Three Planes with One Common Point

Three Plane Equations, One Point

A plane in \(\mathbb{R}^3\) can be written as a linear equation in \(x\), \(y\), and \(z\). When three planes are given, a point belongs to all three exactly when its coordinates satisfy all three equations at once. The geometric question—where the planes meet—therefore becomes a question about the solutions of a system of three linear equations.

For example, a point that lies on the first and second planes but not the third is not a common point of all three. To establish that there is exactly one common point, we must do two things: find a point satisfying every equation, and show that no other point satisfies them all. In the systems below, elimination produces equations that determine each coordinate in turn, which lets us establish both facts.

The equivalence in the definition follows directly from what it means for a point to lie on a plane: its coordinates must make that plane’s equation true. Thus, solving the system is not a separate geometric approximation; it finds the coordinates of the common point itself.

Eliminating Equations Without Changing Their Solutions

To solve a three-equation system, we can add a multiple of one equation to another. The resulting equation may look different, but it imposes the same combined conditions when the unchanged equations are retained. The reason is that this operation is reversible.

Proof. Suppose the two equations involved are \(L_1=d_1\) and \(L_2=d_2\), where each \(L_i\) is a linear expression in the variables. Replace the first equation by \(L_1+cL_2=d_1+cd_2\), keeping \(L_2=d_2\). Any solution of the original system satisfies the replacement, because adding \(c\) times the equality \(L_2=d_2\) to \(L_1=d_1\) gives \(L_1+cL_2=d_1+cd_2\). Conversely, any solution of the new system satisfies \(L_2=d_2\), so subtracting \(cL_2=cd_2\) from the replacement gives \(L_1=d_1\). All other equations are unchanged, so the two systems have exactly the same solutions. \(\square\)

We can also swap two equations without changing which points satisfy all of them, because the order in which conditions are listed does not affect whether they all hold. These observations justify the equation combinations used in the examples. The goal is to simplify the conditions, not to alter the set of points they describe.

Why a Triangular System Determines One Point

In the example, elimination produced equations that could be solved one coordinate at a time. This pattern is called a triangular system: the last equation determines one variable, the equation above it then determines a second, and the remaining equation determines the last. The nonzero coefficients are important, because division by zero cannot determine a variable.

Proof. We first show that a solution exists. Since \(a_{33}\ne0\), set \(z=b_3/a_{33}\); this makes the third equation true. Since \(a_{22}\ne0\), set \(y=(b_2-a_{23}z)/a_{22}\); this makes the second equation true with the chosen \(z\). Since \(a_{11}\ne0\), set \(x=(b_1-a_{12}y-a_{13}z)/a_{11}\); this makes the first equation true with the chosen \(y\) and \(z\). These choices therefore give a solution of all three equations. To show uniqueness, consider any solution. Its \(z\)-coordinate must equal \(b_3/a_{33}\) by the third equation. With \(z\) fixed, its \(y\)-coordinate must equal \((b_2-a_{23}z)/a_{22}\) by the second. With both \(y\) and \(z\) fixed, its \(x\)-coordinate must equal \((b_1-a_{12}y-a_{13}z)/a_{11}\) by the first. Every solution must therefore have the same three coordinates. There is one solution, and no other. \(\square\)

This theorem applies directly when elimination has produced triangular equations. The equation operations must preserve the solution set, as established above; otherwise, the unique solution of the simplified system would not necessarily describe the original planes. The examples keep track of that equivalence while solving.

One More Example: Elimination in a Different Order

The variables do not have to be determined in the same order in every system. Sometimes a combination of equations isolates a variable first, after which the remaining equations determine the others. As long as each operation is reversible and the final equations determine all three coordinates, the same existence-and-uniqueness reasoning applies.

Common Pitfall: Stopping After Two Equations

A common mistake is to solve two of the equations, find a point satisfying them, and declare it the common point without checking the third. The third equation is an additional constraint; it must also hold. Even when elimination has produced a plausible coordinate triple, substituting it into every original equation is an important check.

There is also room for sign errors during elimination. In the first worked example, subtracting the second equation from the first gives \(2y=4\), and the third equation then gives \(x-z=-1\). A mistaken sign could change this last equation to \(x-z=1\). Combining that incorrect equation with \(x+z=4\) would produce \(x=\tfrac52\) and \(z=\tfrac32\). But this proposed point fails the actual third equation: \(\tfrac52+2(2)-\tfrac32=5\), not \(3\). The correction is to preserve the sign in \(x+4-z=3\), which simplifies to \(x-z=-1\), giving \(x=\tfrac32\) and \(z=\tfrac52\). Substitution in all three original equations catches the error.

For a system that has been reduced to triangular form with nonzero coefficients on the successively determined variables, the theorem guarantees exactly one solution. If elimination does not produce that pattern, do not assume uniqueness from the appearance of the original equations alone. The important conclusion in this tutorial is specific: when the equations determine each coordinate and the resulting point satisfies all three plane equations, those planes have exactly one common point.