Question

Find the distance between the parallel planes \(5 x-3 y-2 z=5\) and \(-5 x+3 y+2 z=7\).

Short answer

The distance between the parallel planes is \(\frac{2}{\sqrt{38}}\).

Step-by-step solution

Identify if Planes are Parallel

The given equations of the planes are \(5x - 3y - 2z = 5\) and \(-5x + 3y + 2z = 7\). To check if they are parallel, compare their normal vectors. The normal vector of the first plane is \(\langle 5, -3, -2 \rangle\) and for the second plane is \(\langle -5, 3, 2 \rangle\). The vectors are negatives of each other, indicating the planes are indeed parallel.

Convert One Plane Equation to Normal Form

Choose one plane to convert into the form \(Ax + By + Cz = D\) that simplifies calculations. We'll use the first plane, \(5x - 3y - 2z = 5\), which already is in this form. Its normal vector is \(\mathbf{n} = \langle 5, -3, -2 \rangle\) and is used as \(\langle A, B, C \rangle\).

Calculate Distance Between Parallel Planes

The formula to find the distance \(d\) between two parallel planes \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\) is given by \(d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}}\). For our planes, \(D_1 = 5\) and \(D_2 = -7\). Calculate the magnitude of the normal vector \(\|\mathbf{n}\| = \sqrt{5^2 + (-3)^2 + (-2)^2} = \sqrt{38}\).

Apply Distance Formula

Substitute into the distance formula: \(d = \frac{|7 - 5|}{\sqrt{38}} = \frac{2}{\sqrt{38}}\). Simplify if necessary: if exact values are required, the distance is \(\frac{1}{19}\) following rationalization, but we stop calculation here for clarity.

Key concepts

Parallel Planes

In the world of calculus and geometry, parallel planes are an interesting concept. Two planes in three-dimensional space are considered parallel if they do not intersect at any point. This means they have the same orientation and are an equal distance apart everywhere. However, a key property of parallel planes is that their normal vectors are either identical or negatives of each other. This relationship between normal vectors is what confirms the parallelism of the planes.
For example, consider the planes described by the equations \(5x - 3y - 2z = 5\) and \(-5x + 3y + 2z = 7\). By examining their normal vectors, \( \langle 5, -3, -2 \rangle \) and \( \langle -5, 3, 2 \rangle \), we observe that they are negatives of each other, suggesting that these planes are indeed parallel.

Distance Between Planes

Determining the distance between two parallel planes is important in various calculus problems. The distance refers to the shortest path perpendicular to both planes, essentially capturing how far the planes are apart in space. It is calculated using a straightforward formula that utilizes the plane equations and their corresponding normal vectors.
Given two parallel planes such as \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\), the distance \(d\) between them is found using:

• \( \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \)

This formula efficiently calculates the distance by evaluating the difference of the constant terms and the magnitude of the normal vector. In practical applications and exercises, this might involve geometric visualization or application to ensure a comprehensive understanding.

Normal Vector

The normal vector is a fundamental component in the context of plane equations, particularly for determining properties about planes like parallelism and distance. A normal vector is a perpendicular vector to the plane. It provides critical direction and orientation details for the plane.
For the plane equation \(Ax + By + Cz = D\), the normal vector is \(\langle A, B, C \rangle\). In examining the parallel planes \(5x - 3y - 2z = 5\) and \(-5x + 3y + 2z = 7\), the vectors are \(\langle 5, -3, -2 \rangle\) and \(\langle -5, 3, 2 \rangle\), respectively. Their negativity confirms their parallel nature.
The magnitude of the normal vector, \( \sqrt{A^2 + B^2 + C^2} \), is used in calculations involving distances, highlighting the role of the normal vector in geometric calculations.

Plane Equation

Understanding plane equations is crucial for solving many calculus problems. A plane equation in the form \(Ax + By + Cz = D\) describes a flat, two-dimensional surface in three-dimensional space. Each component, \(A, B,\) and \(C\), relates to the orientation and slope of the plane, with \(D\) affecting its position.
Planes can often be found in their standard form, simplifying mathematical manipulation. For instance, transforming plane equations or comparing their components directly aids in assessing parallelism and utilizing distance formulas effectively. In our example, transformations are not necessary as the given planes already serve their purpose for calculating distances and analyzing normal vectors directly from their given standard forms.

Distance Formula

The distance formula is a powerful tool in geometry, providing a method to find the shortest path between two points, lines, or in our case, planes. This formula is particularly applied when dealing with parallel planes to compute the shortest distance between them, utilizing the characteristics of the plane equations and their normal vectors.
For parallel planes \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\), the formula \(d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}}\) is employed.

• Here, \( |D_2 - D_1| \) is the absolute difference of the constant terms from the plane equations.
• \( \sqrt{A^2 + B^2 + C^2} \) represents the magnitude of the normal vector, ensuring precision in the calculation.

This formula aids in deriving accurate measurements of how far apart parallel planes are in a three-dimensional space. It is instrumental in both theoretical understanding and practical problem-solving scenarios.