Lesson 04

Intersections and Consistent Systems

Every question in this lesson is the same question: solve the equations, and then say what the answer looks like. A unique solution is a point, infinitely many solutions are a line or a plane, and no solution at all means the objects miss each other. The algebra and the geometry are two readings of one page.

By the end of this lesson

  • Classify two lines as intersecting, parallel, coincident or skew.
  • Classify and solve the intersection of a line and a plane.
  • Find the line of intersection of two planes, and recognise when there is none.
  • Use consistent and inconsistent correctly, and say what each case looks like.

Section 1

Two lines

In two-space, two lines either cross once, are parallel and distinct, or are the same line: three cases, and that is all. Three-space adds a fourth that has no two-dimensional analogue. Skew lines are not parallel and yet never meet — they pass one another at different heights, like an overpass and the road beneath it. Skew lines are impossible in R², because two non-parallel lines drawn on the same flat sheet are forced to cross.

Before any algebra, compare the direction vectors. If they are scalar multiples, the lines are parallel, and a single point test settles which of the two parallel cases you have. If they are not scalar multiples, the lines are not parallel, and the only question left is whether they actually meet.

The four possible relationships between two lines in three-space Four small panels. Intersecting: two lines crossing at a marked point. Parallel: two lines that never meet. Coincident: two lines lying exactly on top of one another, the second drawn dashed. Skew: two lines that do not meet and are not parallel, with a break showing one passing behind the other. intersecting parallel coincident skew
fig. 4.1The four cases. In the last one the break shows the front line passing over the back one — they never touch.

The method, and the trap in it

Given \(L_1: \vv{r} = P_1 + s\vv{m}_1\) and \(L_2: \vv{r} = P_2 + t\vv{m}_2\), set the two position vectors equal and solve. The two lines must use different parameters. Writing \(t\) in both equations quietly demands that the two points arrive at the same moment, not merely at the same place, and lines that genuinely intersect will then appear not to.

  1. Are the directions scalar multiples? If yes, test whether \(P_2\) lies on \(L_1\): it does ⇒ coincident; it does not ⇒ parallel and distinct. Stop.
  2. Otherwise write \(P_1 + s\vv{m}_1 = P_2 + t\vv{m}_2\), which is three equations in two unknowns.
  3. Solve any two of them for \(s\) and \(t\).
  4. Substitute into the third. If it holds, the lines intersect, and the point is \(P_1 + s\vv{m}_1\). If it fails, the lines are skew. Skipping this step is how a skew pair gets reported as intersecting.

There is a one-line alternative that settles intersecting-versus-skew without solving anything. Two non-parallel lines meet if and only if they are coplanar, and three vectors are coplanar exactly when their scalar triple product vanishes:

\[\overrightarrow{P_1P_2} \cdot (\vv{m}_1 \times \vv{m}_2) = 0 \quad\Longleftrightarrow\quad \text{coplanar} \quad\Longleftrightarrow\quad \text{the lines meet}\]

The triple product is the volume of the parallelepiped built on those three vectors; zero volume means all three lie in one plane. It is the fastest check to write, but it does not produce the point of intersection — so use it to decide, and the substitution method to answer.

Classification of two lines in R³
CaseDirectionsPoint testSolutionsSystem
Intersectingnot parallelall three equations agreeexactly oneconsistent
Parallel & distinctparallel\(P_1 \notin L_2\)noneinconsistent
Coincidentparallel\(P_1 \in L_2\)infinitely manyconsistent
Skew (R³ only)not parallelthe third equation failsnoneinconsistent, and non-coplanar
Worked · the skew case in full

Classify \(L_1: \vv{r} = (1,1,2)+s(1,2,1)\) and \(L_2: \vv{r} = (4,2,2)+t(2,-1,3)\)

Show solution
  1. Parallel? \(\tfrac{1}{2} \neq \tfrac{2}{-1}\), so the directions are not scalar multiples. Not parallel, not coincident.
  2. Set the two equal, one equation per coordinate: \(x:\ 1+s = 4+2t,\qquad y:\ 1+2s = 2-t,\qquad z:\ 2+s = 2+3t\)
  3. Solve the \(x\)- and \(y\)-equations. From \(x\): \(s = 3+2t\). Into \(y\): \(1 + 2(3+2t) = 2-t \Rightarrow 7+4t = 2-t \Rightarrow 5t = -5 \Rightarrow t = -1\), so \(s = 1\).
  4. Test the \(z\)-equation — the whole point of the method. Left side \(2+s = 3\); right side \(2+3t = -1\). Since \(3 \neq -1\), no pair \((s,t)\) satisfies all three. The lines are skew.
  5. Independent confirmation. \(\overrightarrow{P_1P_2} = (3,1,0)\) and \(\vv{m}_1\times\vv{m}_2 = (7,-1,-5)\), so the triple product is \(21 - 1 + 0 = 20 \neq 0\): not coplanar, therefore skew. ✓

Lesson 05 finishes the story by measuring how far apart they are.

