Problem 36 - University Calculus: Early Transcendentals

Question

Describe the given set with a single equation or with a pair of equations. The plane through the point (3,-1,2) perpendicular to the a. \(x\) -axis b. y-axis c. z-axis

Step-by-Step Solution

Answer

x = 3; y = -1; z = 2.

Step 1: Understanding the Problem

We need to describe the planes that are perpendicular to each of the coordinate axes (x, y, and z axes) and pass through the point (3,-1,2). Each axis will have a corresponding normal vector, and we will use the point-normal form of a plane's equation.

Step 2: Plane Perpendicular to the x-axis

A plane perpendicular to the x-axis means its normal vector \((1, 0, 0)\). Using the point (3, -1, 2), the equation of the plane is \(x = 3\).

Step 3: Plane Perpendicular to the y-axis

A plane perpendicular to the y-axis means its normal vector \((0, 1, 0)\). Using the point (3, -1, 2), the equation of the plane is \(y = -1\).

Step 4: Plane Perpendicular to the z-axis

A plane perpendicular to the z-axis means its normal vector \((0, 0, 1)\). Using the point (3, -1, 2), the equation of the plane is \(z = 2\).

Key Concepts

Point-Normal Form

In geometry, the point-normal form is a way to express the equation of a plane. It is particularly convenient when you know a point on the plane and the normal vector to the plane. The general form of a plane equation in point-normal form is

• \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]

Here, \(x_0, y_0, z_0\) represents a point on the plane, and \(a, b, c\) is the normal vector to the plane. The normal vector essentially "points" perpendicular to the plane. By substituting the specific values for the point and the normal vector components into this formula, we can find a specific plane equation.
In our exercise, the planes we are considering are defined with equations perpendicular to each coordinate axis, so the normal vector components are very simple: one component is 1 and the others are 0.

Coordinate Axes

In three-dimensional space, the coordinate axes are the \(x\), \(y\), and \(z\) axes. They form the basis of the Cartesian coordinate system, where any point in space can be represented as \( (x, y, z) \). Each axis corresponds to one dimension within this space:

• The \(x\)-axis runs horizontally from left to right.
• The \(y\)-axis runs vertically from bottom to top.
• The \(z\)-axis runs perpendicular to the plane formed by \(x\) and \(y\) and can be thought of as running in/out of your page or screen.

Planes that are perpendicular to one of these axes will have a constant value for the variable corresponding to that axis, such as the exercises' planes:

• Plane perpendicular to the \(x\)-axis: \(x = 3\)
• Plane perpendicular to the \(y\)-axis: \(y = -1\)
• Plane perpendicular to the \(z\)-axis: \(z = 2\)

These planes slice through the 3D space defined by the coordinate axes.

Normal Vector

The normal vector is a fundamental concept when dealing with planes in three-dimensional geometry. A vector itself is an entity with both magnitude and direction. Specifically, the normal vector to a plane is directed perpendicular to the plane's surface. In simple terms, it "sticks out" of the plane.
For a given plane, the normal vector \( \mathbf{n} = (a, b, c) \) holds the coefficients in the plane's equation \( ax + by + cz = d \). In the point-normal form, the normal vector's role is crucial because it determines the plane's orientation in space.

• For a plane perpendicular to the \(x\)-axis, the normal vector is \((1, 0, 0)\), meaning the plane is oriented front to back along the \(x\)-axis.
• For a plane perpendicular to the \(y\)-axis, the normal vector is \((0, 1, 0)\), orienting the plane side to side along the \(y\)-axis.
• For a plane perpendicular to the \(z\)-axis, the normal vector is \((0, 0, 1)\), orienting the plane up and down along the \(z\)-axis.

Understanding the normal vector helps in visualizing and setting up geometric problems, particularly those involving planes since it gives insight into the plane's orientation relative to three-dimensional space.