Problem 45 - University Calculus: Early Transcendentals

Question

Find the distance from the plane \(x+2 y+6 z=1\) to the plane \(x+2 y+6 z=10\)

Step-by-Step Solution

Answer

The distance is \( \frac{9}{\sqrt{41}} \) or approximately 1.41.

Step 1: Identify the Normal Vector

Both planes, the first one given by the equation \(x + 2y + 6z = 1\) and the second one \(x + 2y + 6z = 10\), share the same normal vector. The normal vector to these planes can be identified from the coefficients of \(x\), \(y\), and \(z\), thus it is \( \mathbf{n} = \langle 1, 2, 6 \rangle \).

Step 2: Calculate the Perpendicular Distance Formula

The distance \(d\) between two parallel planes, given the same normal vector \( \mathbf{n} = \langle a, b, c \rangle \), and equations \(ax + by + cz = d_1\) and \(ax + by + cz = d_2\) is calculated using the formula: \[ d = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \] Here, \(d_1 = 1\) and \(d_2 = 10\).

Step 3: Substitute the Values

Substitute the values of \(d_1\), \(d_2\), and the components of the normal vector into the formula: \[ d = \frac{|10 - 1|}{\sqrt{1^2 + 2^2 + 6^2}} = \frac{9}{\sqrt{1 + 4 + 36}} = \frac{9}{\sqrt{41}} \].

Step 4: Simplify the Expression

Calculate \(\sqrt{41}\) and simplify the expression: \( \sqrt{41} \approx 6.4 \). Therefore, \( d = \frac{9}{6.4} \approx 1.40625 \). The exact distance can be left as \( \frac{9}{\sqrt{41}} \), or it can be approximated to 3 significant figures as 1.41.

Key Concepts

Normal Vector

When it comes to understanding the geometry of planes in three-dimensional space, the concept of a **normal vector** is pivotal. The normal vector is a vector that is perpendicular to the surface of the plane. For the plane equation of the form:
\[ ax + by + cz = d \]The normal vector can be easily identified by the coefficients of the variables, which are \(a, b,\) and \(c\). Essentially, the normal vector \(\mathbf{n}\) is given by \( \langle a, b, c \rangle \).
Understanding normal vectors helps in determining if two planes are parallel. If two planes share the same normal vector, they are parallel. This concept was used in the original exercise, where both plane equations had the same normal vector \( \langle 1, 2, 6 \rangle \). This indicates that the planes are indeed parallel.

• Identification of normal vectors is key for determining the orientation of a plane.
• Parallel planes share identical normal vectors.

Perpendicular Distance Formula

Calculating the distance between two parallel planes relies on the **perpendicular distance formula**. This formula finds the shortest path connecting these planes, which is a line perpendicular to both.
For planes parallel to each other and given by the equations:
\[ ax + by + cz = d_1 \] \[ ax + by + cz = d_2 \]The formula to find the distance \(d\) between these planes is:
\[d = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}\]The numerator \( |d_2 - d_1| \) represents the absolute difference between the constant terms of the two plane equations. The denominator \( \sqrt{a^2 + b^2 + c^2} \) comes from the magnitude of the normal vector. This formula zeroes in on the shortest distance because it avoids any diagonal measurement by staying perpendicular to both planes.

• The formula can only be applied to parallel planes.
• It utilizes both the coefficients of the plane equations and their constant terms.
• Identifying the right values for \( d_1 \) and \( d_2 \) is crucial.

Plane Equations

**Plane equations** often appear in their standard form, \( ax + by + cz = d \), where the letters represent:

• \(a, b, c\) - coefficients of the variables providing the components of the normal vector.
• \(x, y, z\) - coordinates of any point lying on the plane.
• \(d\) - a constant, unique to the position of the plane in space.

Understanding how these equations shape a plane in space allows you to identify properties like parallelism and orientation quickly. By comparing the coefficients of two plane equations, we can determine if they share the same normal vector, which means they are parallel and can use the perpendicular distance formula.
In the original exercise, the given planes \(x + 2y + 6z = 1\) and \(x + 2y + 6z = 10\) are clearly parallel due to identical coefficients. The only difference is the \(d\) values, which shifts them apart along the direction of the normal vector. Thus, familiarity with plane equations helps in solving geometric problems related to spatial relationships comfortably.