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c059741
NotesMath AA HLTopic 3.14Vector equation of a line
Back to Math AA HL Topics
3.14.12 min read

Vector equation of a line (Math AA HL)

IB Mathematics: Analysis and Approaches • Unit 3

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Contents

  • A point plus a direction: r = a + λd
  • From two points, and the parametric form
Stand at a point, then walk in a fixed direction: Imagine standing at a fixed point a and walking in a straight, fixed direction d.

Every point on the line is a + (some number) × d. That number is the parameter λ (lambda): λ = 0 puts you at a, λ = 1 takes one step along d, λ = −1 steps backwards, and fractions land you in between.

So the line is the set of all positions r = a + λd as λ runs over every real number.
a = position vector of a known point on the line; d = a direction vector; λ ∈ ℝ is the parameter.
Two important freedoms: Any point on the line can be used as a (the line doesn't care which one you start from).

The direction d can be replaced by any non-zero multiple of itself (2d, −d, ½d all point along the same line). So two equations that look different can describe the same line — match them by checking the directions are parallel and a shared point fits both.

IB-style question — write the vector equation

A line passes through the point P(1, −2, 4) and is parallel to the vector d = (3, 0, −1).

Write down a vector equation of the line.

Step by step

  1. Use the point's position vector as a and the given direction as d. Then r = a + λd.
  2. Substitute into r = a + λd.

Final answer

r = (1, −2, 4) + λ(3, 0, −1). (Any parallel direction, e.g. (−3, 0, 1), is equally correct.)

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Two points give you the direction for free: Given two points A and B on the line, the direction is just the displacement from one to the other:

d = AB = b − a (final position minus starting position).

Then take either point as the start. So the line through A and B is r = a + λ(b − a).
The direction vector between two points: subtract the start from the end.

IB-style question — line through two points

A line passes through A(2, 1, 5) and B(4, 5, 3).

Find a vector equation of the line.

Step by step

  1. Direction = B − A (subtract coordinate by coordinate).
  2. Use A as the starting point a.
  3. You may simplify the direction by dividing by 2 (it stays parallel).

Final answer

r = (2, 1, 5) + λ(2, 4, −2), or equivalently with direction (1, 2, −1).

Parametric form: split into coordinates: Reading the vector equation one row at a time gives the parametric equations — one equation per coordinate, all sharing the same λ:

x = a₁ + λd₁, y = a₂ + λd₂, z = a₃ + λd₃.

These are the same line, just written as separate coordinate formulas. They're handy for substituting into other equations or eliminating λ.

IB-style question — write the parametric form

A line has vector equation r = (1, −2, 4) + λ(3, 0, −1).

Write down its parametric equations.

Step by step

  1. Take each row of the vector equation separately, keeping the same λ.
  2. Tidy the y-row (the direction's y-component is 0, so y is constant).

Final answer

x = 1 + 3λ, y = −2, z = 4 − λ.

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A line passes through C(−2, 5) and has gradient (2D direction) (3, −1). Write a vector equation and hence the parametric equations. [2 marks]

Related Math AA HL Topics

Continue learning with these related topics from the same unit:

3.1.1Distance & midpoint (3D)
3.1.2Volume & surface area
3.1.3Angles in 3D
3.1.4Solids in 3D coordinates
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3.13.2Perpendicular & parallel vectors
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Points on a line3.14.2

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