Classify parallel intersecting and skew lines with independent parameters
Compare directions, solve the coordinate equations and check the third dimension before claiming an intersection.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.7. Paper 1 coordinate geometry and algebra are assumed. Work in two or three dimensions as stated. The general section theorem is not included; midpoint geometry is required. The cross product, shortest distance between skew lines and their common perpendicular are not required. Scalar products include angles between lines and the foot of a perpendicular from a point to a line, including 3D solids. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Interpret a vector line equation and test point membership.
- Solve two simultaneous linear equations and substitute into a third.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which vector is proportional to (1, −1, 2)ᵀ?
What you will learn
- Distinguish distinct parallel, coincident, intersecting and skew lines in three dimensions.
- Use independent parameters to find an intersection and separate path crossing from simultaneous meeting in a timed model.
Compare directions before solving for a crossing
For r = a + td and r = b + se, use independent real parameters t and s. The direction vectors are parallel if e = kd for some non-zero constant k; check every component, including zeros. Parallel directions can describe distinct parallel lines or one coincident line. If b − a is also a scalar multiple of d, the anchors belong to the same line and the lines coincide, giving infinitely many common points rather than one intersection.
For a = (1, 2, 3)ᵀ and d = (1, −1, 2)ᵀ, the line with anchor (0, 3, 1) and direction 2d is coincident: its anchor equals a − d. The line with anchor (1, 2, 4) and direction 2d is instead distinct and parallel, because its offset (0, 0, 1) is not proportional to d. Different anchors alone do not prove two parallel lines are distinct.
A projected crossing is not a 3D intersection
For non-parallel directions, solve the coordinate equations for t and s and then verify every remaining coordinate. The lines (1, 0, 2)ᵀ + t(2, 1, −1)ᵀ and (2, 5, −3)ᵀ + s(1, −1, 1)ᵀ give t = 2 and s = 3 from x and y. Both z values are then 0, so the intersection is (5, 2, 0). There is no reason for the two parameters to be equal. They describe separate positions along separate lines.
Changing only the second anchor's z coordinate from −3 to −2 leaves the x and y equations unchanged, but at t = 2 and s = 3 the z coordinates become 0 and 1. These non-parallel lines have no common point, so they are skew. Skew lines occur in 3D; their 2D projections may cross. The syllabus does not require their shortest distance or common perpendicular. If parameters represent elapsed time with a shared start, a spatial crossing at different times does not establish a simultaneous meeting.
Pause and explain
At a candidate intersection, the first line has t = 2 and the second has s = 3. Must the claim fail because t ≠ s?
Worked example
Find the intersection of r = (1, 0, 2)ᵀ + t(2, 1, −1)ᵀ and r = (2, 5, −3)ᵀ + s(1, −1, 1)ᵀ.
Show the worked solution
- Equate x: 1 + 2t = 2 + s; equate y: t = 5 − s.
- Solve to get t = 2 and s = 3, then check z: 2 − 2 = −3 + 3 = 0.
- Substitute into either whole vector equation to obtain the same point (5, 2, 0).
Answer Intersection (5, 2, 0), with t = 2 and s = 3.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Classify r = (1, 2, 3)ᵀ + t(1, −1, 2)ᵀ and r = (0, 3, 1)ᵀ + s(2, −2, 4)ᵀ as distinct parallel or coincident lines.
Give me a hint
Compare directions, then test whether the second anchor lies on the first line.
Compare my reasoning
- The second direction is twice the first, so their directions are parallel.
- The anchor difference is (−1, 1, −2)ᵀ, equal to −1 times the first direction.
- The lines coincide; t = 2s − 1 relates their equations, giving infinitely many common points.
Coincident lines, with parameter relation t = 2s − 1.
Look for these in your work
- Check every direction component with one multiplier.
- Test the anchor offset before claiming the lines are distinct.
Classify r = (0, 0, 0)ᵀ + t(1, 1, 0)ᵀ and r = (1, 0, 1)ᵀ + s(−1, 1, 0)ᵀ, showing all coordinate checks.
Give me a hint
Solve x and y, then compare the constant z values.
Compare my reasoning
- The directions are not proportional. Equating x and y gives t = 1 − s and t = s, hence t = s = 1/2.
- The first line's z is 0 and the second line's z is 1 at every parameter value.
- There is no 3D intersection, so these non-parallel lines are skew despite the projected crossing (1/2, 1/2).
Skew lines; the projected crossing has unequal z coordinates 0 and 1.
Look for these in your work
- Check non-parallel directions and solve independent parameters.
- Use the third coordinate to decide actual 3D intersection.
Two simulated particles start simultaneously. Their positions in metres are p(t) = (0, 0, 0)ᵀ + t(1, 0, 1)ᵀ and q(s) = (2, −1, 1)ᵀ + s(0, 1, 1)ᵀ, where t and s are their elapsed seconds. Do their forward paths intersect, and does the model predict a simultaneous meeting?
Give me a hint
First solve the spatial equations independently; then check the shared-time requirement.
Compare my reasoning
- The x and y equations give t = 2 and s = 1; the z values are both 2.
- The forward paths therefore intersect at (2, 0, 2), reached at different elapsed times.
- A simultaneous meeting requires t = s. The unique crossing has 2 ≠ 1, so these simultaneously started particles never meet in the supplied model.
Paths intersect at (2, 0, 2); no simultaneous meeting because the crossing times are 2 s and 1 s.
Look for these in your work
- Separate spatial intersection from a shared-time condition.
- Check forward parameter ranges and all coordinates before interpreting the model.
Common mistakes
- Independent line parameters need not have equal values at a common point.
- Proportional directions require a separate anchor check to distinguish coincident from distinct parallel lines.
- An x–y crossing can still fail the z equation and give skew lines.
What can you explain now?
Non-parallel 3D lines have a projected crossing, but their z values disagree there. How are they classified?
Compare with the explanation
Skew.
They are non-parallel and have no common 3D point; the projected crossing is insufficient.
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Make it stick
Classify two vector lines by their directions and coordinate equations. Explain coincident and skew cases, independent parameters, and why crossing paths do not always mean a simultaneous meeting.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.