Problem 52 - Essential Calculus Early Transcendentals

Question

\(51-52=\) Find the distance between the given parallel planes. $$6 z=4 y-2 x, \quad 9 z=1-3 x+6 y$$

Step-by-Step Solution

Answer

The distance between the planes is \( \frac{\sqrt{14}}{42} \).

Step 1: Understanding the Plane Equations

The given equations of the planes are:1. For the first plane, rearrange the equation to standard form: \( -2x + 4y - 6z = 0 \)2. For the second plane, rearrange it to: \( -3x + 6y - 9z = 1 \).

Step 2: Check Parallel Planes

To check if the planes are parallel, compare their normal vectors.The normal vector for the first plane is \( \langle -2, 4, -6 \rangle \), and for the second plane it is \( \langle -3, 6, -9 \rangle \).These vectors are proportional, confirming that these planes are parallel.

Step 3: Use Distance Formula for Parallel Planes

Distance between two parallel planes \( ax + by + cz = d_1 \) and \( ax + by + cz = d_2 \) is given by \[ \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \].Rewrite each plane's equation in terms of constant: first, \(-2x + 4y - 6z = 0\), second \(-3x + 6y - 9z = 1 \).

Step 4: Calculate Distance

Using the formula, the normal vector components are proportional: \( \langle -2, 4, -6 \rangle \) and \( \langle -3, 6, -9 \rangle \) can be simplified.Let's express the second plane like the first: divide by -3, \(-x + 2y - 3z = \frac{1}{3}\).Distance is \( \frac{|\frac{1}{3} - 0|}{\sqrt{(-1)^2 + 2^2 + (-3)^2}} \).This simplifies to \( \frac{1}{3} \div \sqrt{1 + 4 + 9} = \frac{1}{3} \div \sqrt{14} = \frac{1}{3\sqrt{14}} \).

Step 5: Simplify Distance

Rationalize the denominator: multiply top and bottom by \( \sqrt{14} \), \[ \frac{1 \times \sqrt{14}}{3\sqrt{14} \times \sqrt{14}} = \frac{\sqrt{14}}{42} \].This is the final simplified distance.

Key Concepts

Normal Vectors

When dealing with planes in three-dimensional geometry, normal vectors play a crucial role. A normal vector is a vector that is perpendicular to a surface. For a plane given by an equation of the form \( ax + by + cz = d \), the normal vector is \( \langle a, b, c \rangle \). This vector defines the orientation of the plane.

In this exercise, we had two plane equations:

• \(-2x + 4y - 6z = 0\)
• \(-3x + 6y - 9z = 1\)

The normal vectors extracted from these equations are \( \langle -2, 4, -6 \rangle \) and \( \langle -3, 6, -9 \rangle \), respectively.

To determine if two planes are parallel, we check if their normal vectors are proportional. In this case, \( \langle -3, 6, -9 \rangle \) is a scalar multiple of \( \langle -2, 4, -6 \rangle \), confirming that the planes are indeed parallel. Understanding normal vectors is critical in solving problems involving planes, as it helps define their position and orientation.

Equation of Planes

The equation of a plane in three-dimensional space is a mathematical way to represent that plane using coordinates. Typically, plane equations are written as \( ax + by + cz = d \). Here, \( x, y, \text{ and } z \) are coordinates of any point on the plane, and \( a, b, \text{ and } c \) are coefficients that correspond to the plane's normal vector.

Planes can be rearranged into the standard format if needed. For instance, the given planes \( 6z = 4y - 2x \) and \( 9z = 1 - 3x + 6y \) were rearranged as:

• First plane: \(-2x + 4y - 6z = 0\)
• Second plane: \(-3x + 6y - 9z = 1\)

By comparing the forms, it's easier to identify the coefficients of the normal vector and understand the geometric relations between planes. The equation of a plane is fundamental in deducing geometric properties, like parallelism or orthogonality, based on the normal vectors.

Distance Formula

To compute the distance between two parallel planes, we use a special distance formula. Once we have the equations of the planes as \( ax + by + cz = d_1 \) and \( ax + by + cz = d_2 \), the formula is:\[ \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \]

This formula helps find the shortest distance between two parallel planes by using the normal vector components \( a, b, \text{ and } c \). Since the planes are parallel, their normal vector remains constant, simplifying the computations.

For our example, first, simplify the normal vector if possible. We adjusted the second plane equation to \(-x + 2y - 3z = \frac{1}{3}\). By applying the formula, the distance is \[ \frac{\left| \frac{1}{3} - 0 \right|}{\sqrt{(-1)^2 + 2^2 + (-3)^2}} = \frac{\frac{1}{3}}{\sqrt{14}} \]

Finally, simplifying further and rationalizing the denominator gives us the final distance as \[ \frac{\sqrt{14}}{42} \]. Understanding this formula is key in geometry, as it allows us to measure how far apart two planes are in space.