What is a vector in the context of Cambridge IGCSE Mathematics?

In Cambridge IGCSE Mathematics, a vector is a mathematical entity that has both magnitude and direction. It is often represented as an arrow in a coordinate plane, where the length of the arrow indicates the magnitude and the direction of the arrow shows the direction. Vectors are used to describe quantities that have both size and direction, such as force or velocity.

How do you add and subtract vectors in Mathematics?

To add vectors, you place the tail of the second vector at the head of the first vector and draw a new vector from the tail of the first to the head of the second. This is known as the 'tip-to-tail' method. For subtraction, you add the negative of the vector you want to subtract. This involves reversing the direction of the vector to be subtracted and then adding it to the first vector using the same 'tip-to-tail' method.

What is the difference between a scalar and a vector?

A scalar is a quantity that has only magnitude, such as temperature or mass, while a vector has both magnitude and direction, such as velocity or force. Scalars are represented by simple numerical values, whereas vectors are represented by arrows or coordinates in a plane. Understanding the distinction between scalars and vectors is crucial in solving problems related to vectors and transformations in Cambridge IGCSE Mathematics.

How do you find the magnitude of a vector?

The magnitude of a vector can be found using the Pythagorean theorem. If a vector is represented by coordinates (x, y) in a 2D plane, its magnitude is calculated as the square root of the sum of the squares of its components: √(x² + y²). This formula gives the length of the vector, which is an essential concept in vector mathematics at the Cambridge IGCSE level.

What are unit vectors and how are they used in vector mathematics?

A unit vector is a vector with a magnitude of 1. It is used to indicate direction without regard to magnitude. In vector mathematics, unit vectors are often used to express a vector in terms of its direction. For example, in a 2D coordinate system, the unit vectors i and j represent the x and y directions, respectively. Unit vectors are fundamental in breaking down vectors into their components, a key concept in the Cambridge IGCSE Mathematics curriculum.

Why is "Computing $\overrightarrow{AB}$ as $\mathbf{a} - \mathbf{b}$" a common mistake in Vectors?

Why it happens: Confusing the order of subtraction. How to avoid it: AB = OB - OA = b - a. End point minus start point. Source: 0580/42 — recurring

Why is "Forgetting the square root in magnitude" a common mistake in Vectors?

Why it happens: Speed. How to avoid it: Magnitude formula has √(\ ) — not just x^2 + y^2.

Why is "Multiplying only one component by the scalar" a common mistake in Vectors?

Why it happens: Forgetting the scalar applies to BOTH components. How to avoid it: k pmatrix a b pmatrix = pmatrix ka kb pmatrix — multiply BOTH.

Why is "Treating $\mathbf{a}$ and $\mathbf{b}$ as scalars in proofs" a common mistake in Vectors?

Why it happens: They look like ordinary letters in algebra. How to avoid it: Vectors are bolded or have arrows. Don't divide by a vector or take square roots of vectors.

Cambridge IGCSE Mathematics (0580)

Vectors

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Short Notes - Vectors

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Detailed Study Notes

Detailed notes on Vectors and Transfomations for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Vectors — Cambridge IGCSE 0580 Maths Extended (2026)

Magnitude AND direction. Add, subtract, scale, find magnitude, and reason geometrically with position vectors. The skill that unlocks Paper 4 "prove these points are collinear" or "find $\overrightarrow{AB}$ " questions.

At a glance

A vector has magnitude AND direction. Written as a column $\begin{pmatrix} x \\ y \end{pmatrix}$ or as a letter (bold $\mathbf{a}$ or arrow $\vec{a}$ ).
Add component-wise: $\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a + c \\ b + d \end{pmatrix}$ .
Subtract similarly: component-wise.
Scalar multiplication: $k \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}$ .
Magnitude: $|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ .
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ (end minus start).
Two vectors parallel ⇔ one is a scalar multiple of the other.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E5.1 — Describe a translation using a vector represented by $\begin{pmatrix} x \\ y \end{pmatrix}$ .
E5.2 — Add and subtract vectors; multiply a vector by a scalar.
E5.3 — Calculate the magnitude of a vector.
E5.4 — Use position vectors to describe paths in geometric problems.

