3.15 Coincident, parallel, intersecting and skew lines · Educerie Mathematics: analysis and approaches

This whole subtopic is higher level. Nothing in it is on an SL paper.

Educerie · IB Diploma · Mathematics: analysis and approaches

Topic 3 Geometry and trigonometry · 3.15 Coincident, parallel, intersecting and skew lines

Level: HL only, the whole subtopic. If you are SL, none of this is on your papers.
Themes (key concepts): space, relationships, equivalence. How two lines sit in space is a relationship you cannot always see, but you can decide it for certain by asking whether a system of equations has one solution, none, or infinitely many.
The question this unit answers: given the equations of two lines, how do you decide for certain whether they are the same line, parallel, meeting at a point or passing each other by, and where exactly do they meet if they do?
Where it is examined: Paper 1 and Paper 2, usually as a 5 to 8 mark part of a Section B lines-and-planes question: "determine whether the lines intersect", "show that the lines are skew", "find the value of k for which the lines intersect". It also appears inside kinematics questions as "do the paths cross?", which is a different question from "do the objects collide?".

What you must be able to do

You must be able to	Level	What it looks like in the exam
Decide whether two lines are parallel by comparing direction vectors	HL	"Show that L₁ and L₂ are parallel" (2 marks)
Distinguish coincident lines from distinct parallel lines	HL	"Determine whether L₁ and L₂ are the same line" (2 to 3 marks)
Find the point of intersection of two lines, or show there is none	HL	"Find the coordinates of the point where L₁ and L₂ meet" (4 to 6 marks)
Show that two lines are skew, with reasons	HL	"Show that the lines are skew" (4 to 6 marks): both conditions stated
Find an unknown so that two lines intersect	HL	"Find the value of k for which the lines intersect" (5 to 6 marks)
Tell crossing paths from a collision in a kinematics context	HL	Paper 2: "Show that the paths cross but the ships do not collide" (4 to 6 marks)

Before you start

You need 3.14: the vector, parametric and Cartesian equations of a line, and how to test whether a point lies on one. You need parallel vectors from 3.12: b₁ and b₂ are parallel when b₁ = kb₂ for some scalar k. And you need to solve two linear equations in two unknowns by hand, which is prior learning. Nothing on this page is in the formula booklet as a formula; it is a method.

A column vector is printed on one line here: (2, −1, 3) means the column with entries 2, −1, 3. A point is written with its letter, P(2, −1, 3).

1The idea in one paragraph

Two lines in the plane can do three things: be the same line, be parallel and never meet, or cross at one point. In space there is a fourth possibility. Two lines can point in different directions and still never meet, because one passes above the other, like a road on a bridge and a road underneath it. Such lines are skew. Deciding which case you have needs two questions, asked in order. First: are the directions parallel? If yes, the lines are either coincident (the same line) or parallel and distinct, and one point settles it. If no, the lines either intersect at one point or are skew, and solving the equations settles it: set the two position vectors equal, with a different parameter for each line, solve two of the three component equations, and check the third. If the third fits, you have the point of intersection. If it fails, the lines are skew.

2Four cases, and why skew is new

Figure 1 draws the four cases.

Figure 1 · Two lines in space: the four cases

Coincident: the two equations describe the same line. Every point of one is a point of the other, so there are infinitely many common points.
Parallel: same direction, different lines. No common point.
Intersecting: different directions, exactly one common point.
Skew: different directions and no common point. This needs three dimensions.

Skew lines are non-parallel lines that do not intersect. Both halves matter: parallel lines also never meet, but they are not skew.

Why can skew lines not happen in the plane? Two non-parallel lines in a plane always cross somewhere, because each one eventually cuts across the other's path. In space, two lines that point in different directions can lie in two different "layers" and slide past each other. A room gives you skew lines everywhere, as Figure 2 shows: the edge where the front wall meets the floor, and the edge where a side wall meets the ceiling. They point in different directions, and no amount of extending them makes them touch.

