Question

Find a condition for four points in space to lie in a plane. Your answer should be in the form a determinant which must be equal to zero. Hint: The equation of a plane is of the form \(a x+b y+c z=d,\) where \(a, b, c, d\) are constants. The four points \(\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right),\) etc., are all to satisfy this equation. When can you find \(a\), \(b, c, d\) not all zero?

Short answer

The determinant for the four points to lie in a plane is \[ \begin{vmatrix} x_1 & y_1 & z_1 & 1 \ x_2 & y_2 & z_2 & 1 \ x_3 & y_3 & z_3 & 1 \ x_4 & y_4 & z_4 & 1 \ \end{vmatrix} = 0 \].

Step-by-step solution

Understand the Plane Equation

The general equation of a plane in three-dimensional space is given by \( a x + b y + c z = d \), where \(a\), \(b\), \(c\), and \(d\) are constants.

Set Up the Equations for Four Points

Let the coordinates of the four points be \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), \( (x_3, y_3, z_3) \), and \( (x_4, y_4, z_4) \). Since all points lie on the plane, they satisfy the plane equation: \( a x_1 + b y_1 + c z_1 = d \), \( a x_2 + b y_2 + c z_2 = d \), \( a x_3 + b y_3 + c z_3 = d \), and \( a x_4 + b y_4 + c z_4 = d \).

Form the Matrix

Represent these equations in matrix form: \[\begin{pmatrix} x_1 & y_1 & z_1 & 1 \ x_2 & y_2 & z_2 & 1 \ x_3 & y_3 & z_3 & 1 \ x_4 & y_4 & z_4 & 1 \ \end{pmatrix} \begin{pmatrix} a \ b \ c \ -d \ \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ \end{pmatrix}\]}

Condition for Consistent System

For this system of linear equations to have a non-trivial solution (where \(a, b, c, d\) are not all zero), the determinant of the coefficient matrix must be zero. The determinant is given by: \[ \begin{vmatrix} x_1 & y_1 & z_1 & 1 \ x_2 & y_2 & z_2 & 1 \ x_3 & y_3 & z_3 & 1 \ x_4 & y_4 & z_4 & 1 \ \end{vmatrix} = 0 \]}

Key concepts

Determinants in Linear Algebra

Determinants play a crucial role in understanding whether a set of points in space lies on a common plane. A determinant is a special number calculated from a square matrix. It provides useful information about the matrix, such as whether it's invertible.
The determinant of a matrix captures the volume scaling factor of the linear transformation described by the matrix. For instance, consider a 2x2 matrix: \ \ \ \ In this problem, we use a 4x4 matrix containing the coordinates of four points \( (x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3), (x_4, y_4, z_4) \) with each fourth entry being 1. When we set up a determinant for these points, we check if the volume of the parallelotope (a multi-dimensional box) they determine is zero.
If the determinant of this 4x4 matrix is zero, this means that the volume is zero, indicating that the points lie on the same plane, making them coplanar.

Plane Equations in 3D Space

The general equation of a plane in three-dimensional space is given by \( ax + by + cz = d \), where \(a, b, c, d \) are constants. This equation describes all the points \((x, y, z) \) that lie on the plane.
Points that lie on the plane must satisfy this equation.
If you have four points \((x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3), (x_4, y_4, z_4)\), they all need to satisfy the equation of the plane. By solving the system of equations obtained by substituting the points into the plane equation, we can verify if they are coplanar. The challenge is to express this using a determinant.
We represent the plane equation for each point and construct a coefficient matrix. If the determinant of this matrix is zero, it indicates a specific relation among the points, confirming that they lie on the same plane.

Matrix Representation

Matrices give us a compact way to handle multiple linear equations. Here, a 4x4 matrix is formed by putting the coordinates of the four points and appending a 1 to each row. This matrix captures the plane equation for each point in a compact form:
\[\begin{pmatrix} x_1 & y_1 & z_1 & 1 \ x_2 & y_2 & z_2 & 1 \ x_3 & y_3 & z_3 & 1 \ x_4 & y_4 & z_4 & 1 \ \end{pmatrix}\begin{pmatrix} a \ b \ c \ -d \ \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ \end{pmatrix}\]
The product of this matrix with \(a, b, c, -d \) equals a zero vector, implying that for a non-trivial solution to exist (where not all constants are zero), the determinant of this matrix must be zero.
Setting the determinant of this matrix to zero provides a neat condition for coplanarity of the points, ensuring that they indeed lie on a common plane. This powerful use of determinants and matrices streamlines complex spatial problems to more manageable algebraic forms.