Vector arithmetic

Add and subtract component-wise. Scalar multiplication scales each component.

Addition. $\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a + c \\ b + d \end{pmatrix}.$

Subtraction. $\begin{pmatrix} a \\ b \end{pmatrix} - \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a - c \\ b - d \end{pmatrix}.$

Scalar multiplication. $k \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}.$

The scalar $k$ multiplies BOTH components.

Worked. $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ , $\mathbf{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$ . Find $2\mathbf{a} - 3\mathbf{b}$ .

$2\mathbf{a} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}$ .
$3\mathbf{b} = \begin{pmatrix} -3 \\ 12 \end{pmatrix}$ .
$2\mathbf{a} - 3\mathbf{b} = \begin{pmatrix} 6 - (-3) \\ -4 - 12 \end{pmatrix} = \begin{pmatrix} 9 \\ -16 \end{pmatrix}$ .

Add and subtract component by component.
Scalar multiplication: scale BOTH components.
Don't divide by a vector — undefined.
Always show the column vectors clearly.

Magnitude (length) of a vector

Pythagoras on the components. $|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ .

Formula. $|\mathbf{v}|$ for $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ : $|\mathbf{v}| = \sqrt{x^{2} + y^{2}}.$

This is just Pythagoras — the magnitude is the length of the vector if you draw it as an arrow.

Worked. $\mathbf{v} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}$ .

$|\mathbf{v}| = \sqrt{25 + 144} = \sqrt{169} = 13$ .

The magnitude is the hypotenuse of a right triangle built from the components.

Magnitude is always positive. And $|\mathbf{0}| = 0$ — only the zero vector has zero magnitude.

$|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ .
Always non-negative.
Pythagoras applied to the components.

Position vectors and $\overrightarrow{AB}$

$\overrightarrow{AB}$ is end minus start. Position vectors point from origin.

A position vector $\overrightarrow{OA}$ points from the origin $O$ to the point $A$ .

Path between two points. $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}.$ End minus start. The $\overrightarrow{}$ notation tells you direction matters.

To travel from A to B, subtract the start position vector from the end one.

Worked. $A(2, 1)$ , $B(7, 5)$ .

$\overrightarrow{AB} = \begin{pmatrix} 7 - 2 \\ 5 - 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}$ .
$\overrightarrow{BA} = \begin{pmatrix} -5 \\ -4 \end{pmatrix}$ (just negate).

Midpoint. If $M$ is the midpoint of $AB$ , $\overrightarrow{OM} = \dfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}).$

Triangle path navigation. $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ . (Chain the trips.)

Cambridge geometry style. Often Cambridge gives you $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ and asks you to express $\overrightarrow{AB}$ , $\overrightarrow{AM}$ , etc. in terms of $\mathbf{a}$ and $\mathbf{b}$ .

Worked. $\overrightarrow{OA} = \mathbf{a}$ , $\overrightarrow{OB} = \mathbf{b}$ . $M$ is the midpoint of $AB$ . Express $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .

$\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ .
Midpoint: $\overrightarrow{OM} = \dfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})$ .
Chain: $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ .
Reverse: $\overrightarrow{BA} = -\overrightarrow{AB}$ .

Parallel and collinear vectors

Two vectors are parallel iff one is a scalar multiple of the other. Three points are collinear iff a vector chain factorises this way.

Parallel test. $\mathbf{u}$ is parallel to $\mathbf{v}$ if and only if there exists a scalar $k \ne 0$ such that $\mathbf{u} = k\mathbf{v}.$

Worked. $\mathbf{u} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ , $\mathbf{v} = \begin{pmatrix} -6 \\ 3 \end{pmatrix}$ . Are they parallel?

Try $k$ : $\dfrac{-6}{4} = -1.5$ and $\dfrac{3}{-2} = -1.5$ . Same $k$ .
Yes, $\mathbf{v} = -1.5\mathbf{u}$ . Parallel.

Collinear points. Three points $A, B, C$ are collinear iff $\overrightarrow{AB}$ is a scalar multiple of $\overrightarrow{AC}$ (they share the same line).

