Equation Of Plane From 3 Points

Okay, here's a comprehensive article on finding the equation of a plane from three points. This will cover the necessary concepts, the derivation, examples, common pitfalls, and more.

Finding the Equation of a Plane from Three Points

The equation of a plane is a fundamental concept in three-dimensional geometry and linear algebra. It describes a flat, two-dimensional surface that extends infinitely in three-dimensional space. Determining the equation of a plane is a common problem, and one frequent scenario is when you are given three points that lie on the plane. In this article, we'll explore how to derive the equation of a plane given three non-collinear points, discuss the underlying principles, and provide practical examples.

Let's dive in.

Introduction

Imagine holding a flat piece of paper in the air. That piece of paper, if infinitely extended, represents a plane. To fully define its orientation and location in space, you need a certain amount of information. One way to define a plane is by specifying three points that reside on it, as long as those points do not lie on the same straight line (i.e., they are non-collinear). This is because three non-collinear points uniquely determine a plane in 3D space.

The equation of a plane is typically expressed in one of two forms:

1. Scalar (or Cartesian) Form: Ax + By + Cz + D = 0
2. Vector Form: r · n = a · n where r is the position vector of a general point on the plane, n is a normal vector to the plane, and a is the position vector of a known point on the plane.

Our goal is to find the values of A, B, C, D (in the scalar form) or the normal vector n (in the vector form) given three points.

The Underlying Principles

Before diving into the step-by-step process, let's understand the core principles that make this possible:

1. Vectors from Points: Given two points, we can create a vector that lies along the line connecting them. If we have three points on a plane, we can create two such vectors that lie within the plane.

2. Normal Vector: A normal vector is a vector that is perpendicular (orthogonal) to the plane. If we have two vectors lying in the plane, their cross product will result in a vector that is normal to both of them, and therefore, normal to the plane.

3. Dot Product: The dot product of two orthogonal vectors is zero. This property is key to defining the relationship between the normal vector, a known point on the plane, and any other general point on the plane.

Step-by-Step Method to Find the Equation of a Plane

Let's outline the process to determine the equation of the plane defined by three points, P1, P2, and P3.

Step 1: Define the Points

Let the three non-collinear points be defined as:

• P1 = (x1, y1, z1)
• P2 = (x2, y2, z2)
• P3 = (x3, y3, z3)

Step 2: Create Two Vectors in the Plane

Create two vectors, v1 and v2, using these points. These vectors will lie in the plane.

• v1 = P2 - P1 = (x2 - x1, y2 - y1, z2 - z1)
• v2 = P3 - P1 = (x3 - x1, y3 - y1, z3 - z1)

(Note: You can also choose P3 - P2, P1 - P2, P1 - P3, etc. The choice doesn't fundamentally change the final result, but it might change the direction of the normal vector, which is still valid.)

Step 3: Find the Normal Vector

The normal vector, n, is found by taking the cross product of v1 and v2.

• n = v1 x v2

The cross product is calculated as follows:

n = ( (y2 - y1)(z3 - z1) - (z2 - z1)(y3 - y1), (z2 - z1)(x3 - x1) - (x2 - x1)(z3 - z1), (x2 - x1)(y3 - y1) - (y2 - y1)(x3 - x1) )

Let's denote the components of the normal vector as n = (A, B, C). So:

• A = (y2 - y1)(z3 - z1) - (z2 - z1)(y3 - y1)
• B = (z2 - z1)(x3 - x1) - (x2 - x1)(z3 - z1)
• C = (x2 - x1)(y3 - y1) - (y2 - y1)(x3 - x1)

Step 4: Form the Equation of the Plane

Now that we have the normal vector n = (A, B, C), we can use the scalar form of the equation of the plane:

Ax + By + Cz + D = 0

To find D, substitute one of the original points (P1, P2, or P3) into this equation. Let's use P1 = (x1, y1, z1):

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

Solve for D:

D = -[A(x1) + B(y1) + C(z1)]

Step 5: Write the Final Equation

Substitute the values of A, B, C, and D into the scalar equation of the plane:

Ax + By + Cz + D = 0

Or, substituting the expression for D:

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

Example

Let's find the equation of the plane containing the points P1(1, 2, 3), P2(3, 2, 1), and P3(2, -1, 0).

