Problem 26 - Thomas Calculus

Question

Find equations for the planes in Exercises 21-26. The plane through \(A(1,-2,1)\) perpendicular to the vector from the origin to \(A\)

Step-by-Step Solution

Answer

The equation of the plane is \( x - 2y + z = 6 \).

Step 1: Identify the Vector

The plane is perpendicular to the vector from the origin to point \( A \). This vector can be represented as \( \mathbf{n} = \langle 1, -2, 1 \rangle \), which we'll use as the normal vector for the plane.

Step 2: Use the Point-Normal Form

The point-normal form of a plane is given by the equation \( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \), where \((a, b, c)\) is the normal vector, and \((x_0, y_0, z_0)\) is a point on the plane. Here, \( a = 1 \), \( b = -2 \), \( c = 1 \), and \((x_0, y_0, z_0) = (1, -2, 1)\).

Step 3: Substitute Values into the Equation

Substitute the values into the point-normal form: \[ 1(x - 1) - 2(y + 2) + 1(z - 1) = 0 \].

Step 4: Simplify the Equation

Distribute and simplify the terms: \[ x - 1 - 2y - 4 + z - 1 = 0 \]. Combine like terms to get \[ x - 2y + z - 6 = 0 \].

Step 5: Write the Final Equation

The equation of the plane in simplified form is \( x - 2y + z = 6 \).

Key Concepts

Vector Calculus

Vector Calculus is a branch of mathematics that extends calculus to functions with vector variables. In dealing with vectors, which have both magnitude and direction, this field allows us to explore physical concepts like force, velocity, and acceleration more comprehensively.
In our exercises, vectors are used to represent directions and magnitudes in space. A vector is often written in the form \( \langle a, b, c \rangle \), signifying its components along the x, y, and z axes.
Key operations in vector calculus include:

• Addition: Combining vectors to find a resultant vector.
• Scalar Multiplication: Scaling a vector by a real number, affecting its magnitude.
• Dot Product: A measure of how much one vector extends in the direction of another, useful for finding angles and checking perpendicularity.
• Cross Product: Results in a vector perpendicular to two given vectors, aiding in determining orientations in 3D space.

These operations form the foundation needed to handle tasks involving vectors in three-dimensional spaces, like solving for planes.

Planes in 3D Space

Planes are flat, two-dimensional surfaces that extend infinitely in three-dimensional (3D) space. They can be thought of as giant sheets stretching outwards from any chosen point. Understanding planes is essential for various applications in physics, engineering, and computer graphics.
In mathematical terms, a plane can be defined by a point and a vector called the normal vector, which is perpendicular to every line on the plane. This normal vector plays a crucial role in determining the plane's orientation in space.
Planes can be described using several forms, with the most common being the point-normal form and the general form. The point-normal form is particularly useful when you know a point on the plane and the normal vector, as it directly relates to the geometry of the plane.

• Point-normal form: \( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \)
• General form: Typically written as \( Ax + By + Cz + D = 0 \), where \( A, B, C, \) and \( D \) are constants.

By understanding these different representations, we can switch between them as needed for solving practical problems.

Normal Vector

A Normal Vector is fundamental in describing the orientation of a plane in 3D space. It is a vector that is perpendicular, or "normal," to every point on the plane, allowing us to define the plane uniquely.
The significance of the normal vector lies in its capacity to simplify the computation of planes and predict their interaction with other geometric entities. It has components \( (a, b, c) \), corresponding to the coefficients in the plane's equation.
Some important properties and uses of normal vectors:

• Defines orientation: Indicates the perpendicular direction to the plane, crucial for calculating reflections, lighting, and shadows in graphics.
• Helps in collision detection: In physics, normal vectors help in determining points of contact and impact forces between objects.
• Part of the point-normal form: Facilitates converting between different forms of the plane equation.

Knowing the normal vector means you can craft an equation for the plane efficiently, as seen in our exercise where the vector from the origin to a point is used.

Equation of a Plane

The Equation of a Plane provides a mathematical representation of a plane in 3D space, which can be crucial for solving geometric and spatial problems. This equation helps us understand where the plane exists relative to the coordinate system and other planes or lines.
The simplest and most commonly used form is the point-normal form, which we utilized in our exercise. It's particularly helpful when you have a point on the plane, \((x_0, y_0, z_0)\), and a normal vector \(\langle a, b, c \rangle\). The equation is

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

Here, substituting the known values gives a specific form of the plane's equation. To further simplify, you can convert it into the general form:

• \( Ax + By + Cz + D = 0 \)

In our case, after substituting and simplifying, the equation became \( x - 2y + z - 6 = 0 \), representing a flat surface in space that fits our specified constraints. Understanding how to derive this equation is crucial for applications ranging from computer graphics to engineering designs.