Linear Algebra, Interactive EditionXYZ Homework Edition

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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+1z = x + 1 and z=y−1z = y - 1, whose intersection is the amber line, together with a third plane z=0.5x+0.5y+dz = 0.5x + 0.5y + d controlled by the slider dd. A second figure, after the Explore steps, replaces the third plane with a tilted one.

Three semi-transparent planes and an amber line marking where the first two planes cross. A slider named d raises and lowers the third plane; when d is 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. The next figure replaces the third plane with a tilted one to show the unique-solution case.Explore in 3D (opens in a new tab)

Three semi-transparent planes and an amber line marking where the first two planes cross. A slider named d raises and lowers the third plane; when d is 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. The next figure replaces the third plane with a tilted one to show the unique-solution case.

Two fixed planes meet along the amber line. The slider dd raises and lowers the third plane; only at d=0d = 0 does it capture the line. The second figure, below the Explore steps, shows the generic one-point case.

Explore the figure

  1. Start at d=0.9d = 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?
  2. Drag dd toward 00 and watch the third plane sweep down onto the amber line. At exactly d=0d = 0, what is the solution set of the system — a point, a line, or nothing?
  3. Push dd below zero. The moment dd leaves 00, where does the third plane sit relative to the amber line?
  4. Turn to the second figure, just below this list, where the tilted plane z=−x−y+dz = -x - y + d replaces the third plane and a small gray ball marks where it crosses the amber line. Drag its slider dd. Does the system ever lose its solution? What shape is the solution set at each dd?
The same two planes as the previous figure, crossing along an amber line, with the third plane replaced by a tilted plane that the slider d slides back and forth. The tilted plane cuts straight across the amber line, and a small gray ball marks the single point where it crosses. At every setting of d the three planes share exactly that one point; moving d only slides the point along the amber line.Explore in 3D (opens in a new tab)

The same two planes as the previous figure, crossing along an amber line, with the third plane replaced by a tilted plane that the slider d slides back and forth. The tilted plane cuts straight across the amber line, and a small gray ball marks the single point where it crosses. At every setting of d the three planes share exactly that one point; moving d only slides the point along the amber line.

The same two planes and their amber line, with the third plane replaced by the tilted plane z=−x−y+dz = -x - y + d. The slider dd slides the tilted plane along the amber line, and the small gray ball marks the one point where it crosses that line.

Why the third plane is special

Write the three equations with the unknowns on one side: x−z=−1x - z = -1, then y−z=1y - z = 1, then 0.5x+0.5y−z=−d0.5x + 0.5y - z = -d. Take half of the first equation plus half of the second: the left side is 0.5x+0.5y−z0.5x + 0.5y - z — exactly the left side of the third equation — while the right side is 0.5(−1)+0.5(1)=00.5(-1) + 0.5(1) = 0. So the first two equations already force 0.5x+0.5y−z=00.5x + 0.5y - z = 0 at every one of their common points. The third equation demands that this same quantity equal −d-d. If d≠0d \ne 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=0d = 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)(t - 1, \; t + 1, \; t).

The tilted variant behaves differently because its left side, −x−y−z-x - y - z shifted to standard form as x+y+z=dx + 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 dd merely moves that point.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Chapters 1–6 follow the topic sequence of Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL). Chapter 0's scope (its topics, their order, and the idea each figure shows) follows the Vectors chapter of Ximera's Linear Algebra: An Interactive Introduction. The prose, worked examples and figures were written for this book, which is not affiliated with or endorsed by the authors of either work. License: CC-BY-NC-SA-4.0.

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