Problem 75 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the distance between the parallel planes \(-3 x+2 y+z=9\) and \(6 x-4 y-2 z=19\).

Step-by-Step Solution

Answer

The distance between the parallel planes is approximately 4.94 units.

Step 1: Identify the Planes

Observe the given equations for the planes. The first plane is given by \(-3x + 2y + z = 9\). The second plane needs to be rewritten. Dividing the entire second equation \(6x - 4y - 2z = 19\) by 2 gives \(3x - 2y - z = 9.5\). Thus, the parallel planes can be expressed as \(-3x + 2y + z = 9\) and \(-3x + 2y + z = -9.5\), clearly showing that they share the same normal vector.

Step 2: Confirm Parallelism

We need to ensure that both planes have normal vectors that are multiples of each other, confirming their parallelism. The normal vector of the first plane is \(\langle -3, 2, 1 \rangle\) and for the second plane is also \(\langle -3, 2, 1 \rangle\). Thus, these planes are parallel.

Step 3: Use the Distance Formula for Parallel Planes

The formula to calculate the distance \(d\) between two parallel planes \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\) is \[ d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}} \]. Here, \(D_1 = 9\) and \(D_2 = -9.5\).

Step 4: Calculate the Distance

Substitute the values into the formula: \[ d = \frac{|9 + 9.5|}{\sqrt{(-3)^2 + 2^2 + 1^2}} = \frac{18.5}{\sqrt{14}} \].

Step 5: Simplify the Expression

Simplifying the expression we compute \( \sqrt{14} \approx 3.741657\), therefore the distance can be approximated as \( d \approx \frac{18.5}{3.741657} \approx 4.94 \).

Key Concepts

Parallel Planes

Parallel planes are two planes that never intersect, no matter how far they are extended in any direction. This is similar to parallel lines, but in a higher dimension. You might picture this as two flat sheets of paper stacked on top of each other, extending infinitely in the plane. For two planes to be parallel, their normal vectors must be scalar multiples of each other. In the given problem, this means that both planes share the same direction in space. Once you have confirmed that the planes are parallel, you can proceed to find the distance between them, which will be constant everywhere.

Normal Vectors

A normal vector to a plane is a vector that is perpendicular to the plane. It defines the 'tilt' or orientation of the plane in three-dimensional space. The equation of a plane is typically written as \(Ax + By + Cz = D\), where \((A, B, C)\) is the normal vector. In our exercise, the normal vector for both planes is \(\langle -3, 2, 1 \rangle\). This shared normal vector confirms that the planes are indeed parallel. Understanding the role of normal vectors is crucial because they help determine not just the parallelism, but also the orientation of the planes.

Distance Formula

The distance formula for parallel planes provides a method to calculate the shortest distance between them. When you know that two planes are parallel, you can use the formula:

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

Here, \(D_1\) and \(D_2\) represent the constants from the plane equations, and \(A, B, C\) are components of the normal vector. Plugging in the specific values, we can compute the exact distance. In this case, it involves determining the absolute difference between the \(D\) values of the planes and dividing it by the magnitude of the normal vector.

Plane Equation

A plane equation represents a flat two-dimensional surface in a three-dimensional space, described mathematically with the general form \(Ax + By + Cz = D\). Each point \((x, y, z)\) on the plane satisfies this equation. The coefficients \(A, B,\) and \(C\) form the components of the normal vector, while \(D\) is a scalar that adjusts the plane's position along the normal vector's direction. In the original exercise, rewriting the second plane equation was essential to confirm that both equations share similar coefficients, leading to a clear interpretation of parallelism. Familiarity with plane equations allows you to transform and manipulate them to understand their interactions in space.