Three Planes Intersecting at a Point (Math Proofs)

To prove that three planes meet at exactly one point, write their equations as a three-variable linear system. Show that the coefficient matrix has a nonzero determinant, solve the system, and substitute the answer into all three plane equations. The determinant proves uniqueness; the substitution proves that the common point exists.

It can be frustrating when a sketch seems to show one shared point, but the equations tell a different story. In three dimensions, planes can meet in several ways, and a picture alone cannot prove which one occurs. I use the same reliable order each time: set up the equations, test the system, solve it, and check the result.

Geometric setup: three planes as a linear system

A plane in (\mathbb R^3) can be written as (ax+by+cz=d). Its normal vector, ((a,b,c)), points perpendicular to the plane. For three planes, place their coefficients into a matrix; then their shared points are exactly the solutions of one system of three linear equations.

Write the equations as

[ \begin{aligned} a_1x+b_1y+c_1z&=d_1,\ a_2x+b_2y+c_2z&=d_2,\ a_3x+b_3y+c_3z&=d_3. \end{aligned} ]

The matching matrix form is

[ A\mathbf{x}=\mathbf{d}, \qquad A= \begin{bmatrix} a_1&b_1&c_1\ a_2&b_2&c_2\ a_3&b_3&c_3 \end{bmatrix}, \quad \mathbf{x}= \begin{bmatrix}x\y\z\end{bmatrix}, \quad \mathbf{d}= \begin{bmatrix}d_1\d_2\d_3\end{bmatrix}. ]

Each row of (A) is the normal vector of one plane. A point belongs to all three planes only if its coordinates satisfy all three equations at once. That last phrase matters: solving only two equations finds points on the intersection of two planes, not necessarily on the third.

Key step: Put the three equations in the same variable order, then use each equation as part of one system.

Core criterion: rank, determinants, and a unique point

The determinant of the coefficient matrix tells whether the three plane normals are independent. If (\det(A)\ne0), the matrix is invertible, so the system has exactly one solution for any right-hand side (\mathbf d). If the determinant is zero, the planes may have no common point or infinitely many.

For a (3\times3) matrix, a nonzero determinant is equivalent to rank 3. Rank is the number of independent rows or columns. In this setting, rank 3 means the three normal vectors point in independent directions; no one is a combination of the others.

The unique-intersection test is therefore

[ \det(A)\ne0 \quad\Longleftrightarrow\quad \operatorname{rank}(A)=3. ]

When this condition holds, (A^{-1}) exists, and

[ \mathbf{x}=A^{-1}\mathbf d. ]

This gives a solution for any (\mathbf d), and the solution is unique. Another way to see uniqueness is to compare two supposed solutions, (\mathbf{x}_1) and (\mathbf{x}_2). Subtracting their equations gives (A(\mathbf{x}_1-\mathbf{x}_2)=\mathbf 0). An invertible matrix has only the zero vector as a solution to this homogeneous system, so (\mathbf{x}_1=\mathbf{x}_2).

A zero determinant does not, by itself, tell you what happens. The rows of (A) are then dependent, but the right-hand sides may still make the full system inconsistent. Compare the rank of (A) with the rank of the augmented matrix ([A\mid\mathbf d]):

Here, ([A\mid\mathbf d]) is made by adding the constants column to (A). Row reduction can reveal these ranks and show whether the equations are consistent.

Key step: A nonzero determinant proves a unique solution. A zero determinant calls for a consistency check; it does not prove that the planes fail to meet.

Proof workflow: solve, verify, and establish uniqueness

A complete proof has two jobs: show that a common point exists and show that no second common point exists. I keep those jobs separate. First solve and check all three equations; then use the nonzero determinant, or an equivalent rank argument, to establish uniqueness.

Consider the planes

[ \begin{aligned} x+y+z&=6,\ 2x-y+z&=3,\ x+2y-z&=2. \end{aligned} ]

Their coefficient matrix and right-hand side are

[ A= \begin{bmatrix} 1&1&1\ 2&-1&1\ 1&2&-1 \end{bmatrix}, \qquad \mathbf d= \begin{bmatrix}6\3\2\end{bmatrix}. ]

Step 1: Test for uniqueness. Expanding the determinant along the first row gives

[ \det(A) =1\bigl((-1)(-1)-1(2)\bigr) -1\bigl(2(-1)-1(1)\bigr) +1\bigl(2(2)-(-1)(1)\bigr) =7. ]

Because (7\ne0), (A) is invertible. If this system is consistent, its solution must be unique; in fact, invertibility also guarantees that it is consistent for this right-hand side.

Step 2: Solve the equations. Subtract the first equation from the second:

[ x-2y=-3. ]

From the first equation, (z=6-x-y). Substitute that expression into the third equation:

[ x+2y-(6-x-y)=2, ]

which simplifies to (2x+3y=8). Using (x=2y-3), we get (7y=14), so (y=2), (x=1), and (z=3).