Figure 2 · Skew lines in a room

One more way to see it. Parallel lines and intersecting lines always lie together in one flat plane. Skew lines never do. That is the reason a question about the plane containing two lines (3.17) only makes sense when they are not skew.

3The method

Figure 3 is the whole method as a decision tree.

Figure 3 · Deciding which case you have

Step 1. Compare directions. Is b₁ a scalar multiple of b₂? Check component by component: the ratios must all be equal. (2, −4, 6) and (−1, 2, −3) are parallel, because (2, −4, 6) = −2(−1, 2, −3). (2, 1, −1) and (1, −2, 3) are not: 2 ÷ 1 = 2 but 1 ÷ (−2) = −1/2.

Step 2a. If parallel: test one point. Take the known point of L₁ and ask whether it lies on L₂ (the method of 3.14). If it does, the lines share a point and a direction, so they are the same line: coincident. If not, they are parallel and distinct.

Step 2b. If not parallel: solve. A common point must be reachable from both equations, so set them equal with separate parameters, λ on L₁ and μ on L₂:

a₁ + λb₁ = a₂ + μb₂ gives three equations in two unknowns. Solve any two for λ and μ, then substitute into the third. It fits: they intersect. It fails: they are skew.

Why three equations and only two unknowns? Two equations in two unknowns (almost always) have a solution, so any two components can be made to agree. The third component is the real test. In the plane there is no third component, which is exactly why non-parallel lines in the plane always meet.

4Parallel or coincident: worked

L₁: r = (1, 2, −1) + λ(2, −4, 6) and L₂: r = (3, −2, 5) + μ(−1, 2, −3).

(2, −4, 6) = −2(−1, 2, −3)directions parallel

is (1, 2, −1) on L₂?

x: 3 − μ = 1 → μ = 2

y: −2 + 2(2) = 2 ✓

z: 5 − 3(2) = −1 ✓same point and same direction: coincident

The two equations look nothing alike, and yet they are the same line. Now take L₃: r = ν(1, −2, 3), through the origin, also parallel to L₁.

is (1, 2, −1) on L₃?

x: ν = 1

y: −2(1) = −2 ≠ 2 ✗parallel, but not the same line

L₁ and L₃ are parallel and distinct. The argument in words is what earns the reasoning mark: "the direction vectors are scalar multiples, so the lines are parallel; the point (1, 2, −1) of L₁ does not lie on L₃, so the lines are not coincident."

5Intersecting: worked

L₁: r = (1, −2, 3) + λ(2, 1, −1) and L₂: r = (3, 4, −5) + μ(1, −2, 3). Step 1 showed the directions are not parallel.

x: 1 + 2λ = 3 + μ → 2λ − μ = 2 (1)

y: −2 + λ = 4 − 2μ → λ + 2μ = 6 (2)

z: 3 − λ = −5 + 3μ → λ + 3μ = 8 (3)

2 × (1) + (2): 5λ = 10 → λ = 2

from (1): μ = 2λ − 2 = 2

check (3): 2 + 3(2) = 8 ✓the third equation fits: they intersect

point: r = (1, −2, 3) + 2(2, 1, −1) = (5, 0, 1)

Always find the point from both lines as a check: L₂ at μ = 2 gives (3 + 2, 4 − 4, −5 + 6) = (5, 0, 1). Same point, so no slip. Figure 4 shows the two lines meeting there.

Figure 4 · Two lines that meet

Notice that λ and μ both came out as 2 here. That is a coincidence of these numbers. In general they differ, and that is why they need different letters.

Which two equations to solve first? Any two, but choose the pair that eliminates easily. If one equation has only one parameter in it (because a direction component is 0), start there.

On Paper 2 you may solve the pair with the GDC's simultaneous-equation solver, but write the three component equations first; they carry the method mark.