Section 2

A line and a plane: the three cases

To intersect a line with a plane, substitute the line's parametric equations into the plane's Cartesian equation and solve for \(t\). Do that in general once, and the three cases stop being three separate things to remember. Writing the line as \(\vv{r} = \vv{r}_0 + t\vv{m}\) and the plane as \(\vv{n}\cdot\vv{r} + D = 0\), substitution gives

\[(\vv{m}\cdot\vv{n})\,t + (\vv{n}\cdot\vv{r}_0 + D) = 0\]

a single linear equation in \(t\). Its behaviour is decided entirely by whether the two brackets vanish, and that is the whole classification.

The three possible relationships between a line and a plane Three panels, each showing a shaded plane. First, a line piercing the plane at one marked point, dashed where it passes behind. Second, a line floating above the plane and parallel to it, with its shadow drawn on the plane below and dashed drop lines connecting the two. Third, a line lying entirely within the plane. one point no points infinitely many
fig. 4.2One point, no points, infinitely many. Everything below is a way of telling which picture you are in without drawing it.
Case 1 The line crosses the plane \(\vv{m}\cdot\vv{n} \neq 0\), so the equation has a unique solution \(t\) and there is exactly one common point. Consistent · 1 solution.
Case 2 Parallel and distinct \(\vv{m}\cdot\vv{n} = 0\) and a point of the line fails the plane's equation. Substitution collapses to \(0 = k,\ k\neq 0\) — a contradiction. Inconsistent · 0 solutions.
Case 3 The line lies in the plane \(\vv{m}\cdot\vv{n} = 0\) and a point of the line satisfies the equation. Substitution collapses to \(0 = 0\) — every \(t\) works. Consistent · ∞ solutions.

Notice the shape of the decision. \(\vv{m}\cdot\vv{n}\) is a pre-test: compute it first, and it tells you immediately whether you are in Case 1 or in one of the two parallel cases. It never distinguishes Cases 2 and 3 — only a point test does that. Students who stop after finding \(\vv{m}\cdot\vv{n}=0\) and announce “no intersection” have answered a question that was not asked.

Case explorer — watch the substitution collapse Interactive

Choose a line and a plane, and the substitution is carried out line by line, ending in a unique \(t\), in \(0=k\), or in \(0=0\) — with the classification stated underneath.

Section 3

Two planes

Compare the normals first. If \(\vv{n}_1\) and \(\vv{n}_2\) are not parallel, the planes are not parallel, and two non-parallel planes in R³ always meet — in a line, never in a point. That line's direction is \(\vv{d} = \vv{n}_1 \times \vv{n}_2\), because it must lie in both planes and so be perpendicular to both normals.

If the normals are parallel, one point test separates the two remaining cases exactly as it did for lines: the planes are coincident if their equations are proportional throughout, and parallel and distinct if only the left-hand sides are.

To produce the line, eliminate one variable between the two equations, then set one of the remaining variables equal to a parameter. Always check your answer by substituting the parametrised point back into both plane equations and confirming that the parameter cancels.

Two planes meeting in a line Two shaded sheets crossing one another. Their common points form a single straight line, drawn in the accent colour and labelled with its vector equation. π2 π1 r = r0 + td
fig. 4.3Non-parallel normals mean the planes cut one another in a line. Its direction is the cross product of the two normals.
Two parallel and distinct planes Two shaded sheets, one floating above the other, sharing a single normal vector drawn perpendicular to both. They never meet, so the system of their two equations has no solution. π1 π2 n
fig. 4.4Parallel and distinct: one normal, two constants, and no point in common.

The parallel-and-distinct case is the geometric meaning of an inconsistent system, and it is worth being able to describe in words as well as symbols: two flat sheets sharing a normal arrow, one floating above the other, never touching. There is no clever algebra hiding a solution — there is genuinely nowhere for a common point to be.

That description is a Communication answer in itself. When a question says “explain what it means geometrically”, the marks are for the picture, not for repeating that the equations have no solution.

Two-plane solver Interactive

Enter two plane equations and watch the elimination worked out symbolically, ending in a vector equation for the line of intersection — or in a statement that the planes are parallel.

Section 4

Consistency, in one table

A system of equations is consistent if it has at least one solution and inconsistent if it has none. Note what that does not say: infinitely many solutions is a perfectly consistent outcome. The only inconsistent situations in this topic are the ones where two objects run alongside each other without ever touching.