Step 3: Verify the point. Substitute ((1,2,3)) into each original equation:

[ 1+2+3=6,\qquad 2(1)-2+3=3,\qquad 1+2(2)-3=2. ]

All three equalities hold. This proves that the point lies on every plane. The nonzero determinant proves that no other point can satisfy all three equations.

A good written proof states both results. Finding ((1,2,3)) establishes existence, while (\det(A)=7) establishes uniqueness. Neither part should be left for the reader to infer.

Key step: Check the candidate in the original equations, not just in equations formed during elimination.

What pairwise intersections do not prove

Pairwise intersection means that each chosen pair of planes shares points. It does not guarantee that one point belongs to all three planes. To prove a common intersection, test the full three-equation system, not just three separate pairs.

For example, take

[ x=0,\qquad y=0,\qquad x+y=1. ]

The first and second planes intersect along the (z)-axis. The first and third intersect where (x=0) and (y=1), also a line. The second and third intersect where (y=0) and (x=1), again a line.

Yet no point lies on all three: the first two equations require (x+y=0), while the third requires (x+y=1). These conditions cannot both hold. This example shows why pairwise checks are not a substitute for solving the whole system.

Key step: A shared point must satisfy all three equations at the same time.

When the determinant is zero

A zero determinant means the normal vectors are dependent, so the coefficient matrix cannot have rank 3. The planes may still meet, but the result is not a unique point. Row-reduce the augmented matrix to decide whether the system has no solution or infinitely many.

For instance, the equations (x=0), (y=0), and (x+y=1) are inconsistent, as shown above. By contrast, (x=0), (y=0), and (x+y=0) all hold along the (z)-axis. There are infinitely many common points in the second case.

The rank test makes the distinction precise:

• If the augmented system produces a row such as ([0\ 0\ 0\mid 1]), it says (0=1). The system is inconsistent, so there is no common point.
• If there is no contradiction and fewer than three independent equations remain, at least one variable is free. The system has infinitely many solutions.
• If there are three independent equations, the solution is unique.

Key step: When (\det(A)=0), row-reduce ([A\mid\mathbf d]) before drawing a geometric conclusion.

Applications of plane intersections in geometry and modeling

Plane equations let you turn a geometric question into a system of linear equations. The same method applies whenever three linear constraints describe a point in three-dimensional space. The algebra tells you whether those constraints identify one point, conflict, or leave several possible points.

For example, in an analytic geometry problem, three plane equations may represent conditions that a point must meet. In a mathematical model, they may express three linear constraints on three unknown coordinates. The proof method stays the same: form the coefficient matrix, test its rank or determinant, solve, and verify.

A sketch can help you picture the planes, but it cannot replace these checks. In particular, a drawing may not show whether a small gap or a shared line is present. The equations provide the testable evidence.

Next step: If you are given three plane equations, write down (A) first. Its determinant tells you whether a unique intersection is possible.

FAQ

These answers summarize the tests used to decide whether three planes share exactly one point. The key distinction is between finding a candidate point and proving that it is the only one. Use the determinant or rank to test uniqueness, and use all three equations to check that the candidate is common to the planes.

How do you prove that three planes meet at exactly one point?
Write their equations as (A\mathbf{x}=\mathbf d). Show (\det(A)\ne0), solve the system, and verify the solution in all three original equations.

What does a nonzero determinant mean here?
It means the coefficient matrix is invertible and has rank 3. The system has exactly one solution, so the planes share exactly one point.

Does a zero determinant mean the planes do not intersect?
No. They may have no common point or infinitely many. Compare the ranks of the coefficient and augmented matrices to decide.

Is solving two of the plane equations enough?
No. Two equations can describe a line of points. You must check that a point also satisfies the third equation.

Why must I verify the point in all three equations?
The substitution confirms that the point lies on every plane. Checking fewer than three equations does not prove a common intersection.

Can three planes intersect pairwise but have no common point?
Yes. For (x=0), (y=0), and (x+y=1), every pair intersects in a line, but no point satisfies all three equations.

What is the role of the normal vectors?
The coefficient rows are the plane normal vectors. If they are linearly independent, the determinant is nonzero and the system has a unique solution.

Can I use row reduction instead of a determinant?
Yes. Row-reduce the augmented matrix. Three independent coefficient rows give a unique solution; a contradiction means no solution; fewer independent rows with no contradiction mean infinitely many solutions.

Does a diagram prove a unique intersection?
No. A diagram can suggest the geometry, but the algebra proves consistency and uniqueness. Use the equations and the determinant or rank test.

What should I write as the final conclusion?
State the verified point, note that it satisfies all three equations, and state that (\det(A)\ne0) proves it is the only common point.

(This article was written by one of our staff writers, Alan Westbrook. Visit our Meet the Team page.)