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Mathematics 05807.2

Vectors

Column vectors, magnitude, and vector geometry.

Review these first

Learning objectives

What you need to be able to do

Teacher-mapped phrasing — check against the official Cambridge syllabus for exact wording.

  • 7.2.1Add, subtract and multiply vectors by a scalar, and find magnitude.
  • 7.2.2Use vectors in geometric proofs, including parallel and collinear points.Supplement

6 minute read

Vectors

A vector has both magnitude and direction, unlike a scalar which has magnitude only.

Column vectors

Written (x above y): x is movement right, y is movement up, with negatives for left and down.

  • Adding: add the components. Geometrically, follow one vector then the other.
  • Subtracting: a − b = a + (−b), where −b is the same length in the opposite direction.
  • Scalar multiple: 3a is three times as long, in the same direction. −2a is twice as long in the opposite direction.

Magnitude

|a| = √(x² + y²) — Pythagoras applied to the components.

Route notation

AB (with the arrow) means the vector from A to B, and BA = −AB. To travel from A to C via B: AC = AB + BC. Any journey can be broken into steps this way, which is the core skill in vector geometry.

Parallel and collinear — the key insight

If one vector is a scalar multiple of another, the two are parallel. If they are parallel and share a common point, the points are collinear (lie on the same straight line). So to prove three points A, B, C are collinear:

  1. Find AB and BC in terms of the base vectors.
  2. Show BC = k × AB for some number k.
  3. State that they are parallel and share the point B, therefore A, B and C lie on a straight line.

Step 3 is essential — parallel alone does not prove collinear, since parallel lines can be separate.

Think of it like this

A vector is an instruction, not a place: "three streets east, two north" works from wherever you start. That is why the same vector can appear in different parts of a diagram.

Worked examples

Method, step by step

Given a = (3 above 4), find |a|.

  1. 1Magnitude uses Pythagoras on the components.
  2. 2|a| = √(3² + 4²) = √(9 + 16).
  3. 3= √25 = 5.

5

OA = a and OB = b. M is the midpoint of AB. Express OM in terms of a and b.

  1. 1Travel from O to A, then halfway along AB: OM = OA + ½AB.
  2. 2AB = OB − OA = b − a.
  3. 3So OM = a + ½(b − a) = a + ½b − ½a.
  4. 4Simplify: OM = ½a + ½b, or ½(a + b).

OM = ½(a + b)

Common misconceptions

  • Concluding that parallel vectors mean collinear points. A shared point is also required.
  • Forgetting that `BA = −AB`, which reverses signs when a route is traversed backwards.
  • Adding vectors by multiplying components, or finding magnitude by adding them instead of using Pythagoras.

In the exam

  • Write every route as a chain of steps through known vectors — `AC = AB + BC` — rather than trying to see the answer directly.
  • For a collinearity proof, always finish with the sentence naming the common point. That sentence is a mark.