Lesson 04

Intersections and Systems

Every question in this lesson asks for two things. Solve the equations, 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 picture always agree, so you can check one against the other.

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, four cases

In two-space, two lines either cross once, are parallel and distinct, or are the same line — there are only those three cases. Three-space adds a fourth that has no flat version. Skew lines are not parallel and they still 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 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.
Classification of two lines in three-space
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

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. That is three equations in two unknowns, so solve any two of them for \(s\) and \(t\), then substitute into the third. If it holds, the lines intersect at \(P_1 + s\vv{m}_1\); if it fails, they are skew.

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 is zero: \(\overrightarrow{P_1P_2} \cdot (\vv{m}_1 \times \vv{m}_2) = 0 \iff \text{coplanar} \iff \text{the lines meet}\). The triple product is the volume of the parallelepiped built on those three vectors, and 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 the case and then substitute to find the point.

Section 2

A line and a plane, in one substitution

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.

Substituting a line into a plane

\(\vv{r} = \vv{r}_0 + t\vv{m} \quad\text{into}\quad \vv{n}\cdot\vv{r} + D = 0\)
The line, written into the plane
\(\vv{n}\cdot(\vv{r}_0 + t\vv{m}) + D = 0\)
Expand the dot product
\((\vv{m}\cdot\vv{n})\,t + (\vv{n}\cdot\vv{r}_0 + D) = 0\)
One linear equation in \(t\) — the two brackets decide every case
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 \(0 = 0\), so every \(t\) works. Consistent · ∞ solutions.

The order of the two tests matters. \(\vv{m}\cdot\vv{n}\) is a pre-test: compute it first and it tells you whether you are in Case 1 or in one of the two parallel cases. It never separates Cases 2 and 3, so you still need a point test. Students who stop at \(\vv{m}\cdot\vv{n}=0\) 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, and the line they share

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 — they can never meet at a single 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 just 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 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. 4.2Non-parallel normals mean the planes cut one another in a line. Its direction is the cross product of the two normals.
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

Parallel and distinct, in words

The parallel-and-distinct case is the geometric meaning of an inconsistent system, and you should be able to describe it in words as well as in symbols: two flat sheets sharing a normal arrow, one floating above the other, never touching. No clever algebra is hiding a solution. There is 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. Saying again that the equations have no solution earns nothing.

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. 4.3Parallel and distinct: one normal, two constants, and no point in common.

Section 5

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 still a consistent outcome. The only inconsistent situations in this topic are the ones where two objects run alongside each other without ever touching.

Read the table the other way round as well. Every row pairs an algebraic collapse with a picture, and a question is usually asking for both.

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, lineskewthe third 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

Practice

Worked examples

Example 1 · routine

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

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

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

  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. Cross-check the direction against the normals: \(\vv{n}_1\times\vv{n}_2 = (0,3,3) \sim (0,1,1)\ \checkmark\), and the point in both planes: \(\pi_1:\ 2+1+0=3\ \checkmark\qquad \pi_2:\ 1-1-0=0\ \checkmark\)
\(\vv{r} = (1,-1,0) + t(0,1,1)\)
Example 3 · demanding

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

  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 \(y\): \(1 + 2(3+2t) = 2-t \Rightarrow 7+4t = 2-t \Rightarrow t = -1\), so \(s = 1\).
  4. Test the \(z\)-equation — this is the step that decides the answer. Left side \(2+s = 3\); right side \(2+3t = -1\). Since \(3 \neq -1\), no pair \((s,t)\) satisfies all three.
  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. ✓
The lines are skew — no solution, so the system is inconsistent, and unlike the parallel case the lines are not even coplanar. Lesson 05 finishes the story by measuring how far apart they are.

Watch out

Common mistakes

  • Reusing one parameter for two lines. Two lines need \(s\) and \(t\). Writing \(t\) twice asks whether the lines meet at the same instant, which is a different and much rarer event than meeting at the same place.
  • Stopping before 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.
  • Reading \(\vv{m}\cdot\vv{n}=0\) as “no intersection”. It means parallel. Whether the line misses the plane or lies inside it is settled by a point test, never by the dot product alone.
  • Calling infinitely many solutions inconsistent. Infinitely many solutions is consistent. Only a contradiction such as \(0=k,\ k \neq 0\) makes a system inconsistent.