Paper 3 only. In a plane two lines either cross or run parallel. In space there is a third possibility: lines that pass each other at different heights, like a road on a bridge and the river beneath it.
First compare the directions: if one is a multiple of the other, the lines are parallel (or the same line). If not, set the two general points equal, solve two component equations for the two parameters, and test the third: it holds when the lines intersect and fails when they are skew.
Fig. 1: on a cuboid the edge AB runs along the length and the edge CD runs up the back; they point in different directions yet never meet, so they are skew
Section 01
Three possible relationships.
Picture the edges of a box (Fig. 1). Two edges along the length are parallel; two edges meeting at a corner intersect; an edge along the length and an edge up the far side are skew. Every pair of lines in space falls into exactly one of these cases (or the two equations describe the same line).
Directions
Common point?
Relationship
parallel (\(\mathbf{b}_2 = \lambda\mathbf{b}_1\))
yes
the same line
parallel
no
parallel, distinct
not parallel
yes (exactly one)
intersecting
not parallel
no
skew
Worked example Parallel or the same line?
\(l_{1}\): \(\mathbf{r} = (1, 2, 3) + s(1, -1, 2)\) and \(l_{4}\): \(\mathbf{r} = (0, 4, 1) + u(-2, 2, -4)\). Show that \(l_{1}\) and \(l_{4}\) are parallel and distinct.
\((-2, 2, -4) = -2(1, -1, 2)\), so the directions are parallel.
Test whether \((0, 4, 1)\) lies on \(l_{1}\): \(1 + s = 0\) gives \(s = -1\).
Then \(y = 2 - (-1) = 3 \ne 4\), so the point is not on \(l_{1}\).
Answer: Parallel directions but no common point: parallel, distinct lines.
Examiner trap
In two dimensions, non-parallel lines must meet, so students skip the check. In three dimensions non-parallel is not enough: you must show the lines share a point before calling them intersecting.
Section 02
Finding the intersection.
Write each line as a general point, using a DIFFERENT parameter for each line, and set them equal. That gives three equations in two unknowns. Solve any two, then use the third as the test.
Worked example Lines that intersect
\(l_{1}\): \(\mathbf{r} = (1, 2, 3) + s(1, -1, 2)\) and \(l_{2}\): \(\mathbf{r} = (5, 1, 4) + t(2, 1, -3)\). Show that the lines intersect and find the point of intersection.
Directions \((1, -1, 2)\) and \((2, 1, -3)\) are not multiples of each other, so the lines are not parallel.
Add the \(x\) and \(y\) equations to eliminate \(s\): \(3 = 6 + 3t\), so \(t = -1\). Then \(2 - s = 0\), so \(s = 2\).
Test \(z\): left \(3 + 2(2) = 7\), right \(4 - 3(-1) = 7\) ✓. All three equations hold.
Point: put \(s = 2\) into \(l_{1}\): \((3, 0, 7)\). Check with \(t = -1\) in \(l_{2}\): \((5 - 2, 1 - 1, 4 + 3) = (3, 0, 7)\) ✓.
Answer: The lines intersect at \((3, 0, 7)\).
Worked example Lines that are skew
Show that \(l_{1}\): \(\mathbf{r} = (1, 2, 3) + s(1, -1, 2)\) and \(l_{3}\): \(\mathbf{r} = (5, 1, 5) + t(2, 1, -3)\) are skew.
Directions as before: not parallel.
\(x\): \(1 + s = 5 + 2t\); \(y\): \(2 - s = 1 + t\). These are the same two equations, so again \(t = -1\), \(s = 2\).
Test \(z\): left \(3 + 2(2) = 7\), right \(5 - 3(-1) = 8\). \(7 \ne 8\), so there is no common point.
Not parallel and not intersecting.
Answer: \(l_{1}\) and \(l_{3}\) are skew.
Your turn
Show that the lines \(\mathbf{r} = (1, 0, 2) + s(1, 1, -1)\) and \(\mathbf{r} = (1, 3, -1) + t(2, -1, 1)\) intersect, and find the point of intersection.
