1.2 Three Planes: Point, Line, or Nothing

Add a third equation and the story sharpens. Three planes in space usually pin down a single point: the first two intersect in a line, and the third plane crosses that line in one spot. But "usually" is doing real work in that sentence. This section is about the exceptional cases — and about seeing exactly how special they are.

The figure below shows the planes $z = x + 1$ and $z = y - 1$ , whose intersection is the dashed amber line, together with a third plane $z = 0.5 x + 0.5 y + d$ controlled by the slider $d$ . A fourth plane, a tilted replacement for the third, is included but hidden.

Three semi-transparent planes and a dashed amber line marking where the first two planes cross. A slider named d raises and lowers the third plane; at d = 0 the third plane swallows the amber line and all three planes share it, while at any other d the third plane misses the line and the system has no common point. A hidden tilted variant of the third plane can be revealed to show the unique-solution case. — Two fixed planes meet along the dashed amber line. The slider $d$ raises and lowers the third plane; only at $d equals 0$ does it capture the line. Toggle the hidden tilted variant to see the generic one-point case.

Explore the figure

Start at $d equals 0.9$ . Orbit and check each pair of planes: every pair crosses in a line. Now look for a single point on all three planes. Can you find one?
Drag $d$ toward $0$ and watch the third plane sweep down onto the amber line. At exactly $d equals 0$ , what is the solution set of the system — a point, a line, or nothing?
Push $d$ below zero. The moment $d$ leaves $0$ , where does the third plane sit relative to the amber line?
Hide the third plane and reveal the hidden tilted variant $z = - x - y + d$ . Now drag $d$ again. Does the system ever lose its solution? What shape is the solution set at each $d$ ?

Why the third plane is special

Write the three equations with the unknowns on one side: $x - z = - 1$ , then $y - z = 1$ , then $0.5 x + 0.5 y - z = - d$ . Take half of the first equation plus half of the second: the left side is $0.5 x + 0.5 y - z$ — exactly the left side of the third equation — while the right side is $0.5 (- 1) + 0.5 (1) = 0$ . So the first two equations already force $0.5 x + 0.5 y - z = 0$ at every one of their common points. The third equation demands that this same quantity equal $- d$ . If $d not equal to 0$ the demands clash and no point can satisfy all three: the system is inconsistent, even though every pair of planes meets happily. If $d equals 0$ the third equation is redundant — it adds no new information — and the solution set stays the entire amber line $(t - 1, t + 1, t)$ .

The tilted variant behaves differently because its left side, $- x - y - z$ shifted to standard form as $x + y + z = d$ , is not a combination of the first two left sides. Three independent conditions in three unknowns cut the solution down to a single point, and sliding $d$ merely moves that point.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.