Vectors
A vector has magnitude (size) and direction. The arithmetic is easy; the marks are in vector geometry — finding a route through a diagram, and proving that points are collinear.
1. Column vectors
A column vector ⎛x⎞ over ⎝y⎠ means x across and y up.
Written inline as (x, y) with x on top: positive x is right, negative x is left; positive y is up, negative y is down.
Notation: vectors appear as a, or as AB with an arrow, or underlined by hand (a̲). Always underline yours — it distinguishes a vector from a length.
Arithmetic — component by component:
If a = (3, 2) and b = (1, −4):
- a + b = (3+1, 2−4) = (4, −2)
- a − b = (3−1, 2+4) = (2, 6)
- 3a = (9, 6)
- −a = (−3, −2) — same length, opposite direction
Add the tops and the bottoms separately. Never cross-multiply or combine x with y.
A negative vector reverses the direction. −a is the same arrow pointing the other way, so BA = −AB. This one fact does most of the work in vector geometry.
Multiplying by a scalar multiplies both components, changing the length but not the direction (unless negative).
2. Magnitude
|a| = √(x² + y²) — Pythagoras.
For a = (3, 4): |a| = √(9 + 16) = √25 = 5
Magnitude is a length, so it is never negative. For (−3, −4) the magnitude is still 5.
3. Vector geometry — finding a route
This is the part worth the most marks. Given OA = a and OB = b, you are asked for other vectors in the diagram.
The key rule: to go from X to Y, travel BACKWARDS along the first vector and FORWARDS along the second. AB = AO + OB = −a + b = b − a
Think of it as a journey. There is no direct arrow from A to B, so go A → O (which is −a, against the arrow) then O → B (which is +b).
| Vector | Route | Result |
|---|---|---|
| AB | A→O→B | b − a |
| BA | B→O→A | a − b |
| OA | direct | a |
AB = b − a, not a − b. The order catches people out constantly. You subtract the start point from the end point — the same rule as coordinates.
Any path works. If a diagram gives you a longer route, use it — the answer will simplify to the same thing. Tutors specifically advised finding two different routes to the same vector as a check.
4. Midpoints and ratios
M is the midpoint of AB:
- AM = ½AB = ½(b − a)
- OM = OA + AM = a + ½(b − a) = ½(a + b)
Get to the point in two stages: reach the line first, then travel along it. OM is “go to A, then half way along AB” — this two-step habit handles every ratio question.
With a ratio. If P divides AB in the ratio 2:3, then P is 2/5 of the way along:
- AP = ⅖AB = ⅖(b − a)
- OP = a + ⅖(b − a)
Convert the ratio to a fraction of the WHOLE. For 2:3 the total is 5 parts, so P is 2/5 along — not 2/3. Misreading the ratio was a documented error.
Always simplify at the end: OP = a + ⅖b − ⅖a = ⅗a + ⅖b
5. Parallel vectors and collinearity
Two vectors are parallel if one is a scalar multiple of the other.
2a + 4b and a + 2b are parallel, because 2a + 4b = 2(a + 2b).
To prove three points are collinear (on the same straight line), show that one vector is a multiple of another and that they share a common point.
Example: show A, B and C are collinear given AB = 2p + 4q and BC = 3p + 6q.
- BC = 3p + 6q = 1.5(2p + 4q) = 1.5 AB
- So BC is parallel to AB
- They share the point B
- Therefore A, B and C are collinear
Both parts are needed for the marks. Parallel alone is not enough — parallel lines can be far apart. The common point is what forces them onto the same line, and it was the half students most often omitted.
You can also state the ratio: BC = 1.5 AB means AB : BC = 2 : 3.
6. Method for harder questions
- Mark the known vectors on the diagram with arrows
- Write down what you’re asked for as a journey between points
- Build it from vectors you already know, reversing sign where you travel against an arrow
- Simplify by collecting like terms
- Check with a second route where possible
Equating coefficients: if a question gives you the same vector expressed two ways, set the a parts equal and the b parts equal to form two equations. Equating only some terms, or the wrong ones, was a recorded error — a and b are independent, so each must balance separately.
Answer in terms of the vectors given. If the question uses p and q, the answer must be in p and q, fully simplified.
7. Mistakes that cost marks
Writing AB as a − b instead of b − a.
Forgetting to reverse the sign when travelling against an arrow.
Combining x and y components.
Giving a negative magnitude.
Reading a ratio 2:3 as 2/3 instead of 2/5.
Not simplifying the final expression.
Proving vectors are parallel but not stating the common point for collinearity.
Equating coefficients carelessly.
Not underlining vectors, so a length and a vector look the same.
Answering in the wrong letters.
Frequently asked questions
What is a column vector? A vector written with the x-component on top and the y-component below: x across, y up.
How do I add vectors? Add the components separately — tops together, bottoms together.
What does −a mean? The same vector in the opposite direction.
How do I find the magnitude? √(x² + y²) — Pythagoras. It’s a length, so never negative.
How do I find AB from OA and OB? AB = b − a — end point minus start point.
Why is it b − a and not a − b? Because you travel backwards along a (giving −a) then forwards along b.
How do I find the midpoint vector? OM = ½(a + b), or go to A then half way along AB.
How do I handle a ratio like 2:3? The total is 5 parts, so the point is 2/5 of the way along.
How do I show two vectors are parallel? Show one is a scalar multiple of the other.
How do I prove three points are collinear? Show two vectors are parallel and that they share a common point.
Quick revision checklist
- I can read and write column vectors
- I add and subtract components separately
- I know −a reverses direction, and BA = −AB
- I can multiply a vector by a scalar
- I can find magnitude with Pythagoras
- I know AB = b − a
- I can build any vector as a journey through the diagram
- I reverse the sign when travelling against an arrow
- I can find midpoint vectors
- I convert ratios to fractions of the whole
- I simplify my final answer in the given letters
- I can show two vectors are parallel
- I prove collinearity with parallel + common point
- I check with a second route where I can
- I underline my vectors
These notes cover vectors in the Cambridge IGCSE Mathematics (0580) syllabus, and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Maths lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and formula list for your own exam series.
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