Solution counts and consistency for every case in this lesson
ObjectsGeometryAlgebra collapses toSolutionsSystem
Line, lineintersectinga unique \((s,t)\)1consistent
Line, linecoincidentevery equation an identity∞consistent
Line, lineparallel distincta contradiction0inconsistent
Line, lineskewthird equation fails0inconsistent
Line, planecrossinga unique \(t\)1consistent
Line, planeline in plane\(0=0\)∞consistent
Line, planeparallel distinct\(0=k,\ k\neq0\)0inconsistent
Plane, planemeeting in a lineone free parameter∞consistent
Plane, planecoincidenttwo free parameters∞consistent
Plane, planeparallel distincta contradiction0inconsistent

Two planes can never meet in exactly one point. If a question appears to ask for that, either a third plane is hiding somewhere or the answer is “impossible” — and saying so earns the mark.

Practice

Worked examples

Example 1 · routine

Where a line meets a plane

Find the intersection of \(\vv{r} = (2,-1,0)+t(1,2,-1)\) with \(x+y+z=5\).

Show solution
  1. Pre-test: \(\vv{m}\cdot\vv{n} = 1+2-1 = 2 \neq 0\), so expect exactly one point.
  2. Parametric: \(x = 2+t,\ y = -1+2t,\ z = -t\).
  3. Substitute: \((2+t) + (-1+2t) + (-t) = 5 \Rightarrow 1 + 2t = 5 \Rightarrow t = 2\).
  4. The point is \((4,\,3,\,-2)\). Check: \(4+3-2=5\ \checkmark\) — one solution, consistent.
Example 2 · standard

Twin problems: \(0=k\) against \(0=0\)

Take the line \(\vv{r} = (1,1,1)+t(2,-1,1)\) against two planes that differ only in their constant: \(x+y-z=3\) and \(x+y-z=1\).

Show solution
  1. Both planes share the normal \(\vv{n}=(1,1,-1)\), and \(\vv{m}\cdot\vv{n} = 2-1-1 = 0\). The line is parallel to both planes, so neither can give a single crossing point.
  2. Test the line's point \((1,1,1)\) in the first plane: \(1+1-1 = 1 \neq 3\). Substituting the whole line gives \(0\cdot t = 2\) — a contradiction. Parallel and distinct; no solution; inconsistent.
  3. Test the same point in the second plane: \(1+1-1 = 1\ \checkmark\). Substituting gives \(0\cdot t = 0\). The line lies in the plane; infinitely many solutions; consistent.
  4. The only difference between the two problems was the number on the right. That is how thin the line between “no solutions” and “infinitely many” is — and why the point test is not optional.
Example 3 · standard

The line of intersection of two planes

Find where \(\pi_1: 2x-y+z=3\) and \(\pi_2: x+y-z=0\) meet.

Show solution
  1. The normals \((2,-1,1)\) and \((1,1,-1)\) are not parallel, so the planes meet in a line.
  2. Add the two equations to eliminate both \(y\) and \(z\) at once: \(3x = 3 \Rightarrow x = 1\).
  3. Substitute into \(\pi_2\): \(1 + y - z = 0 \Rightarrow y = z-1\). Let \(z = t\).
  4. \(\therefore\ \vv{r} = (1,-1,0) + t(0,1,1)\).
  5. Cross-check the direction: \(\vv{n}_1\times\vv{n}_2 = (0,3,3) \sim (0,1,1)\ \checkmark\)
  6. Check the point in both planes: \(\pi_1:\ 2+1+0=3\ \checkmark\qquad \pi_2:\ 1-1-0=0\ \checkmark\)
Example 4 · demanding

Working backwards to find a constant

For which value of \(k\) is the system \(2x+ky-z=3\), \(4x+6y-2z=1\) inconsistent? Is there a value making the planes coincident?

Show solution
  1. Two planes are inconsistent only when they are parallel and distinct, so first make the normals parallel: \((4,6,-2) = 2(2,3,-1)\), so we need \((2,k,-1) \parallel (2,3,-1) \Rightarrow k = 3\).
  2. With \(k=3\), doubling the first equation gives \(4x+6y-2z=6\), but the second says the same left-hand side equals \(1\). Since \(6 \neq 1\), the planes are parallel and distinct: inconsistent when \(k=3\).
  3. For coincidence the constants would have to scale by the same factor 2, needing \(2\times 3 = 1\). They do not, so no value of \(k\) makes the planes coincident.
  4. For every other \(k\) the normals are not parallel, so the planes meet in a line and the system is consistent with infinitely many solutions.

Watch out

Four ways to lose a mark here
  • One parameter per line. Two lines need \(s\) and \(t\). Reusing \(t\) asks whether the lines meet at the same instant, which is a different — and much rarer — event than meeting at the same place.
  • Always test the third equation. Any two of the three coordinate equations can be solved; it is the one you did not use that decides between intersecting and skew.
  • \(\vv{m}\cdot\vv{n}=0\) does not mean “no intersection”. It means parallel. Whether the line misses the plane or lies inside it is settled by a point test, and never by the dot product alone.
  • Infinitely many solutions is consistent. Only a genuine contradiction — \(0=k,\ k\neq 0\) — makes a system inconsistent.