6Skew: worked

Now change one number. L₃: r = (3, 4, −4) + μ(1, −2, 3) is L₂ lifted one unit in the z-direction, so it is still not parallel to L₁.

x: 1 + 2λ = 3 + μ → 2λ − μ = 2

y: −2 + λ = 4 − 2μ → λ + 2μ = 6

these give λ = 2, μ = 2, exactly as before

z: 3 − λ = −4 + 3μ ?

left: 3 − 2 = 1; right: −4 + 6 = 2

1 ≠ 2no common point

The directions are not parallel and there is no common point, so L₁ and L₃ are skew. Figure 5 shows what happened: above the point (5, 0, 1) on L₁, the line L₃ passes one unit higher, at (5, 0, 2).

Figure 5 · Change one number and the lines go skew

A full-marks conclusion states both conditions:

"The direction vectors are not scalar multiples, so the lines are not parallel. The values of λ and μ that satisfy the x- and y-equations do not satisfy the z-equation, so the lines do not intersect. Hence the lines are skew."

Leaving out "not parallel" is the most common way to drop the last mark: lines that never meet could still be parallel.

7Finding an unknown so that lines meet

When a line contains a letter, the question usually asks for the value that makes the lines intersect. Use two equations that do not contain the letter to find λ and μ, then put them into the third to find the letter.

L₁: r = (2, 1, 0) + λ(1, −1, 2) and L₂: r = (4, 5, k) + μ(1, 1, −1).

x: 2 + λ = 4 + μ → λ − μ = 2

y: 1 − λ = 5 + μ → λ + μ = −4

add: 2λ = −2 → λ = −1, μ = −3

z: 0 + 2λ = k − μ

−2 = k + 3 → k = −5

point: (2 − 1, 1 + 1, 0 − 2) = (1, 2, −2)

check on L₂: (4 − 3, 5 − 3, −5 + 3) = (1, 2, −2) ✓

For any other k the same λ and μ fail the z-equation, so the lines are skew (their directions are not parallel).

8Two dimensions, and paths versus collisions

In the plane. With only two components, two non-parallel lines always meet: two equations, two unknowns, nothing left to check. For r = (1, 3) + λ(2, −1) and r = (0, −2) + μ(1, 1):

1 + 2λ = μ and 3 − λ = −2 + μ

3 − λ = −2 + 1 + 2λ → 3λ = 4 → λ = 4/3

point: (1 + 8/3, 3 − 4/3) = (11/3, 5/3)

Paths that cross versus objects that collide. In kinematics (3.14), a position like r = a + tv describes both a path and a timetable. The two questions need different methods.

Do the paths cross? That is this page: give each line its own parameter, λ and μ, and solve.
Do the objects collide? They must be at the same point at the same time, so use one t for both.

Drone A flies r_A = (1, 0, 2) + t(2, 2, 1) and drone B flies r_B = (3, 8, 4) + t(4, −2, 1), with t in seconds.

paths: 1 + 2λ = 3 + 4μ, 2λ = 8 − 2μ, 2 + λ = 4 + μ

third gives λ = 2 + μ; into the second: 4 + 2μ = 8 − 2μ → μ = 1, λ = 3

first: 1 + 6 = 7 and 3 + 4 = 7 ✓the paths cross at (7, 6, 5)

A is there at t = 3; B is there at t = 1different times: no collision

If the values had been equal, the drones would collide. If the path equations had failed, the paths themselves would be skew and a collision would be impossible.

9Where marks are lost

Using the same parameter for both lines. Writing a₁ + λb₁ = a₂ + λb₂ asks whether two objects are at the same place at the same "time", not whether the lines meet. It can wrongly conclude "no intersection". Use λ and μ.

Solving two equations and stopping. Two equations in two unknowns can always be solved. Without the check in the third equation you have shown nothing, and the answer "they intersect" earns no reasoning mark.

Calling lines skew because they do not meet. Parallel lines do not meet either. Skew needs "not parallel" as well, and the reasoning mark needs both stated.

