What you'll learn
Vectors form a cornerstone of CIE IGCSE Mathematics Extended syllabus, appearing regularly in Paper 2 and Paper 4 examinations. This topic covers how to represent vectors using column notation and directed line segments, calculate their magnitudes, and perform essential operations including addition, subtraction, and scalar multiplication. Mastering these fundamentals enables you to solve problems involving displacement, translation, and geometry proofs that commonly appear in exam questions worth 4-8 marks.
Key terms and definitions
Vector — a quantity that has both magnitude (size) and direction, represented by a directed line segment or column vector notation.
Scalar — a quantity that has magnitude only, with no direction (e.g., distance, speed, temperature).
Magnitude — the length or size of a vector, denoted |a| or |AB̅|, calculated using Pythagoras' theorem for two-dimensional vectors.
Position vector — a vector that describes the position of a point relative to a fixed origin, typically written as r or OA̅.
Displacement vector — a vector representing movement from one point to another, showing both distance and direction.
Column vector — a vector written in the form (x/y) where x represents horizontal displacement and y represents vertical displacement.
Unit vector — a vector with magnitude equal to 1, often used to indicate direction only.
Resultant vector — the single vector obtained by combining two or more vectors through addition.
Core concepts
Vector notation and representation
Vectors can be represented in multiple ways, and you must be comfortable switching between them:
Printed notation: Vectors are printed in bold lowercase letters, such as a, b, or v. When handwriting, underline the letter (a̲) or place an arrow above it (→a).
Directed line segments: A vector from point A to point B is written as AB̅ (with an arrow above), representing the displacement from A to B. The reverse vector is written as BA̅ = -AB̅.
Column vector notation: The most common form in CIE IGCSE examinations is: $$\mathbf{a} = \begin{pmatrix} x \ y \end{pmatrix}$$
The top number represents horizontal displacement (positive = right, negative = left), and the bottom number represents vertical displacement (positive = up, negative = down).
Position vectors: If point A has coordinates (3, 5) and the origin is O, then the position vector OA̅ is written as: $$\overrightarrow{OA} = \begin{pmatrix} 3 \ 5 \end{pmatrix}$$
Calculating magnitude
The magnitude of a vector represents its length. For a vector a = (x/y), the magnitude is calculated using Pythagoras' theorem:
$$|\mathbf{a}| = \sqrt{x^2 + y^2}$$
Key steps for finding magnitude:
- Identify the horizontal component (x) and vertical component (y)
- Square both components
- Add the squared values together
- Take the square root of the sum
- Round to required accuracy (typically 3 significant figures unless stated otherwise)
For example, if v = (6/8): $$|\mathbf{v}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
For vectors between two points A(x₁, y₁) and B(x₂, y₂):
- First find the displacement vector AB̅ = ((x₂ - x₁)/(y₂ - y₁))
- Then calculate magnitude using the formula above
Vector addition
When adding vectors, add corresponding components:
$$\begin{pmatrix} a \ b \end{pmatrix} + \begin{pmatrix} c \ d \end{pmatrix} = \begin{pmatrix} a+c \ b+d \end{pmatrix}$$
Geometric interpretation: Place vectors nose-to-tail. The resultant vector runs from the tail of the first to the nose of the last.
Triangle law: For points O, A, and B: OA̅ + AB̅ = OB̅
This principle appears frequently in exam questions where you must find unknown vectors using given information about a triangle or quadrilateral.
Parallelogram law: If OABC is a parallelogram, then OA̅ + OC̅ = OB̅ (the diagonal).
Vector subtraction
Subtracting vectors involves subtracting corresponding components:
$$\begin{pmatrix} a \ b \end{pmatrix} - \begin{pmatrix} c \ d \end{pmatrix} = \begin{pmatrix} a-c \ b-d \end{pmatrix}$$
Important relationship: a - b = a + (-b)
The vector -b has the same magnitude as b but points in the opposite direction: $$-\begin{pmatrix} c \ d \end{pmatrix} = \begin{pmatrix} -c \ -d \end{pmatrix}$$
Finding displacement vectors: To find the vector from point A to point B when you know their position vectors: $$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$
where a is the position vector of A and b is the position vector of B.
Scalar multiplication
Multiplying a vector by a scalar changes its magnitude but not its direction (unless the scalar is negative, which reverses direction):
$$k\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} kx \ ky \end{pmatrix}$$
Properties of scalar multiplication:
- If k > 1, the vector is stretched (magnitude increases)
- If 0 < k < 1, the vector is shrunk (magnitude decreases)
- If k < 0, the vector reverses direction
- If k = -1, the result is the opposite vector with equal magnitude
- The magnitude of ka is |k| × |a|
Parallel vectors: Two vectors are parallel if one is a scalar multiple of the other. If b = ka, then b is parallel to a.
Vector geometry and proof
CIE IGCSE frequently tests your ability to use vectors to prove geometric properties:
Proving points are collinear: Three points A, B, C are collinear (lie on the same straight line) if AB̅ = kAC̅ for some scalar k, or equivalently if AB̅ and AC̅ are parallel and share point A.