Not parallel. \(x\): \(1 + s = 1 + 2t\) so \(s = 2t\); \(y\): \(s = 3 - t\), so \(2t = 3 - t\), \(t = 1\), \(s = 2\). \(z\): \(2 - 2 = 0\) and \(-1 + 1 = 0\) ✓. Point \((3, 2, 0)\).
Examiner trap
Using the same letter for both parameters (say \(t\) and \(t\)) forces the two points to have equal parameters, which they almost never do. You then wrongly conclude 'skew'. Always use \(s\) and \(t\), or \(\lambda\) and \(\mu\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Two lines have directions \(\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\) and \(-2\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}\). What can you conclude?
Quick check+10 XP
What does it mean for two lines to be skew?
Quick check+10 XP
Two non-parallel lines are written with parameters \(s\) and \(t\). The \(x\) and \(y\) equations give \(s = 2\), \(t = -1\), but these values do not satisfy the \(z\) equation. The lines are
Quick check+10 XP
When you equate the general points of two lines in three dimensions, how many equations do you get in the two parameters?
Section 03
Lines with an unknown.
If you are told the lines intersect, the test equation becomes an extra condition. Solve the two equations that do not contain the unknown first, then use the third to find it.
Worked example Find the constant
The lines \(\mathbf{r} = (1, 2, 3) + s(1, -1, 2)\) and \(\mathbf{r} = (4, a, 0) + t(1, 2, -1)\) intersect. Find \(a\) and the point of intersection.
The \(x\) and \(z\) equations do not involve \(a\): \(1 + s = 4 + t\) and \(3 + 2s = -t\).
From the first, \(s = 3 + t\). Substitute: \(3 + 6 + 2t = -t\), so \(3t = -9\), \(t = -3\) and \(s = 0\).
Now the \(y\) equation: \(2 - s = a + 2t\) becomes \(2 = a - 6\).
So \(a = 8\), and the point is \(s = 0\) on the first line: \((1, 2, 3)\).
Check with \(t = -3\) on the second line: \((4 - 3,\ 8 - 6,\ 0 + 3) = (1, 2, 3)\) ✓.
Answer: \(a = 8\); the lines meet at \((1, 2, 3)\).
Worked example Flight paths in context
Relative to an air-traffic control tower, the path of aircraft \(A\) is \(\mathbf{r} = (-4, 1, 5) + t(3, 1, 0)\) and the path of aircraft \(B\) is \(\mathbf{r} = (6, -9, 1) + s(-1, 3, 1)\), with distances in km and \(t\), \(s\) the times in minutes after noon. Show that the paths cross, and explain whether the aircraft collide.
\(z\): \(5 = 1 + s\), so \(s = 4\).
\(x\): \(-4 + 3t = 6 - 4 = 2\), so \(t = 2\).
Test \(y\): \(1 + 2 = 3\) and \(-9 + 3(4) = 3\) ✓. The paths cross at \((2, 3, 5)\).
But \(A\) is there at \(t = 2\) and \(B\) at \(s = 4\): two minutes apart.
Answer: The paths meet at \((2, 3, 5)\), but the aircraft pass that point at different times, so they do not collide.
Examiner trap
Intersecting PATHS do not mean a collision. The parameters are separate; a collision needs both objects at the point at the same time, which means equal times, not just equal positions.
Stage 2 · Exam
Exam-style questions
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Exam style+20 XP
The lines \(\mathbf{r} = (2, 0, 1) + s(1, 2, -1)\) and \(\mathbf{r} = (1, 3, 1) + t(3, 1, -2)\) intersect. Find the point of intersection.
Exam style+20 XP
Describe the relationship between \(l_{1}\): \(\mathbf{r} = (1, 2, 3) + s(1, -1, 2)\) and \(l_{2}\): \(\mathbf{r} = (3, 1, 0) + t(-2, 2, -4)\).