Write vector equations of lines from points and directions
Interpret r = a + td, construct a line from two points, and distinguish a full line from a bounded segment.
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
- Find the displacement between two points.
- Solve linear equations and interpret scalar multiples of a direction vector.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A and B are distinct points. Which direction vector can define their line?
What you will learn
- Explain the position vector, anchor, non-zero direction vector and scalar parameter in a line equation.
- Construct a line from a point and direction or two distinct points, and check membership using every coordinate.
One parameter changes every coordinate together
In r = a + td, r is the position of a general point on the line, a is the position of a fixed anchor point, d is a non-zero direction vector and t is a real scalar. At t = 0 the point is a; increasing t by 1 moves by d. A full line allows all real t, while a segment or physical path may restrict its range. The parameter need not represent time unless the question says that it does.
For a = (1, −2, 3)ᵀ and d = (2, 1, −1)ᵀ, the equation gives x = 1 + 2t, y = −2 + t and z = 3 − t. The point (5, 0, 1) belongs at t = 2. The point (5, 0, 2) does not: the first two coordinates imply t = 2 but the third would have to be 1. All coordinates must come from one common parameter value for a point on a single line.
Different equations can describe the same line
Given distinct points A and B, use d = b − a and r = a + t(b − a). The segment AB corresponds to 0 ≤ t ≤ 1 in this particular parametrisation. Any non-zero scalar multiple of d gives the same full line, with a rescaled parameter; its old bounds cannot be copied without checking. A different anchor on the same line is also valid. A zero direction vector gives one point, not a line.
The diagram uses r = (1, 1)ᵀ + t(2, 1)ᵀ. Its marked positions at t = −1, 0, 1 and 2 are (−1, 0), (1, 1), (3, 2) and (5, 3). Each unit parameter step adds the same vector (2, 1)ᵀ. Replacing the direction by (4, 2)ᵀ leaves the full line unchanged but halves the parameter needed to reach the same point from the same anchor. Separate the geometry of the line from the choice of parameter.
Pause and explain
A candidate point matches x and y at t = 2 but fails z at t = 2. Is it on that 3D line?
Worked example
For r = (1, −2, 3)ᵀ + t(2, 1, −1)ᵀ, test P = (5, 0, 1) and Q = (5, 0, 2).
Show the worked solution
- The x coordinate requires t = 2 for both points.
- At t = 2 the line has y = 0 and z = 1, so P belongs.
- Q has z = 2 and fails the third coordinate despite matching x and y.
Answer P belongs at t = 2; Q does not belong.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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.
Write the full line through A = (1, 2, −1) and B = (5, 0, 3), then give the bounds for segment AB in your parametrisation.
Give me a hint
Use B − A as the direction and A as the anchor.
Compare my reasoning
- The direction is (4, −2, 4)ᵀ.
- A valid line is r = (1, 2, −1)ᵀ + t(4, −2, 4)ᵀ for t ∈ ℝ.
- A occurs at t = 0 and B at t = 1, so segment AB uses 0 ≤ t ≤ 1.
r = (1, 2, −1)ᵀ + t(4, −2, 4)ᵀ; full line t ∈ ℝ, segment 0 ≤ t ≤ 1.
Look for these in your work
- Use a non-zero between-point direction.
- Check both endpoints and distinguish the full line from the segment.
For r = (2, −1, 4)ᵀ + t(−1, 2, 0)ᵀ, test P = (−1, 5, 4) and Q = (−1, 5, 5). Show that r = P + u(2, −4, 0)ᵀ gives the same full line.
Give me a hint
Check the constant third coordinate and find the relation between t and u.
Compare my reasoning
- The first two coordinates require t = 3; z is always 4, so P belongs and Q does not.
- The new anchor P is on the original line and the new direction is −2 times the original direction.
- Using t = 3 − 2u makes both equations identical for all real u, proving the same full line.
P belongs at t = 3; Q does not; the second equation is equivalent with t = 3 − 2u.
Look for these in your work
- Check constant coordinates as well as the changing ones.
- Verify an anchor and a non-zero proportional direction or an exact parameter relation.
A straight pipe segment runs from A = (1, 1, 0) m to B = (5, 3, 2) m. Parametrise its centre line using A + t(B − A), find the position at t = 3/4, and decide whether (7, 4, 3) lies on the pipe segment.
Give me a hint
The physical segment has 0 ≤ t ≤ 1; a point on its infinite extension may lie outside those bounds.
Compare my reasoning
- The equation is r = (1, 1, 0)ᵀ + t(4, 2, 2)ᵀ with 0 ≤ t ≤ 1.
- At t = 3/4 the position is (4, 5/2, 3/2) metres.
- The candidate (7, 4, 3) corresponds to t = 3/2, so it is on the infinite extension but outside the pipe segment.
Segment r = (1, 1, 0)ᵀ + t(4, 2, 2)ᵀ, 0 ≤ t ≤ 1; at 3/4: (4, 5/2, 3/2) m; candidate is outside the segment.
Look for these in your work
- Use the stated physical endpoints to set parameter bounds.
- Check every coordinate and the range before accepting membership of the segment.
Common mistakes
- Use one parameter value for all coordinates of one line.
- A direction vector is a displacement, not an arbitrary second point's position.
- Equivalent full-line equations may have different parameter values and segment bounds.
What can you explain now?
What point does r = a + td describe at t = 0?
Compare with the explanation
The anchor with position a.
The displacement term is zero, leaving the fixed position vector.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain every symbol in a vector line equation, construct one from two points, and test a point against all coordinates and any parameter bounds.
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.