When Three Planes Have No Common Point
Three planes in \(\mathbb{R}^3\) may meet at one point, share a line, or even coincide. They may also have no point in common. That last possibility has more than one geometric form: two of the planes may be distinct parallel planes, or the planes may meet pairwise in lines that do not all pass through the same point. In either case, their three equations form an inconsistent system: there is no single triple of coordinates satisfying all three equations.
A plane in \(\mathbb{R}^3\) has an equation \(ax+by+cz=d\), where \(a,b,c,d\in\mathbb{R}\) and at least one of \(a,b,c\) is nonzero. The vector \((a,b,c)\) gives a normal direction to the plane. Two planes are parallel when their normal vectors are proportional. If their normal vectors are proportional but the constants do not have the same proportionality, the planes are distinct and disjoint; if the constants do have that proportionality, the equations describe the same plane.
For instance, the equations \(x+2y-z=1\) and \(2x+4y-2z=5\) have proportional normal vectors, but doubling the first equation would give \(2x+4y-2z=2\), not \(5\). They therefore describe distinct parallel planes. A third plane cannot make those two planes intersect: any point common to all three would have to lie on both of the disjoint planes.
Two Ways the Common Intersection Can Fail
The first arrangement is identified by a parallel pair. If one plane equation is multiplied by a nonzero number to match the other equation’s left-hand side, but the right-hand sides do not match, the two planes are distinct and parallel. This already rules out a common point for all three planes.
The second arrangement can be less obvious. Each pair of planes may intersect in a line, while the three lines are parallel and distinct. Each pair then has points in common, but no point lies on all three planes. The word pairwise matters: an intersection point for one pair need not be an intersection point for another pair.
Proof. Let \(L_{12}\) be the intersection line of planes 1 and 2, and let \(L_{13}\) be the intersection line of planes 1 and 3. Both lines lie in plane 1. If they met at a point, that point would lie on planes 1, 2, and 3, contradicting the assumption that there is no common point. They are therefore disjoint. Two disjoint lines in the same plane are parallel, so \(L_{12}\) and \(L_{13}\) are distinct and parallel. The same reasoning in plane 2 shows that \(L_{12}\) and \(L_{23}\) are distinct and parallel. Hence all three pairwise intersection lines are parallel and distinct. \(\square\)
This result describes the geometry once all three pairs do intersect. If a pair consists of distinct parallel planes, that pair itself prevents any common point. If instead every pair meets and there is no common point, the theorem explains why the pairwise intersection lines do not come together at one location.
Turning the Geometry into a Contradiction
A direct algebraic way to prove inconsistency is to combine equations so that the variable terms cancel but the constants do not. Such a contradiction cannot hold at any point.
Proof. If a point satisfied every equation in the system, then it would also satisfy any linear combination of those equations: multiply each equation by its chosen scalar and add the resulting equalities. But the resulting equation \(0=c\), with \(c\ne0\), is false for every choice of the variables. No point can therefore satisfy all the original equations. \(\square\)
This is a sufficient test: finding a contradiction proves inconsistency. The examples below use it in two different geometric arrangements. In each example, the equations are checked against the claimed intersections or contradiction.
Worked Example: A Distinct Parallel Pair
Consider the three planes
The first two normal vectors are \((1,2,-1)\) and \((2,4,-2)\). The second is twice the first, so the planes are parallel. However, doubling the first equation gives \(2x+4y-2z=2\), whereas the second requires \(2x+4y-2z=5\). The planes are distinct. In particular, no point can satisfy both equations.
The contradiction can also be displayed by subtracting twice the first equation from the second. The left-hand sides cancel, while the right-hand sides give
Since \(0=3\) is false, the system is inconsistent. The third plane cannot change that conclusion. As a check that the third plane does meet the first, set \(z=0\). The third equation gives \(x=y\), and the first then gives \(3y=1\), so \((x,y,z)=(\tfrac13,\tfrac13,0)\) lies on both. Substitution confirms the first left-hand side is \(\tfrac13+2(\tfrac13)=1\), and the third is \(\tfrac13-\tfrac13=0\). At this point the second left-hand side is \(\tfrac23+\tfrac43=2\), not \(5\), as expected.