Step 1: We already have the points:

• P1 = (1, 2, 3)
• P2 = (3, 2, 1)
• P3 = (2, -1, 0)

Step 2: Create the vectors v1 and v2:

• v1 = P2 - P1 = (3 - 1, 2 - 2, 1 - 3) = (2, 0, -2)
• v2 = P3 - P1 = (2 - 1, -1 - 2, 0 - 3) = (1, -3, -3)

Step 3: Find the normal vector n:

n = v1 x v2 = ( (0)(-3) - (-2)(-3), (-2)(1) - (2)(-3), (2)(-3) - (0)(1) ) = (-6, 4, -6)

So, A = -6, B = 4, C = -6. We can simplify this by dividing by -2, resulting in the normal vector (3, -2, 3), and A = 3, B = -2, C = 3. This simplification will make the subsequent calculations easier, and it does not change the plane defined by the equation.

Step 4: Find D using point P1(1, 2, 3):

D = -[A(x1) + B(y1) + C(z1)] = -[3(1) + (-2)(2) + 3(3)] = -[3 - 4 + 9] = -8

Step 5: Write the final equation:

3x - 2y + 3z - 8 = 0

Therefore, the equation of the plane is 3x - 2y + 3z = 8.

Vector Form of the Equation

Let's revisit the example above and express the equation in vector form. We already have the normal vector n = (3, -2, 3). We also have a point on the plane, let's use P1 = (1, 2, 3), which has a position vector a = <1, 2, 3>. A general point on the plane has a position vector r = <x, y, z>.

The vector form of the equation is r · n = a · n.

Calculate a · n:

a · n = (1)(3) + (2)(-2) + (3)(3) = 3 - 4 + 9 = 8

Therefore, the vector form of the equation is:

<x, y, z> · <3, -2, 3> = 8

This simplifies to 3x - 2y + 3z = 8, which is consistent with the scalar form we derived earlier.

Common Pitfalls and Considerations

• Collinear Points: If the three points are collinear (lie on the same line), the vectors v1 and v2 will be parallel, and their cross product will be the zero vector. This indicates that the points do not uniquely define a plane.

• Arithmetic Errors: The cross product calculation involves several steps, increasing the chance of arithmetic errors. Double-check your calculations to ensure accuracy.

• Choosing the Correct Vectors: While the choice of vectors (P2 - P1, P3 - P1, etc.) doesn't change the final plane, it might change the direction of the normal vector. A normal vector in the opposite direction is also valid since Ax + By + Cz + D = 0 and -Ax - By -Cz - D = 0 represent the same plane.

• Simplifying the Equation: After finding the equation, check if you can simplify it by dividing all coefficients (A, B, C, and D) by a common factor. This makes the equation easier to work with.

• Checking Your Answer: A good way to verify your equation is to substitute the coordinates of the original three points into the equation. All three points should satisfy the equation Ax + By + Cz + D = 0.

Applications

Finding the equation of a plane from three points has numerous applications in various fields:

• Computer Graphics: Determining the orientation of surfaces in 3D models.

• Game Development: Collision detection and physics simulations.

• Engineering: Calculating forces and moments on planar structures.

• Navigation: Defining surfaces for path planning and obstacle avoidance.

• Linear Algebra: Understanding the properties of vector spaces and linear transformations.

Advanced Considerations

• Parametric Equation of a Plane: While we focused on scalar and vector forms, another representation is the parametric equation of a plane:

  r = a + s*v1 + t*v2

  where r is a general point on the plane, a is a known point, v1 and v2 are vectors in the plane, and s and t are parameters.

• Distance from a Point to a Plane: Once you have the equation of the plane, you can calculate the shortest distance from any point to the plane using the formula:

  distance = |Ax0 + By0 + Cz0 + D| / sqrt(A^2 + B^2 + C^2)

  where (x0, y0, z0) are the coordinates of the point.

Conclusion

Finding the equation of a plane from three points is a fundamental skill in 3D geometry with widespread applications. By understanding the underlying principles of vectors, normal vectors, and the dot product, you can confidently derive the equation of a plane in either scalar or vector form. Remember to pay attention to potential pitfalls like collinear points and arithmetic errors. The ability to perform this calculation accurately opens doors to solving more complex problems in various fields, from computer graphics to engineering.

How do you think this method could be optimized for computational efficiency in a program that generates many planes? And, have you encountered situations where using a different form of the plane equation (e.g., parametric) proved more advantageous for specific tasks?