Finding Plane Equation Parallel To Vectors And Passing Through A Point

This article delves into the method of determining the equation of a plane given two parallel vectors and a point it passes through. This is a fundamental concept in three-dimensional geometry, with applications spanning computer graphics, physics, and engineering. This comprehensive guide provides a step-by-step approach to solving this type of problem, ensuring clarity and understanding. We will explore the theoretical underpinnings and apply them to a concrete example, illustrating the process in detail. This problem is a classic example of vector algebra and its application in defining geometric objects in 3D space. Understanding the concepts of normal vectors, dot products, and plane equations is crucial for success in this area. By carefully working through the steps, you'll gain a solid grasp of how to define a plane using vectors and points.

Understanding the Fundamentals of Plane Equations

Before diving into the specifics of the problem, let's establish a firm grasp of the fundamental concepts. A plane in three-dimensional space can be uniquely defined by two key elements: a normal vector and a point on the plane. The normal vector, often denoted as n, is a vector perpendicular to the plane. It dictates the plane's orientation in space. Imagine a flagpole standing upright on a flat surface; the flagpole represents the normal vector, and the surface represents the plane. The point, denoted as P₀(x₀, y₀, z₀), anchors the plane in space. It provides a specific location that the plane must pass through. With the normal vector and a point, we can write the equation of the plane in several forms. One common form is the scalar equation:

A(x - x₀) + B(y - y₀) + C(z - z₀) = 0

Where (A, B, C) are the components of the normal vector n, and (x₀, y₀, z₀) are the coordinates of the point P₀. This equation expresses the condition that any vector lying in the plane is orthogonal to the normal vector. Another way to represent the plane is using the general form:

Ax + By + Cz + D = 0

Where A, B, and C are the components of the normal vector, and D is a constant related to the position of the plane. The normal vector (A, B, C) plays a crucial role in both forms of the equation. It defines the plane's orientation, and the point (x₀, y₀, z₀) fixes its position in space. To find the equation of a plane, our primary task is often to determine the normal vector. In this article, we'll explore how to find the normal vector when given two vectors parallel to the plane.

Finding the Normal Vector from Two Parallel Vectors

When we are given two vectors, let's call them u and v, that are parallel to the plane, we can find the normal vector n by taking their cross product. The cross product of two vectors results in a new vector that is perpendicular to both original vectors. This is precisely what we need for the normal vector of the plane. Let u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃). The cross product n = u × v is calculated as follows:

n = (u₂v₃ - u₃v₂, u₃v₁ - u₁v₃, u₁v₂ - u₂v₁)

The resulting vector n is orthogonal to both u and v, and thus it is normal to the plane containing u and v. This is a fundamental property of the cross product and a cornerstone of solving problems involving planes in 3D space. Once we have the normal vector n and a point P₀ on the plane, we can readily write the equation of the plane using the scalar or general form. The cross product provides a powerful tool for translating information about vectors parallel to the plane into the essential normal vector needed to define the plane's equation. Understanding this relationship is key to mastering plane equations in three-dimensional geometry. In the following sections, we will apply these concepts to a specific problem, demonstrating the step-by-step process of finding the plane equation.

Problem Statement and Solution

Let's address the problem at hand: Consider the plane parallel to the vectors (-3, 4, 5) and (2, -6, -7), passing through the point (-8, 0, 1). Our goal is to find a normal vector to the plane and then write the equation of the plane. This problem exemplifies the application of the concepts discussed earlier, providing a concrete example of how to find a plane's equation given parallel vectors and a point. We will proceed in a structured manner, first finding the normal vector using the cross product and then using the normal vector and the given point to construct the equation of the plane. This step-by-step approach will highlight the key steps involved in solving such problems and reinforce the understanding of the underlying principles.

Finding the Normal Vector

(a) Find a normal vector ñ to the plane.

As discussed earlier, we can find the normal vector n by taking the cross product of the two vectors parallel to the plane. Let u = (-3, 4, 5) and v = (2, -6, -7). The cross product n = u × v is calculated as follows:

n = (u₂v₃ - u₃v₂, u₃v₁ - u₁v₃, u₁v₂ - u₂v₁) n = ((4)(-7) - (5)(-6), (5)(2) - (-3)(-7), (-3)(-6) - (4)(2)) n = (-28 + 30, 10 - 21, 18 - 8) n = (2, -11, 10)

Therefore, the normal vector n to the plane is (2, -11, 10). This vector is perpendicular to the plane and will be used in the subsequent steps to define the plane's equation. The cross-product calculation is a critical step, and careful attention to the arithmetic is essential to obtain the correct normal vector. With the normal vector in hand, we are now prepared to determine the equation of the plane.

Writing the Equation of the Plane

(b) Write the equation of the plane.

Now that we have the normal vector n = (2, -11, 10) and the point P₀(-8, 0, 1) on the plane, we can write the equation of the plane. We can use the scalar equation of a plane:

A(x - x₀) + B(y - y₀) + C(z - z₀) = 0

Where (A, B, C) are the components of the normal vector n, and (x₀, y₀, z₀) are the coordinates of the point P₀. Plugging in the values, we get:

2(x - (-8)) - 11(y - 0) + 10(z - 1) = 0 2(x + 8) - 11y + 10(z - 1) = 0 2x + 16 - 11y + 10z - 10 = 0 2x - 11y + 10z + 6 = 0

Thus, the equation of the plane is 2x - 11y + 10z + 6 = 0. This equation represents all the points (x, y, z) that lie on the plane parallel to the given vectors and passing through the given point. We have successfully determined the equation of the plane by first finding the normal vector and then utilizing the point-normal form of the plane equation. This process demonstrates the power of vector algebra in solving geometric problems in three-dimensional space. The final equation provides a concise mathematical description of the plane, allowing us to analyze its properties and relationships with other geometric objects.

Conclusion: Mastering Plane Equations in 3D Space

In this comprehensive guide, we have explored the method of finding the equation of a plane parallel to two vectors and passing through a given point. We began by establishing the fundamental concepts of plane equations, including the role of the normal vector and the point on the plane. We then delved into the technique of finding the normal vector using the cross product of two vectors parallel to the plane. Finally, we applied these concepts to a concrete problem, demonstrating the step-by-step process of finding the normal vector and writing the equation of the plane. This problem-solving approach reinforces the understanding of the theoretical underpinnings and provides a practical framework for tackling similar problems. Mastering plane equations in 3D space is essential for various applications in mathematics, physics, engineering, and computer graphics. The ability to define and manipulate planes using vectors and equations is a powerful tool for solving geometric problems and modeling real-world phenomena. By understanding the concepts and techniques presented in this article, you can confidently approach problems involving planes in three-dimensional space and apply them to a wide range of practical scenarios. The key takeaways from this article include the importance of the normal vector in defining a plane, the use of the cross product to find the normal vector from two parallel vectors, and the application of the point-normal form to write the equation of the plane. These concepts and techniques form the foundation for further exploration of three-dimensional geometry and its applications.