Calling parallel lines coincident without a point check. Parallel directions only say the lines point the same way. One point of L₁ must be shown to lie on L₂.

Comparing position vectors to test for parallel lines. Whether lines are parallel depends on the direction vectors alone. (1, 2, −1) and (3, −2, 5) being different says nothing.

Arithmetic slips in the point of intersection. Substitute λ into L₁ and μ into L₂. If the two points differ, there is an error to find before you move on.

Thinking skew lines are impossible because "they must meet somewhere". That intuition is true in the plane and false in space. Trust the third equation.

10Work it right

Write both direction vectors and state whether one is a scalar multiple of the other, with the multiple if it is.
If parallel, test a point of one line in the other and conclude "coincident" or "parallel, not coincident".
If not parallel, write the three component equations with λ and μ, one per line.
Solve two of them, showing the elimination, then substitute into the third and say whether it fits.
For an intersection, find the point from one line and check it with the other.
For skew lines, write the two-part conclusion: not parallel, and no common point.
For an unknown, solve the two equations free of it first, then use the third to find it.
In kinematics, say which question you are answering: paths (two parameters) or collision (one t).

11Try it

Marks in brackets. Q1 to Q3 are Paper 1 style, no calculator. Q4 and Q5 are Paper 2 style, with a GDC.

Q1. The lines L₁ and L₂ have equations L₁: r = (1, 0, 3) + λ(1, 2, −1) and L₂: r = (2, 5, 0) + μ(1, −1, 2). Determine whether L₁ and L₂ are parallel, intersecting or skew. If they intersect, find the point of intersection. 6 marks

Q2. The lines L₁: r = (2, 1, 0) + λ(1, −1, 2) and L₂: r = (0, 3, k) + μ(1, 1, −1) intersect. Find the value of k and the coordinates of the point of intersection. 6 marks

Q3. The line L₁ has Cartesian equation (x − 1)/2 = (y + 3)/(−1) = z/4, and L₂ has equation r = (5, −5, 8) + μ(−4, 2, −8). Show that L₁ and L₂ are the same line. 4 marks

Q4. Two straight tunnels are being bored through a hill. Relative to a survey point, with distances in metres, their centre lines are T₁: r = λ(4, 3, −1) and T₂: r = (10, −5, 2) + μ(−1, 5, −1).

(a) Show that the tunnels' centre lines are skew. 5 marks

(b) Find the acute angle between the two centre lines. 3 marks

Q5. Two drones move in straight lines. Drone A's position after t seconds is r_A = (2, −1, 0) + t(1, 2, 1) and drone B's is r_B = (8, 1, 0) + t(−1, 3, 2), in metres.

(a) Show that the flight paths cross, and find where. 4 marks

(b) Determine whether the drones collide. 2 marks

12In one breath

Two lines in space are coincident, parallel, intersecting or skew, and skew lines, non-parallel lines that never meet because they lie in different layers, only exist in three dimensions. Decide the case in two steps. First compare direction vectors: if one is a scalar multiple of the other the lines are parallel, and testing one point of L₁ on L₂ tells you whether they are the same line or two distinct parallel lines. If the directions are not parallel, set a₁ + λb₁ = a₂ + μb₂ with a different parameter for each line, solve two of the three component equations, and check the third: if it fits, substitute back for the point of intersection and check it on both lines; if it fails, the lines are skew, and the conclusion must say both "not parallel" and "no common point". An unknown in one line is found by solving the two equations that do not contain it and using the third. In kinematics, crossing paths need two parameters, but a collision needs the same t for both objects.

