Distance Between Two Parallel Planes (Vector Form)

Understanding the distance between two parallel planes in vector form is an important concept in 3D geometry, particularly in fields such as physics, engineering, and computer graphics. In this article, we will explore the concept in depth, including formulas and examples.

Parallel Planes

Before we delve into the distance between them, let's define parallel planes. Two planes are said to be parallel if they do not intersect, no matter how far they are extended. In vector form, two planes are parallel if their normal vectors are scalar multiples of each other.

Equation of a Plane

The general equation of a plane in 3D space can be given in vector form as:

$$ \vec{r} \cdot \vec{n} = d $$

where:

• $\vec{r}$ is the position vector of any point on the plane,
• $\vec{n}$ is the normal vector to the plane,
• $d$ is the scalar representing the plane's distance from the origin along the normal vector.

Distance Between Two Parallel Planes

The distance between two parallel planes can be found using the following formula:

$$ D = \frac{|d_2 - d_1|}{|\vec{n}|} $$

where:

• $D$ is the distance between the two planes,
• $d_1$ and $d_2$ are the scalar distances from the origin to the first and second planes, respectively,
• $\vec{n}$ is the normal vector to the planes (since they are parallel, both planes have the same normal vector).

Steps to Calculate the Distance

1. Find the normal vector $\vec{n}$ of the planes.
2. Obtain the scalar distances $d_1$ and $d_2$ from the plane equations.
3. Substitute $\vec{n}$, $d_1$, and $d_2$ into the distance formula.

Example

Let's consider two parallel planes with the following equations:

Plane 1: $2x - 3y + z = 5$
Plane 2: $4x - 6y + 2z = 14$

The normal vector $\vec{n}$ for both planes is the same and can be represented as $\vec{n} = \langle 2, -3, 1 \rangle$.

The scalar distances from the origin are $d_1 = 5$ and $d_2 = 14$ for Plane 1 and Plane 2, respectively.

Now, let's calculate the distance between these two planes:

$$ D = \frac{|14 - 5|}{\sqrt{2^2 + (-3)^2 + 1^2}} = \frac{|9|}{\sqrt{4 + 9 + 1}} = \frac{9}{\sqrt{14}} = \frac{9}{\sqrt{14}} \approx 2.40 $$

So, the distance between the two parallel planes is approximately 2.40 units.

Table of Differences and Important Points

Feature	Description
Normal Vector	The normal vector $\vec{n}$ is the same for both parallel planes.
Scalar Distance	The scalar distances $d_1$ and $d_2$ are unique to each plane and represent the distance from the origin to the plane along the normal vector.
Distance Formula	The distance $D$ between two parallel planes is given by the formula $D = \frac{
Non-intersecting	Parallel planes never intersect, which is why the distance between them is constant.

Conclusion

The distance between two parallel planes is a straightforward calculation once you have the normal vector and the scalar distances from the origin to each plane. This concept is widely used in various applications, and understanding it is crucial for solving problems in 3D geometry.