Use vector letters and calculate a non-negative magnitude
Distinguish v from |v| · Calculate length from components · Recognise equal lengths without assuming equal vectors
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Vector components
- Square roots
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Distinguish v from |v|
- Calculate length from components
- Recognise equal lengths without assuming equal vectors
A vector and its length are different objects
A letter such as v names a vector; |v| names its magnitude, a non-negative scalar length. For v = [a; b], |v| = √(a²+b²), from the right triangle of horizontal and vertical displacements. The signs are squared, so opposite vectors have equal lengths.
Length does not fix direction
[3; 4] and [−3; −4] both have length 5 but point in opposite directions. [0; 5] also has length 5 and a different direction. Only [0; 0] has magnitude zero; its direction is undefined. Keep exact root forms unless a decimal is requested.
Sources for this lesson
Pause and explain
What is the magnitude of [−3; 4]?
Worked example
For v = [−6; 8], find |v|. Does a negative component make the magnitude negative?
Show the worked solution
- |v| = √(36+64) = 10.
- Magnitude is a non-negative length, not a signed component.
Answer |v| = 10.
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.
Find the magnitude of [5; 12].
Give me a hint
Use the sum of squares.
Compare my reasoning
- 5²+12² = 169.
- √169 = 13.
13.
Look for these in your work
- Square both components
- Select the non-negative root
Find the exact magnitudes of [−2; −3] and [0; 0].
Give me a hint
Signs disappear when squared.
Compare my reasoning
- The first length is √(4+9) = √13.
- The zero vector has length 0 and no direction.
√13; 0 with undefined direction.
Look for these in your work
- Keep an exact root
- State the zero-vector condition
Vectors [3; 4] and [0; 5] have equal magnitudes. Are the vectors equal? Explain.
Give me a hint
Compare components as well as lengths.
Compare my reasoning
- Both lengths are 5.
- Their horizontal and vertical components differ, so their directions and vectors differ.
No: equal length 5 does not make their vectors equal.
Look for these in your work
- Separate length from displacement
- Use component evidence
Common mistakes
- Do not add component magnitudes to obtain vector length.
- Equal magnitude does not imply equal direction.
What can you explain now?
What is the magnitude of [−3; 4]?
Compare with the explanation
5.
The squared components give √(9+16).
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Make it stick
What is the magnitude of [−3; 4]?
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.