Answers

Q1. (1, 2, −1) and (1, −1, 2) are not scalar multiples (1 ÷ 1 = 1 but 2 ÷ (−1) = −2), so the lines are not parallel. Equate: x: 1 + λ = 2 + μ; y: 2λ = 5 − μ; z: 3 − λ = 2μ. Adding the first two after rearranging (λ − μ = 1 and 2λ + μ = 5) gives 3λ = 6, so λ = 2, μ = 1. Check z: 3 − 2 = 1 and 2μ = 2, and 1 ≠ 2. The third equation fails, so there is no common point: the lines are skew. R1 for showing the directions are not parallel, M1 for equating the lines with two different parameters, M1 for solving two equations, A1 for λ = 2 and μ = 1, M1 for substituting into the third equation, R1 for the conclusion "skew" with both reasons. A conclusion without "not parallel" loses the final R1.

Q2. x: 2 + λ = μ; y: 1 − λ = 3 + μ. Substituting μ = 2 + λ into the second: 1 − λ = 5 + λ, so λ = −2 and μ = 0. z: 2λ = k − μ gives −4 = k − 0, so k = −4. L₁ at λ = −2 gives (2 − 2, 1 + 2, −4) = (0, 3, −4), and L₂ at μ = 0 gives (0, 3, −4) as a check. M1 for equating with two different parameters, M1 for solving the x- and y-equations, A1 for λ = −2 and μ = 0, M1 for substituting into the z-equation, A1 for k = −4, A1 for (0, 3, −4).

Q3. From the Cartesian form, L₁ has direction (2, −1, 4) and passes through (1, −3, 0). L₂'s direction (−4, 2, −8) = −2(2, −1, 4), so the lines are parallel. Test (5, −5, 8) in L₁: (5 − 1)/2 = 2, (−5 + 3)/(−1) = 2, 8/4 = 2. All three are equal, so (5, −5, 8) lies on L₁. Parallel lines with a common point are the same line. A1 for the direction of L₁, R1 for showing the directions are parallel, M1 for testing a point of L₂ in L₁ (all three parts), R1 for the conclusion.

Q4. (a) (4, 3, −1) and (−1, 5, −1) are not scalar multiples (4 ÷ (−1) = −4 but 3 ÷ 5 = 0.6), so not parallel. Equate: 4λ = 10 − μ, 3λ = −5 + 5μ, −λ = 2 − μ. From the third, μ = 2 + λ; into the first, 4λ = 8 − λ, so λ = 1.6 and μ = 3.6. Check the second: 3(1.6) = 4.8, but −5 + 5(3.6) = 13. They are not equal, so there is no common point, and the centre lines are skew. R1 for not parallel, M1 for three equations with two parameters, A1 for λ = 1.6 and μ = 3.6, M1 for the check in the remaining equation, R1 for the two-part conclusion.

(b) b₁ · b₂ = −4 + 15 + 1 = 12, |b₁| = √26, |b₂| = √27. cos θ = 12 ÷ √702 = 0.4529…, so θ = 63.1° (3 s.f.). M1 for the scalar product of the direction vectors, M1 for dividing by the product of magnitudes, A1 for 63.1°.

Q5. (a) Paths, with parameters λ for A and μ for B: 2 + λ = 8 − μ, −1 + 2λ = 1 + 3μ, λ = 2μ. Into the first: 2 + 2μ = 8 − μ, so μ = 2 and λ = 4. Check the second: −1 + 8 = 7 and 1 + 6 = 7 ✓. The paths cross at (6, 7, 4). M1 for equating with two different parameters, A1 for μ = 2 and λ = 4, R1 for the check in the remaining equation, A1 for (6, 7, 4).

(b) A reaches (6, 7, 4) at t = 4 and B at t = 2. The times differ, so the drones do not collide. R1 for comparing the two times, A1 for the conclusion.

(c) At t = 2, B is at (6, 7, 4) and A is at (2 + 2, −1 + 4, 0 + 2) = (4, 3, 2). The distance is |(2, 4, 2)| = √24 = 4.90 m (3 s.f.). M1 for A's position at t = 2, M1 for the magnitude of the difference, A1 for 4.90 m.

Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 3.15 Coincident, parallel, intersecting and skew lines. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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