Equation Of Plane From 3 Points

Finding the Equation of a Plane from Three Points

Determining the equation of a plane given three points in three-dimensional space is a fundamental concept in linear algebra and vector calculus. This process involves leveraging the concept of normal vectors and the vector equation of a plane. This article will guide you through the process step-by-step, explaining the underlying mathematics and providing a practical example. Understanding this process is crucial for various applications in computer graphics, physics, and engineering.

Understanding the Fundamentals

Before we delve into the calculations, let's refresh some key concepts:

• Vector: A quantity possessing both magnitude and direction. Represented as <x, y, z>.
• Normal Vector: A vector perpendicular to a plane. This vector is crucial for defining the plane's orientation.
• Equation of a Plane: The general equation of a plane is given by Ax + By + Cz + D = 0, where <A, B, C> represents the normal vector and D is a constant. Alternatively, the vector equation is often used: r ⋅ n = a ⋅ n, where r = <x, y, z> is a position vector, n is the normal vector, and a is a known point on the plane.

Steps to Find the Equation of a Plane:

1. Find Two Vectors in the Plane: Given three points, A(x₁, y₁, z₁), B(x₂, y₂, z₂), and C(x₃, y₃, z₃), we first form two vectors lying within the plane. These can be found by subtracting the coordinates of the points:

   • AB = B - A = <x₂ - x₁, y₂ - y₁, z₂ - z₁>
   • AC = C - A = <x₃ - x₁, y₃ - y₁, z₃ - z₁>

2. Find the Normal Vector: The normal vector, n, is perpendicular to both AB and AC. We find this by computing the cross product of AB and AC:

   • n = AB x AC = <(y₂ - y₁)(z₃ - z₁) - (z₂ - z₁)(y₃ - y₁), (z₂ - z₁)(x₃ - x₁) - (x₂ - x₁)(z₃ - z₁), (x₂ - x₁)(y₃ - y₁) - (y₂ - y₁)(x₃ - x₁)>

3. Determine the Equation of the Plane: Using one of the points (A, B, or C) and the normal vector n = <A, B, C>, we can substitute into the equation of a plane:

   • Ax + By + Cz + D = 0

   Substitute the coordinates of point A (x₁, y₁, z₁) and the components of the normal vector into the equation to solve for D:

   • A(x₁) + B(y₁) + C(z₁) + D = 0 => D = -A(x₁) - B(y₁) - C(z₁)

Therefore, the equation of the plane is:

• Ax + By + Cz - A(x₁) - B(y₁) - C(z₁) = 0

Example:

Let's find the equation of the plane passing through the points A(1, 0, 0), B(0, 1, 0), and C(0, 0, 1).

1. Vectors AB and AC:

   • AB = <0 - 1, 1 - 0, 0 - 0> = <-1, 1, 0>
   • AC = <0 - 1, 0 - 0, 1 - 0> = <-1, 0, 1>

2. Normal Vector:

   • n = AB x AC = <(1)(1) - (0)(0), (0)(-1) - (-1)(1), (-1)(0) - (1)(-1)> = <1, 1, 1>

3. Equation of the Plane: Using point A(1, 0, 0) and n = <1, 1, 1>:

   • 1(x) + 1(y) + 1(z) + D = 0
   • 1(1) + 1(0) + 1(0) + D = 0 => D = -1

Therefore, the equation of the plane is: x + y + z - 1 = 0.

This detailed explanation, coupled with a clear example, should equip you with the knowledge to confidently determine the equation of a plane given three points. Remember to carefully perform each step, double-checking your calculations to ensure accuracy. This skill is fundamental for tackling more complex problems in three-dimensional geometry.