Problem 64 - Thomas Calculus

Question

Use the component form to generate an equation for the plane through \(P_{1}(4,1,5)\) normal to \(\mathbf{n}_{1}=\mathbf{i}-2 \mathbf{j}+\mathbf{k} .\) Then generate another equation for the same plane using the point \(P_{2}(3,-2,0)\) and the normal vector \(\mathbf{n}_{2}=-\sqrt{2} \mathbf{i}+2 \sqrt{2} \mathbf{j}-\sqrt{2} \mathbf{k}\)

Step-by-Step Solution

Answer

The plane's equation through either point is \(x - 2y + z = 7\).

Step 1: Equation of Plane using Point-Normal Form

The equation of a plane with a normal vector \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) through a point \((x_0, y_0, z_0)\) is given by \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\). Here, \(P_1 = (4, 1, 5)\) and the normal vector \(\mathbf{n}_1 = \mathbf{i} - 2\mathbf{j} + \mathbf{k}\) gives components \(a = 1\), \(b = -2\), \(c = 1\). Substituting, we have: \(1(x - 4) - 2(y - 1) + 1(z - 5) = 0\). Simplify this to: \(x - 4 - 2y + 2 + z - 5 = 0\), or \(x - 2y + z - 7 = 0\).

Step 2: Simplify Equation with Point P1

Combining and arranging terms from the previous step: \(x - 2y + z - 7 = 0\) is the equation of the plane derived from \(P_1\) and \(\mathbf{n}_1\).

Step 3: Calculate the Equation Using P2 and n2

The process is similar with \(P_2 = (3, -2, 0)\) and normal vector \(\mathbf{n}_2 = -\sqrt{2} \mathbf{i} + 2 \sqrt{2} \mathbf{j} - \sqrt{2} \mathbf{k}\). This gives \(a = -\sqrt{2}\), \(b = 2\sqrt{2}\), \(c = -\sqrt{2}\). Using the point-normal form: \[-\sqrt{2}(x - 3) + 2\sqrt{2}(y + 2) - \sqrt{2}(z - 0) = 0\]which simplifies:\[-\sqrt{2}x + 3\sqrt{2} + 2\sqrt{2}y + 4\sqrt{2} - \sqrt{2}z = 0\]Combining terms, we get:\[-\sqrt{2}x + 2\sqrt{2}y - \sqrt{2}z + 7\sqrt{2} = 0\].

Step 4: Simplify and Analyze the Second Equation

Thus, the equation for the plane using \(P_2\) and \(\mathbf{n}_2\) is \(-\sqrt{2}x + 2\sqrt{2}y - \sqrt{2}z + 7\sqrt{2} = 0\). To match the form of the first equation, divide by \(-\sqrt{2}\) to yield: \[x - 2y + z = 7\].

Step 5: Check Solutions Consistency

Both plane equations ultimately reduce to the same form, confirming they represent the same plane: \(x - 2y + z = 7\).

Key Concepts

Planes in Three-Dimensional Space

In three-dimensional space, a plane can be thought of as an infinite flat surface that extends in all directions. Planes are fundamental in vector calculus as they define two-dimensional surfaces within three-dimensional geometry. To specify a plane, we typically need a point on the plane and a direction that is perpendicular to it, called a normal vector. This information allows us to describe the orientation and position of the plane fully.

Understanding planes in three-dimensional space is crucial because they help in solving various geometric problems. Planes can intersect lines, other planes, or remain parallel or perpendicular to other geometric entities. This makes them highly versatile in mathematical analysis and applications like computer graphics, physics, and engineering.

The mathematical representation of a plane involves equations which can be manipulated and used to understand relationships and solve geometric problems. Recognizing how to form and interpret these equations is key to mastering the concepts in plane-based exercises.

Normal Vector

A normal vector is essential to define the orientation of a plane in three-dimensional space. It is a vector that is perpendicular to every line on the plane, pointing directly away from the plane's surface.

The importance of the normal vector lies in its role in the point-normal form equation, which enables one to write the equation of a plane effectively. A normal vector is usually expressed in its component form as \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), where \(\mathbf{i}, \mathbf{j},\) and \(\mathbf{k}\) are unit vectors along the Cartesian axes.

Normal vectors can vary in magnitude and direction, but will always remain perpendicular to the plane they define. By controlling the normal vector, we can adjust the plane's orientation, making its understanding central to exercises involving planes. The magnitude of the normal vector does not affect the position of the plane, only its direction does.

Point-Normal Form Equation

The point-normal form is a specific equation representation that makes it easy to express a plane using a known point and a normal vector. This form is handy because it provides a straightforward way to move from geometric properties to an algebraic formula.

The general form of the point-normal 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 \(a, b, c\) are the components of the normal vector. This formula concisely expresses how the point and normal vector determine the position and orientation of the plane.

This approach simplifies the task of describing planes, as it links geometrical concepts directly with algebraic expressions. Once established, the point-normal form equation can be altered or simplified to find intersections, or check plane relations with other geometric or spatial entities.

Component Form of Vectors

Vectors are a cornerstone in mathematics, allowing us to model dynamics, forces, and spatial relationships. The component form of a vector is one of the most intuitive ways to express it in vector calculus.

A vector's component form breaks it down into its effects in each of the space's dimensions. For instance, the vector \( \mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) consists of its components \(a, b,\) and \(c\), which correspond to its projections along the \(x\), \(y\), and \(z\) axes, respectively.

This representation makes it easier to perform addition, subtraction, and scalar multiplication of vectors, and it is especially useful when working with equations of planes. By understanding and utilizing the component form, you can effectively solve numerous problems in vector calculus, such as finding dot products, cross products, and decomposing complex vectors into simpler components.