Problem 20 - Thomas Calculus

Question

In Exercises \(19-28\) , 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 \(\quad\) b. \(y\) -axis \(\quad\) c. \(z\) -axis

Step-by-Step Solution

Answer

a. \(x = 3\); b. \(y = -1\); c. \(z = 2\)

Step 1: Understanding the Plane Equation

To describe a plane in 3D space, we need a point on the plane and a normal vector to the plane. The general equation of a plane is given by: \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\), where \((x_0, y_0, z_0)\) is a point on the plane and \((a, b, c)\) is the normal vector to the plane.

Step 2: Plane Perpendicular to the x-axis

For a plane to be perpendicular to the x-axis, the normal vector must be parallel to the x-axis. Therefore, the normal vector can be \((1, 0, 0)\). Substituting the normal vector and point \((3, -1, 2)\) into the equation of the plane, we get: \(1(x-3) + 0(y+1) + 0(z-2) = 0\), or simply \(x = 3\).

Step 3: Plane Perpendicular to the y-axis

For a plane to be perpendicular to the y-axis, the normal vector must be parallel to the y-axis. Therefore, the normal vector can be \((0, 1, 0)\). Substituting the normal vector and point \((3, -1, 2)\) into the equation, we have: \(0(x-3) + 1(y+1) + 0(z-2) = 0\), which simplifies to \(y = -1\).

Step 4: Plane Perpendicular to the z-axis

For a plane to be perpendicular to the z-axis, the normal vector must be parallel to the z-axis. Thus, the normal vector is \((0, 0, 1)\). Substituting the normal vector and point \((3, -1, 2)\) into the plane equation, we have: \(0(x-3) + 0(y+1) + 1(z-2) = 0\), simplifying to \(z = 2\).

Key Concepts

Normal Vector

In 3D geometry, a plane is defined using a normal vector and a point on the plane. The normal vector is a crucial concept because it is perpendicular to the plane. To visualize a normal vector, imagine it as an arrow sticking straight up from a flat surface. This vector influences how we write the equation of the plane.

The general equation of a plane is \[a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\]Here, \((x_0, y_0, z_0)\) is a specific point on the plane, while \((a, b, c)\) represents the normal vector. The coordinates of the normal vector—\(a\), \(b\), and \(c\)—help determine the plane's orientation in the 3D space.

• When the normal vector is parallel to the x-axis, it is \((1, 0, 0)\).
• If it's parallel to the y-axis, the vector is \((0, 1, 0)\).
• For the z-axis, it is \((0, 0, 1)\).

By adjusting the normal vector, you can change how the plane is positioned in 3D space.

3D Geometry

3D geometry extends the principles of geometry into three dimensions, adding depth to the familiar two-dimensional concepts. Instead of just dealing with length and width, we now consider height as well. This shift introduces more complexity in calculations and visualizations.

In 3D geometry, points are located using coordinates \((x, y, z)\), representing a position in space. These coordinates are combined to define various geometric entities like lines, surfaces, and solids. Planes are flat surfaces extending indefinitely in all directions within the 3D space. Each plane can be described mathematically by an equation, where the coefficients of the equation relate directly to the properties of the normal vector, indicating the plane’s orientation.

Understanding these concepts can help students visualize problems in physics, engineering, and architecture where 3D spaces are ubiquitous. Including the third dimension lets us describe the real world much more accurately.

Perpendicular to Axis

When a plane is described as being perpendicular to one of the coordinate axes (x, y, or z), it means that its normal vector is parallel to that axis. This condition greatly simplifies the equation of the plane.

Let's look at each case in more detail:

• Perpendicular to the x-axis: The normal vector is \((1, 0, 0)\). The plane moves vertically with respect to the x-axis and is described with the equation \(x = \text{constant}\).
• Perpendicular to the y-axis: For this, the normal vector is \((0, 1, 0)\). The plane extends along the xz-plane and is detailed by \(y = \text{constant}\).
• Perpendicular to the z-axis: With the normal vector \((0, 0, 1)\), the plane is horizontal, paralleling the xy-plane, defined by \(z = \text{constant}\).

Understanding how planes relate to the axes helps us grasp how they divide space and provide boundaries within three-dimensional contexts.