Worked example structure. "Show $A, B, C$ are collinear" → compute $\overrightarrow{AB}$ and $\overrightarrow{AC}$ , factor out a common scalar.

Parallel: $\mathbf{u} = k\mathbf{v}$ .
Collinear: $\overrightarrow{AB} = k \cdot \overrightarrow{AC}$ .
$k$ can be negative or fractional.
Both components must give the SAME $k$ .

Quick recap

Add/subtract: component by component.
Scalar multiplication: scales BOTH components.
Magnitude: $\sqrt{x^{2} + y^{2}}$ .
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ .
Midpoint: $\dfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})$ .
Parallel: scalar multiple. Collinear: $\overrightarrow{AB} = k \overrightarrow{AC}$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Vector — quantity with magnitude and direction.
Magnitude — length of the vector: $|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ .
Position vector — vector from origin to a point.
Parallel vectors — one is a scalar multiple of the other.
Collinear points — points on the same straight line.

How it’s examined

Vectors appear every Paper 4 as a 5-7 mark question — typically position-vector geometry with $\mathbf{a}$ and $\mathbf{b}$ given. Paper 2 has 2-3 mark single-step items. Examiner reports flag sign errors in $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (writing $\mathbf{a} - \mathbf{b}$ instead).

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E5.1-5.4); 0580/42 Oct/Nov 2024 — Q15 (position-vector geometry); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Vectors

Step-by-step solutions to past-paper-style questions on vectors, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Vectors

Almost every vectors exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find, work out, express $\overrightarrow{AB}$ in terms of … — acting on given column vectors or position vectors, with one method to apply.

How to approach it

Pick the matching rule: combine components for addition/subtraction/scalar multiples, use $\sqrt{x^{2}+y^{2}}$ for magnitude, and $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ for a path. Show each line of working.

Common trap

Examiner reports flag subtracting in the wrong order ( $\overrightarrow{OA}-\overrightarrow{OB}$ ), multiplying only one component by a scalar, and omitting the square root in the magnitude formula.

Multi-step problem

Recognise it by

A ratio division of a line, a midpoint, or a parallelogram/quadrilateral diagram where $\overrightarrow{OC}$ or $\overrightarrow{OM}$ must be built by chaining several vectors.

How to approach it

Build a route from the origin: travel $\overrightarrow{OA}$ then add the fraction of $\overrightarrow{AB}$ implied by the ratio. Convert a ratio $AC:CB = 2:1$ into the fraction $AC = \tfrac{2}{3}AB$ before substituting.

Common trap

Examiner reports flag the wrong ratio fraction (using $\tfrac{1}{2}$ for a $2:1$ split) and mis-ordering parallelogram vertices, which swaps which sides are equal.

Show that / prove

Recognise it by

The words show that, prove or justify — proving vectors are parallel, points are collinear, or a quadrilateral is a parallelogram. The conclusion is already stated.

How to approach it

Express one vector as a scalar multiple of another to show parallelism; for collinearity also state the shared point; for a parallelogram show two opposite-side vectors are equal (not just parallel). End with an explicit conclusion line.

Common trap

Examiner reports flag stopping at 'parallel' without naming the common point for collinearity, and proving sides are only parallel (a trapezium) when equality of vectors is needed for a parallelogram.

