Problem 23 - Essential Calculus Early Transcendentals

Question

\(21-32=\) Find an equation of the plane. The plane through the point \((1,-1,-1)\) and parallel to the plane \(5 x-y-z=6\)

Step-by-Step Solution

Answer

The equation of the plane is \(5x - y - z = 7\).

Step 1: Identify the Given Plane Equation

We are given the plane equation \(5x - y - z = 6\). This equation is in the standard form \(Ax + By + Cz = D\), where \(A = 5\), \(B = -1\), and \(C = -1\).

Step 2: Determine the Normal Vector of the Given Plane

The coefficients of x, y, and z in the plane equation give the normal vector of the plane. For the equation \(5x - y - z = 6\), the normal vector is \(\mathbf{n} = \langle 5, -1, -1 \rangle\).

Step 3: Use the Normal Vector for the New Plane

Since the new plane must be parallel to the given plane, it will have the same normal vector \(\mathbf{n} = \langle 5, -1, -1 \rangle\).

Step 4: Substitute the Given Point into the Plane Equation

We are asked to find a plane that passes through the point \((1, -1, -1)\). We use the equation form \(5x - y - z = D\) and substitute the point into it: \(5(1) - (-1) - (-1) = D\).

Step 5: Simplify to Find the Value of D

Calculate the value of \(D\) by performing the operations: \(5 + 1 + 1 = 7\). Therefore, \(D = 7\).

Step 6: Write the Equation of the Plane

The equation of the plane that is parallel to the given plane and passes through the point \((1, -1, -1)\) is \(5x - y - z = 7\).

Key Concepts

Normal Vector

In the context of geometry, especially in three-dimensional space, the normal vector is an essential component used to define the orientation of a plane. The normal vector is a perpendicular vector to the surface of the plane. Imagine it as a straight arrow sticking directly out from a flat sheet of paper.

For a plane given by the equation \(Ax + By + Cz = D\), the normal vector is represented by \(\mathbf{n} = \langle A, B, C \rangle\).

• "Normal" implies perpendicular or orthogonal, meaning the vector is at a right angle to the surface.
• The direction of the normal vector dictates the tilt or slant of the plane in space.

Understanding the normal vector is crucial because it allows us to determine parallelism between planes. For two planes to be parallel, their normal vectors must be identical.

Parallel Planes

Planes in three-dimensional space can either intersect or be parallel. Parallel planes are ones that never meet, no matter how far they are extended. This is akin to two railroad tracks that run side by side without touching.

To assess whether two planes are parallel, we observe their normal vectors. Two planes are parallel if their normal vectors are scalar multiples of one another.

• For instance, knowing the plane equation \(5x - y - z = 6\), the normal vector is \(\mathbf{n} = \langle 5, -1, -1 \rangle\).
• Any plane with the same normal vector, such as another plane with equation \(5x - y - z = D\), will be parallel to the one above.

This concept of recognizing parallelism via normal vectors is fundamental and frequently applied in solving geometry problems.

Plane in 3D Geometry

A plane in 3D geometry is essentially a flat, two-dimensional surface extending infinitely in three-dimensional space. When described mathematically, a plane can be specified by an equation of the form \(Ax + By + Cz = D\).

There are several key components to understanding a plane in 3D space:

• The normal vector, as mentioned, defines the plane's orientation.
• The coefficients \(A\), \(B\), and \(C\) give insight into the tilt of the plane in relation to the 3D axes.
• To find the specific position of the plane, you often use a point through which the plane passes. For example, the point \((1,-1,-1)\) is used to define the specific orientation of a plane parallel to \(5x - y - z = 6\).

Understanding these elements allows one to describe and manipulate planes within three-dimensional spaces effectively, making it a vital skill in fields such as physics, computer graphics, and engineering.