Use position vectors to locate points and connecting arrows
Read a position vector relative to a chosen origin · Find a connecting vector by subtracting positions · Combine position and displacement to locate a point
Section Paper 2 · Historical Extended · non-calculator
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, addition and scalar multiples
- Cartesian points
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
- Read a position vector relative to a chosen origin
- Find a connecting vector by subtracting positions
- Combine position and displacement to locate a point
From one fixed origin
With origin O, the position vector a = →OA locates A. If b = →OB, then →AB = b−a because a+(b−a) = b. A position vector depends on the chosen origin; a connecting displacement does not.
Locate an intermediate point
A point M halfway from A to B has position a+(1/2)(b−a) = (a+b)/2. More generally, if a question places M a stated fraction t along →AB, its position is a+t(b−a). Use the stated fraction and direction; t outside [0,1] locates a point on an extension rather than inside the segment.
Sources for this lesson
Pause and explain
If a = →OA and b = →OB, what is →AB?
Worked example
a = [2; 1], b = [6; 5]. Find →AB and the midpoint position.
Show the worked solution
- →AB = b−a = [4; 4].
- Midpoint position = (a+b)/2 = [4; 3].
Answer →AB = [4; 4]; midpoint position [4; 3].
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.
a = [1; −2], b = [4; 3]. Find →AB.
Give me a hint
Subtract the starting position.
Compare my reasoning
- 4−1 = 3.
- 3−(−2) = 5.
[3; 5].
Look for these in your work
- Keep the shared origin
- Use end minus start
A has position [−2; 1] and displacement [5; −3] carries it to B. Find B's position and point coordinates.
Give me a hint
Add the displacement to the starting position.
Compare my reasoning
- [−2; 1]+[5; −3] = [3; −2].
- Thus B is point (3,−2) relative to O.
Position vector [3; −2]; B(3,−2).
Look for these in your work
- Distinguish point and vector notation
- Check the connecting displacement
a = [0; 1], b = [8; 5]. M is one quarter of the way from A to B. Find its position; is it the midpoint?
Give me a hint
Use t = 1/4, not 1/2.
Compare my reasoning
- b−a = [8; 4]; one quarter is [2; 1].
- a+[2; 1] = [2; 2], while midpoint position is [4; 3].
M has position [2; 2]; it is not the midpoint.
Look for these in your work
- Use the stated fraction
- Measure from the correct start
Common mistakes
- A position vector needs a stated origin.
- Do not confuse the connecting vector with the endpoint position.
What can you explain now?
If a = →OA and b = →OB, what is →AB?
Compare with the explanation
b−a.
Following →OA then →AB must end at B.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
If a = →OA and b = →OB, what is →AB?
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.