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IGCSE · Cambridge (CIE) · 0980

Vector magnitude and unit directions

Calculate vector length from perpendicular components and recognise that equal lengths do not imply equal vectors.

Section Paper 4 · Extended · Calculator

Transformations and vectors pathway · Cambridge 0980 (9–1) · E7.3

Extended only. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.

Before you begin

  • Use Pythagoras.
  • Interpret signed vector components.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Can a vector magnitude be negative?

What you will learn

  • Use |a| = √(x² + y²).
  • Compare vector direction as well as magnitude.

Components form a right triangle

The magnitude of vector (x,y) is √(x² + y²). A vector (−6,8) therefore has length √(36 + 64) = 10 units. Signs indicate direction but disappear in the squared length. Magnitude is a non-negative scalar, not another column vector.

Two vectors can have equal magnitude while differing in direction: (3,4) and (−3,4) both have length five, but they are not equal. Equal vectors require both equal direction and equal magnitude, which means equal components in the same coordinate system.

Scaling length preserves the square-root relationship

The magnitude of ka is |k| times the magnitude of a. If a = (3,4), then −2a = (−6,−8) has magnitude ten. Do not multiply magnitude by a negative number and call the result a negative length.

An optional useful direction representation is a/|a| for a nonzero vector, which has magnitude one. For (3,4), this is (3/5,4/5). The zero vector cannot be normalised by division because its magnitude is zero. The syllabus requires magnitude; this unit-direction idea explains scaling rather than introducing another required formula.

Pause and explain

Find the magnitude of (−6,8).

Put the idea to work

Worked example

Find the magnitude of (−6,8).

Show the worked solution
  1. Square the perpendicular components: 36 and 64.
  2. Add them to obtain one hundred.
  3. Take the positive square root: ten units.

Answer 10 units.

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.

1 · Guided

Find |(5,12)|.

Give me a hint

Apply Pythagoras to the components.

Compare my reasoning
  1. 5² + 12² = 169.
  2. √169 = 13.
  3. The vector magnitude is thirteen units.

13 units.

Look for these in your work

  • Squared before adding.
  • Took the square root.
2 · Independent

Find the magnitude of (−2,−3) exactly.

Give me a hint

Keep the final square root exact.

Compare my reasoning
  1. The component squares are four and nine.
  2. Their sum is thirteen.
  3. The magnitude is √13 units, regardless of both negative signs.

√13 units.

Look for these in your work

  • Used both components.
  • Kept an exact magnitude.
3 · Transfer

Vectors (3,4) and (4,3) both have length five. Are they equal? Explain.

Give me a hint

Equality requires the same displacement direction.

Compare my reasoning
  1. Both squared magnitudes equal twenty-five.
  2. Their horizontal and vertical components differ.
  3. They are not equal vectors even though their magnitudes are equal.

No; equal magnitude, different direction and components.

Look for these in your work

  • Compared components.
  • Distinguished vector from scalar equality.

Common mistakes

  • Adding components instead of their squares.
  • Reporting squared length.
  • Equating vectors solely by magnitude.
Recall without your notes

What can you explain now?

What is the magnitude of −3a if |a| = 4?

Compare with the explanation

12.

Magnitude uses the absolute scale factor three; the minus sign reverses direction.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Tomorrow, give two unequal vectors of the same length and justify their difference.

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.

Keep learning Mathematics (9-1)

Course scope and assessment

StudyForA maps the 72 Core and Extended section entries in Cambridge IGCSE Mathematics (9–1) 0980 to original teaching and staged practice. This route covers the 2025–2027 syllabus and the 2028–2030 Version 2 syllabus; Cambridge says the 2028 update makes no significant changes affecting teaching, while its assessment overview is updated. The course keeps the 9–1 grading and paper rules separate from 0580. Questions and written work are self-checked, not Cambridge-graded; this is not a timed mock or an official grade, and subject-teacher review remains pending. 0980 is available in June and November in Administrative Zone 3 and for centres in Oman and the UAE. Confirm the exam series and your centre's entry eligibility with Cambridge and your exams officer.

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