpendicular distance between two parallel planes can be 0 if they are the same plane. In this case, the planes are not only parallel, but also coincident, meaning they have the same equation and therefore the same distance between them.

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