How to Find the Equation of a Plane Passing Through Three Given Points

Understanding how to find the equation of a plane passing through three given points is a fundamental concept in 3D geometry. This article will guide you through the process step-by-step using a practical example.

Introduction

In three-dimensional space, a plane can be uniquely determined by three non-collinear points. The equation of the plane can be derived using various methods, including vector operations and the normal vector concept.

Step-by-Step Guide

Let's consider the three given points: 4, -2, 121, and -3, 0, 3. We will follow these steps to find the equation of the plane passing through these points:

Step 1: Find Two Vectors in the Plane

The first step is to find two vectors that lie in the plane. We can do this by subtracting the coordinates of the points. Let's define three vectors:

Now, we can find the vectors AB and AC by subtracting the vectors:

AB B - A (1 - 4, 2 - (-2), 1 - 121) (-3, 4, -120)

AC C - A (-3 - 4, 0 - (-2), 3 - 121) (-7, 2, -118)

Step 2: Find the Normal Vector to the Plane

The normal vector n to the plane can be found by taking the cross product of the two vectors AB and AC. The cross product is calculated as follows:

n AB times; AC)

The cross product in determinant form is:

n

Expanding this determinant, we get:

n i(begin{vmatrix} 4 -120 2 -118 end{vmatrix}) - j(begin{vmatrix} -3 -120 -7 -118 end{vmatrix}) k(begin{vmatrix} -3 4 -7 2 end{vmatrix})

Calculating the determinants:

begin{vmatrix} 4 -120 2 -118 end{vmatrix} 4(-118) - (-120)(2) -472 240 -232

begin{vmatrix} -3 -120 -7 -118 end{vmatrix} (-3)(-118) - (-120)(-7) 354 - 840 -486

begin{vmatrix} -3 4 -7 2 end{vmatrix} (-3)(2) - (4)(-7) -6 28 22

Thus, the normal vector is:

n -232i - (-486)j 22k -232, 486, 22

Step 3: Equation of the Plane

The general form of the equation of a plane is given by:

n_x(x - x_0) n_y(y - y_0) n_z(z - z_0) 0

Where n_x, n_y, n_z are the components of the normal vector and (x_0, y_0, z_0) is a point on the plane. We can use point (4, -2, 121) for simplicity:

-232(x - 4) 486(y 2) 22(z - 121) 0

Expanding and simplifying:

-232x 928 486y 972 22z - 2662 0

-232x 486y 22z - 762 0

This is the equation of the plane in its simplest form, and it can be further simplified if desired.

Conclusion

By following these steps, you can determine the equation of a plane passing through three given points. This method is widely used in computer graphics, physics, and engineering, among other fields.

Additional Resources

For deeper understanding and additional practice, refer to your class notes or textbooks on 3D geometry. Websites such as Khan Academy and MIT OpenCourseWare provide extensive tutorials and exercises on the topic.