1Add and subtract column vectors
CoreDirect calculation• addition
Question
Given $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$ , find $\mathbf{a} + 2\mathbf{b}$ .
Step-by-step solution
1. Step 1
  Compute $2\mathbf{b}$ .
  $2\mathbf{b} = \begin{pmatrix} -2 \\ 8 \end{pmatrix}$
2. Step 2
  Add component-wise.
  $\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 3 + (-2) \\ -2 + 8 \end{pmatrix} = \begin{pmatrix} 1 \\ 6 \end{pmatrix}$
Answer
$\begin{pmatrix} 1 \\ 6 \end{pmatrix}$
2Find the magnitude of a vector
CoreDirect calculation• magnitude
Question
Find $|\mathbf{v}|$ where $\mathbf{v} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}$ .
Step-by-step solution
1. Step 1
  $|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ .
  $= \sqrt{25 + 144} = \sqrt{169} = 13$
Answer
$|\mathbf{v}| = 13$
3Use position vectors to find a path
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q16• position vector
Question
$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . Express $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Step-by-step solution
1. Step 1
  Use $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ .
  $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
Answer
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
4Position vector of a midpoint
ExtendedDirect calculation• midpoint
Question
$M$ is the midpoint of $AB$ . Express $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Step-by-step solution
1. Step 1
  Midpoint position vector is the average.
  $\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$
Answer
$\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$
5Show two vectors are parallel
ExtendedShow that / prove• parallel
Question
$\mathbf{u} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} -6 \\ 3 \end{pmatrix}$ . Show that $\mathbf{u}$ is parallel to $\mathbf{v}$ .
Step-by-step solution
1. Step 1
  $\mathbf{v} = -\dfrac{3}{2}\mathbf{u}$ since $-\dfrac{3}{2} \times 4 = -6$ and $-\dfrac{3}{2} \times -2 = 3$ .
2. Step 2
  One is a scalar multiple of the other → parallel.
Answer
$\mathbf{v} = -\dfrac{3}{2}\mathbf{u}$ , hence parallel.
6Scalar multiplication of a column vector
ExtendedDirect calculation• scalar multiple
Question
Given $\mathbf{p} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}$ , find $-4\mathbf{p}$ .
Step-by-step solution
1. Step 1
  Multiply both components by $-4$ .
  $-4\mathbf{p} = \begin{pmatrix} -4 \times -3 \\ -4 \times 5 \end{pmatrix} = \begin{pmatrix} 12 \\ -20 \end{pmatrix}$
Answer
$\begin{pmatrix} 12 \\ -20 \end{pmatrix}$
Examiner tip
The examiner report flags candidates who multiply only one component by the scalar. The scalar applies to BOTH components.
7Find $\overrightarrow{AB}$ from two column vectors
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2023 Q15• AB, subtraction
Question
Given $\overrightarrow{OA} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} -4 \\ 5 \end{pmatrix}$ , find $\overrightarrow{AB}$ .
Step-by-step solution
1. Step 1
  $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ .
2. Step 2
  Subtract component-wise.
  $\overrightarrow{AB} = \begin{pmatrix} -4 - 2 \\ 5 - (-3) \end{pmatrix} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}$
Answer
$\overrightarrow{AB} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}$
Examiner tip
The examiner report flags candidates who compute $\overrightarrow{OA} - \overrightarrow{OB}$ instead. Always: end point minus start point.
8Distance between two points using vectors
ExtendedDirect calculation• distance, magnitude
Question
Points $P(1, -2)$ and $Q(5, 4)$ are given. Find the distance $PQ$ .
Step-by-step solution
1. Step 1
  $\overrightarrow{PQ} = \begin{pmatrix} 5 - 1 \\ 4 - (-2) \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$ .
2. Step 2
  Magnitude.
  $|\overrightarrow{PQ}| = \sqrt{4^{2} + 6^{2}} = \sqrt{52} = 2\sqrt{13}$
Answer
$PQ = 2\sqrt{13} \approx 7.21$
9Find $k$ so two vectors are parallel
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2023 Q17• parallel, find k
Question
Vectors $\mathbf{a} = \begin{pmatrix} 3 \\ k \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 9 \\ -6 \end{pmatrix}$ are parallel. Find $k$ .
Step-by-step solution
1. Step 1
  Parallel ⟹ one is a scalar multiple of the other. From the $x$ -components: $\mathbf{b} = 3\mathbf{a}$ .
2. Step 2
  Apply the same factor to $y$ -components.
  $-6 = 3k \implies k = -2$
Answer
$k = -2$
