Cambridge IGCSE Mathematics 0580

Vectors

A vector is a quantity with both magnitude and direction. In Cambridge IGCSE 0580 you meet vectors two ways: as column vectors $\binom{x}{y}$ describing movements on a grid, and as letter vectors like $\mathbf{a}$ and $\mathbf{b}$ in geometry questions, where you build routes through a shape. The second kind is where Extended marks are won and lost — Cambridge examiner reports note multi-step vector routes are “generally not well answered”.

Updated 20 August 2026

Column vectors and magnitude

$\binom{3}{-2}$ means 3 right and 2 down. Add vectors by adding components; multiply by a number by scaling both components:

\binom{3}{-2} + \binom{-1}{5} = \binom{2}{3}, \qquad 3\binom{2}{-1} = \binom{6}{-3}

The magnitude (length) of a vector is Pythagoras on its components:

\left|\binom{x}{y}\right| = \sqrt{x^2 + y^2}

Magnitude answers are lengths — leave them as simplified surds or round as instructed, and never drop the square root.

Vector routes through a shape

Geometry questions define a shape with two base vectors — say $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ — and ask for other vectors in terms of $\mathbf{a}$ and $\mathbf{b}$ . The method never changes: walk from the start letter to the end letter along edges you know, writing each leg with its sign. Walking against an arrow negates it:

\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}

Ratios scale the legs. If $M$ lies on $OA$ with $OM:MA = 2:1$ , then $\overrightarrow{OM} = \tfrac{2}{3}\mathbf{a}$ — the fraction is part over whole, 2 parts of 3.

Position vectors and midpoints

The position vector of a point is its vector from the origin $O$ . The midpoint $M$ of $AB$ has position vector

\overrightarrow{OM} = \tfrac{1}{2}\left(\mathbf{a} + \mathbf{b}\right)

— the average of the endpoints’ position vectors, exactly like a coordinate midpoint.

Worked example

OACB is a quadrilateral with OA = a, OB = b, and BC parallel to OA with BC = ¾·OA. M is the midpoint of AC. Find AM in terms of a and b, in simplest form.

Step 1 — get to C. Walk O → B → C: $\overrightarrow{OC} = \mathbf{b} + \tfrac{3}{4}\mathbf{a}$ (the leg $BC$ is parallel to $OA$ and $\tfrac34$ of its length, same direction).

Step 2 — the target route. $\overrightarrow{AM} = \overrightarrow{AC} \times \tfrac12$ won’t work directly until we know $\overrightarrow{AC}$ . Walk A → O → C:

\overrightarrow{AC} = -\mathbf{a} + \left(\mathbf{b} + \tfrac{3}{4}\mathbf{a}\right) = -\tfrac{1}{4}\mathbf{a} + \mathbf{b}

Step 3 — halve it. $M$ is the midpoint of $AC$ , so

\overrightarrow{AM} = \tfrac{1}{2}\overrightarrow{AC} = -\tfrac{1}{8}\mathbf{a} + \tfrac{1}{2}\mathbf{b}

Mark schemes award the intermediate vector ( $\overrightarrow{OC}$ or $\overrightarrow{AC}$ ) even when the final simplification slips — write the route down before simplifying.

Proving parallel and collinear

Two vectors are parallel when one is a scalar multiple of the other: $\mathbf{u} = k\mathbf{v}$ . To prove three points $A, B, C$ are collinear (on one straight line): show $\overrightarrow{AB} = k\overrightarrow{BC}$ for some number $k$ , then say the two vectors are parallel and share the point $B$ . Both halves of that sentence are needed for the mark — parallel alone doesn’t put the points on one line.

The mistakes that lose the marks

1.Dropping the minus when walking against an arrow
$\overrightarrow{AO} = -\mathbf{a}$ , not $\mathbf{a}$ . Before combining anything, write the route letter-by-letter and attach signs leg by leg — don’t do it in your head.
2.Order-of-operations slips in expressions like a − 2b
Scale first, then add: $\mathbf{a} - 2\mathbf{b}$ means $\mathbf{a} + (-2\mathbf{b})$ . Examiners specifically note candidates mangling exactly this shape.
3.Ratio fractions using part-over-part
$OM:MA = 2:1$ puts $M$ at $\tfrac{2}{3}$ of the way along — 2 parts out of 3 total, never $\tfrac{2}{1}$ or $\tfrac{1}{2}$ .
4.Claiming collinearity from parallelism alone
Finish the sentence: the vectors are parallel and pass through a common point. Without the shared point the proof mark is withheld.
5.Magnitude without the square root
$|\binom{6}{-8}| = \sqrt{36+64} = 10$ — candidates regularly stop at 100. If your “length” looks suspiciously huge, you forgot the root.

Common questions

What does a bold letter like a mean in a vector question?

A named vector — in print it’s bold ( $\mathbf{a}$ ), in handwriting you underline it. It stands for a fixed movement, e.g. $\overrightarrow{OA}$ , and your answers should be combinations like $2\mathbf{a} - \tfrac{1}{2}\mathbf{b}$ .

Are vectors on the Core papers?

Column vectors, addition and scalar multiplication are Core. Magnitude, position vectors and the geometric proof-style questions in letters are Extended.

How do you know which route to walk?

Any route works if every leg is known — that’s the freedom of the method. Prefer routes through the origin or through labelled points, and if you stall, write down every vector you can express first; the target is usually one join away.

Practise vectors against real mark schemes

Quanta has real Cambridge IGCSE Maths 0580 past-paper questions, broken into skill checkpoints, marked criterion by criterion the way examiners mark — and it tracks which skills you’re missing. Free for individual students.

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