Worked Example: Three Pairwise Intersections, No Common Point
Consider
The first two planes intersect where \(x=0\) and \(y=0\); \(z\) can take any value. The first and third intersect where \(x=0\) and \(y=1\), again with any \(z\). The second and third intersect where \(y=0\) and \(x=1\), also with any \(z\). Thus every pair has an intersection line.
But no point lies on all three. The first two equations together force \(x+y=0\), while the third requires \(x+y=1\). Adding the first two equations and comparing with the third produces the contradiction \(0=1\).
Check: A point on the first and second planes has the form \((0,0,z)\), and its value of \(x+y\) is \(0\), not \(1\). A point on the first and third has the form \((0,1,z)\), which does not satisfy \(y=0\). A point on the second and third has the form \((1,0,z)\), which does not satisfy \(x=0\). The three intersection lines are distinct and parallel to the \(z\)-axis.
Pairwise Intersection Lines That Are Parallel
The same pattern occurs even when no two of the planes are parallel. Consider three planes whose equations constrain \(x+y\) and \(z\) in different ways. Each pair can meet, but the pairwise intersection lines can still be parallel and distinct.
Worked Example: No Parallel Pair, Still Inconsistent
Consider
Adding the first two equations gives \(2x+2y=0\), or \(x+y=0\). The third equation requires \(x+y=1\). Subtracting these requirements gives \(0=1\), so there is no common point.
To check the pairwise geometry, the first and second equations together give \(z=0\) and \(x+y=0\). The first and third give \(x+y=1\) and \(z=-1\). The second and third give \(x+y=1\) and \(z=1\). Each pair therefore intersects in a line, and each line has direction \((1,-1,0)\), since changing \(x\) by an amount and \(y\) by its negative preserves \(x+y\). These are distinct parallel lines. The planes themselves are not parallel: their normal vectors \((1,1,1)\), \((1,1,-1)\), and \((1,1,0)\) are not scalar multiples of one another.
Check: The point \((0,0,0)\) satisfies the first two equations but not the third. The point \((0,1,-1)\) satisfies the first and third, because \(0+1-1=0\) and \(0+1=1\), but it does not satisfy the second, whose left-hand side is \(0+1-(-1)=2\). The point \((1,0,1)\) satisfies the second and third, because \(1+0-1=0\) and \(1+0=1\), but it does not satisfy the first, whose left-hand side is \(1+0+1=2\).
Common Pitfalls
Pairwise intersections do not guarantee a common point. In the second example, \((0,0,0)\) lies on the first two planes, \((0,1,0)\) lies on the first and third, and \((1,0,0)\) lies on the second and third. It would be wrong to conclude that the planes have a common point just because each pair intersects. Those are three different points, and none lies on all three planes.
Proportional normal vectors do not by themselves mean coincident planes. For example, \(x+2y-z=1\) and \(2x+4y-2z=5\) have proportional normal vectors, but their constants do not have the same factor of \(2\). They are distinct parallel planes, not two equations for the same plane. Check the constants as well as the coefficients of the variables.
When checking a proposed common point, substitute its coordinates into every original equation. A point that satisfies only two equations establishes an intersection of those two planes, not a solution of the three-equation system.
Check Your Understanding
For each question, distinguish the geometric claim from the algebraic test.
- Are \(x-y+z=2\) and \(2x-2y+2z=4\) coincident or distinct parallel planes? Explain how the constants affect your answer.
- Show that \(x+z=1\), \(y-z=2\), and \(x+y=5\) cannot all hold by combining equations.
- For \(x=0\), \(y=0\), and \(x+y=1\), write the intersection line for each pair of planes.
- Give a reason that three pairwise intersection points need not be one common point of all three planes.
- What geometric arrangement is forced if three planes intersect pairwise in lines but have no common point?