10Find a position vector from a partial path
ExtendedMulti-step problem• position vector
Question
$\overrightarrow{OA} = \mathbf{a}$ , $\overrightarrow{OB} = \mathbf{b}$ . Point $C$ lies on $AB$ such that $AC : CB = 2 : 1$ . Express $\overrightarrow{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Step-by-step solution
1. Step 1
  $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ .
2. Step 2
  $AC = \dfrac{2}{3}AB$ since $AC : CB = 2 : 1$ .
  $\overrightarrow{AC} = \dfrac{2}{3}(\mathbf{b} - \mathbf{a})$
3. Step 3
  $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC}$ .
  $\overrightarrow{OC} = \mathbf{a} + \dfrac{2}{3}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}$
Answer
$\overrightarrow{OC} = \dfrac{1}{3}\mathbf{a} + \dfrac{2}{3}\mathbf{b}$
Examiner tip
The examiner report flags candidates who use the wrong fraction (e.g. $\dfrac{1}{2}$ ). $AC : CB = 2 : 1$ means $C$ is two-thirds of the way from $A$ to $B$ , so $AC = \dfrac{2}{3}AB$ .
11Position-vector word problem
ChallengeMulti-step problem• Adapted from 0580/42 Oct/Nov 2024 Q15• position vector, word problem
Question
$OABC$ is a parallelogram. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$ . $M$ is the midpoint of $AB$ . Express $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{c}$ .
Step-by-step solution
1. Step 1
  In a parallelogram $OABC$ with $O \to A \to B \to C \to O$ , we have $\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}$ .
2. Step 2
  $M$ is the midpoint of $AB$ , so $\overrightarrow{AM} = \dfrac{1}{2}\mathbf{c}$ .
3. Step 3
  Build the path from $O$ to $M$ .
  $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \dfrac{1}{2}\mathbf{c}$
Answer
$\overrightarrow{OM} = \mathbf{a} + \dfrac{1}{2}\mathbf{c}$
Examiner tip
The examiner report flags candidates who confuse the order of vertices. In a parallelogram $OABC$ , opposite sides give equal vectors: $\overrightarrow{OA} = \overrightarrow{CB}$ and $\overrightarrow{OC} = \overrightarrow{AB}$ .
12Show that three points are collinear
ChallengeShow that / prove• Adapted from 0580/42 Feb/Mar 2024 Q15• collinear, proof
Question
$\overrightarrow{OA} = \mathbf{a}$ , $\overrightarrow{OB} = \mathbf{b}$ . Point $P$ has position vector $\dfrac{1}{2}(\mathbf{a} + \mathbf{b})$ and point $Q$ has position vector $\dfrac{3}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b}$ . Show that $O$ , $P$ and $Q$ are collinear.
Step-by-step solution
1. Step 1
  Express $\overrightarrow{OQ}$ in terms of $\overrightarrow{OP}$ .
  $\overrightarrow{OQ} = \dfrac{3}{4}\mathbf{a} + \dfrac{3}{4}\mathbf{b} = \dfrac{3}{2}\times \dfrac{1}{2}(\mathbf{a} + \mathbf{b}) = \dfrac{3}{2}\overrightarrow{OP}$
2. Step 2
  $\overrightarrow{OQ}$ is a scalar multiple of $\overrightarrow{OP}$ , and both have the common point $O$ .
3. Step 3
  Therefore the three points $O$ , $P$ , $Q$ lie on a common straight line through $O$ — they are collinear.
Answer
$\overrightarrow{OQ} = \dfrac{3}{2}\overrightarrow{OP}$ , so $O$ , $P$ , $Q$ lie on the same straight line.
Examiner tip
The examiner report flags candidates who stop at "parallel" without stating that there is a COMMON POINT. Parallel vectors with a shared point ⟹ collinear; without a shared point, they are just parallel.
13Prove a quadrilateral is a parallelogram using vectors
ChallengeShow that / prove• parallelogram, proof
Question
Points $P(1, 2)$ , $Q(5, 3)$ , $R(7, 7)$ , $S(3, 6)$ are given. Use vectors to prove that $PQRS$ is a parallelogram.
Step-by-step solution
1. Step 1
  Find $\overrightarrow{PQ}$ .
  $\overrightarrow{PQ} = \begin{pmatrix} 5 - 1 \\ 3 - 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$
2. Step 2
  Find $\overrightarrow{SR}$ .
  $\overrightarrow{SR} = \begin{pmatrix} 7 - 3 \\ 7 - 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$
3. Step 3
  $\overrightarrow{PQ} = \overrightarrow{SR}$ , so $PQ$ is both equal in magnitude and parallel to $SR$ — therefore $PQRS$ has one pair of opposite sides equal and parallel, which is sufficient for a parallelogram.
Answer
$\overrightarrow{PQ} = \overrightarrow{SR} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ , hence $PQRS$ is a parallelogram.
Examiner tip
The examiner report flags candidates who only show $\overrightarrow{PQ}$ is parallel to $\overrightarrow{SR}$ . Equality of vectors (same magnitude AND same direction) is required to conclude parallelogram — parallel-only would just give a trapezium.

Key Formulae — Vectors

The formulae you need to memorise for vectors on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Magnitude of a 2D vector
$\left|\begin{pmatrix} x \\ y \end{pmatrix}\right| = \sqrt{x^{2} + y^{2}}$
When to use
Length of a vector or distance between two points (via vector form).
Vector addition / subtraction
$\begin{pmatrix} a \\ b \end{pmatrix} \pm \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a \pm c \\ b \pm d \end{pmatrix}$
When to use
Combine vectors component-wise.
Path between two points
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$
When to use
When given position vectors and asked for the vector between two points.
Parallel vectors
$\mathbf{u} \parallel \mathbf{v} \iff \mathbf{v} = k\mathbf{u}$
$k$
non-zero scalar
When to use
To prove two vectors are parallel.

Key Definitions and Keywords — Vectors

Definitions to memorise and the exact keywords mark schemes credit for vectors answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Vector
Examiner keyword
A quantity with both magnitude and direction. Often written as a column vector $\begin{pmatrix} x \\ y \end{pmatrix}$ .
Scalar
A pure number (no direction).
Magnitude
Examiner keyword
The length of a vector, written $|\mathbf{v}|$ .
Position vector
Examiner keyword
A vector from a fixed origin $O$ to a point $A$ , written $\overrightarrow{OA}$ or $\mathbf{a}$ .
Parallel vectors
Two vectors are parallel iff one is a scalar multiple of the other.

Common Mistakes and Misconceptions — Vectors

The traps other students keep falling into on vectors questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Computing $\overrightarrow{AB}$ as $\mathbf{a} - \mathbf{b}$
0580/42 — recurring
Why it happens
Confusing the order of subtraction.
How to avoid it
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$ . End point minus start point.
✕Forgetting the square root in magnitude
Why it happens
Speed.
How to avoid it
Magnitude formula has $\sqrt{\ }$ — not just $x^{2} + y^{2}$ .
✕Multiplying only one component by the scalar
Why it happens
Forgetting the scalar applies to BOTH components.
How to avoid it
$k \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}$ — multiply BOTH.
✕Treating $\mathbf{a}$ and $\mathbf{b}$ as scalars in proofs
Why it happens
They look like ordinary letters in algebra.
How to avoid it
Vectors are bolded or have arrows. Don't divide by a vector or take square roots of vectors.

Vectors — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Vectors

Short Notes - Vectors

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Vectors

1Add and subtract column vectors

2Find the magnitude of a vector

3Use position vectors to find a path

4Position vector of a midpoint

5Show two vectors are parallel

6Scalar multiplication of a column vector

7Find AB→\overrightarrow{AB}AB from two column vectors

8Distance between two points using vectors

9Find kkk so two vectors are parallel

10Find a position vector from a partial path

11Position-vector word problem

12Show that three points are collinear

13Prove a quadrilateral is a parallelogram using vectors

Magnitude of a 2D vector

Vector addition / subtraction

Path between two points

Parallel vectors

Vector

Scalar

Magnitude

Position vector

Parallel vectors

✕Computing AB→\overrightarrow{AB}AB as a−b\mathbf{a} - \mathbf{b}a−b

✕Forgetting the square root in magnitude

✕Multiplying only one component by the scalar

✕Treating a\mathbf{a}a and b\mathbf{b}b as scalars in proofs

Vectors — frequently asked questions

7Find $\overrightarrow{AB}$ from two column vectors

9Find $k$ so two vectors are parallel

✕Computing $\overrightarrow{AB}$ as $\mathbf{a} - \mathbf{b}$

✕Treating $\mathbf{a}$ and $\mathbf{b}$ as scalars in proofs