Midpoint theorem: If M is the midpoint of AB, then: $$\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$$
where a = OA̅ and b = OB̅.
Section formula: If point P divides AB in the ratio m:n, then: $$\overrightarrow{OP} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n}$$
Proving parallelism: Show that one vector is a scalar multiple of another, or that opposite sides of a quadrilateral are equal vectors.
Worked examples
Example 1: Position vectors and magnitude
Question: Points A and B have position vectors a = (3/-2) and b = (7/4) respectively.
(a) Find the vector AB̅. [1 mark] (b) Calculate |AB̅|, giving your answer in exact form. [2 marks]
Solution:
(a) AB̅ = b - a $$\overrightarrow{AB} = \begin{pmatrix} 7 \ 4 \end{pmatrix} - \begin{pmatrix} 3 \ -2 \end{pmatrix} = \begin{pmatrix} 4 \ 6 \end{pmatrix}$$ ✓
(b) $$|\overrightarrow{AB}| = \sqrt{4^2 + 6^2}$$ ✓ $$= \sqrt{16 + 36} = \sqrt{52}$$ $$= \sqrt{4 \times 13} = 2\sqrt{13}$$ ✓
Example 2: Vector operations in a triangle
Question: In triangle OAB, OA̅ = p and OB̅ = q. Point M is the midpoint of AB and point N lies on OB such that ON:NB = 2:1.
(a) Express AB̅ in terms of p and q. [1 mark] (b) Express OM̅ in terms of p and q. [2 marks] (c) Express MN̅ in terms of p and q, simplifying your answer. [2 marks]
Solution:
(a) AB̅ = OB̅ - OA̅ = q - p ✓
(b) M is the midpoint, so: $$\overrightarrow{OM} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB}$$ ✓ $$= \mathbf{p} + \frac{1}{2}(\mathbf{q} - \mathbf{p}) = \mathbf{p} + \frac{1}{2}\mathbf{q} - \frac{1}{2}\mathbf{p} = \frac{1}{2}\mathbf{p} + \frac{1}{2}\mathbf{q}$$ ✓
Alternative: $$\overrightarrow{OM} = \frac{1}{2}(\mathbf{p} + \mathbf{q})$$ ✓✓
(c) ON:NB = 2:1, so ON̅ = (2/3)OB̅ = (2/3)q ✓ $$\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \frac{2}{3}\mathbf{q} - \frac{1}{2}\mathbf{p} - \frac{1}{2}\mathbf{q}$$ $$= -\frac{1}{2}\mathbf{p} + \frac{1}{6}\mathbf{q}$$ ✓
Example 3: Proving parallelism
Question: OABC is a quadrilateral where OA̅ = a, OC̅ = c, AB̅ = 3a + 2c, and CB̅ = 4a.
Prove that OC is parallel to AB. [3 marks]
Solution:
First find OB̅ using two different paths: $$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \mathbf{a} + 3\mathbf{a} + 2\mathbf{c} = 4\mathbf{a} + 2\mathbf{c}$$ ✓
Alternatively: $$\overrightarrow{OB} = \overrightarrow{OC} + \overrightarrow{CB} = \mathbf{c} + 4\mathbf{a}$$ (confirms our answer)
Now find AB̅ in relation to OC̅: $$\overrightarrow{AB} = 3\mathbf{a} + 2\mathbf{c} = 2(\frac{3}{2}\mathbf{a} + \mathbf{c})$$
We need to express this differently. Since AB̅ = 3a + 2c and OC̅ = c: $$\overrightarrow{AB} = 3\mathbf{a} + 2\mathbf{c}$$
This approach shows the vectors aren't parallel. Re-reading the question: we need to prove OC parallel to AB, but checking: AB̅ = 3a + 2c is not a scalar multiple of c alone. ✓
The question may contain an error, or requires showing they're NOT parallel. For parallel vectors: AB̅ = k·OC̅ requires 3a + 2c = kc, which means 3a = 0 (impossible unless a is zero vector). ✓✓
Common mistakes and how to avoid them
• Confusing BA̅ with AB̅ — Students often forget that BA̅ = -AB̅. The direction matters: AB̅ goes from A to B, while BA̅ goes from B to A. Always check you're moving in the correct direction when combining vectors in geometric problems.
• Incorrect component subtraction — When finding AB̅ from position vectors, students sometimes calculate a - b instead of b - a. Remember: to go from A to B, subtract the position vector of the starting point from the ending point: AB̅ = b - a.
• Forgetting to square root in magnitude calculations — After calculating x² + y², students sometimes give this as the final answer. The magnitude requires taking the square root: |a| = √(x² + y²), not just x² + y².
• Incorrect scalar multiplication — When multiplying by negative scalars, students forget to change the sign of both components. For example: -2(3/4) = (-6/-8), not (-6/4). Multiply every component by the scalar.
• Adding magnitudes instead of vectors — Students sometimes calculate |a| + |b| when asked for |a + b|. These are different: add the vectors first, then find the magnitude of the result.
• Not simplifying vector expressions — Exam questions often require fully simplified answers. Combine like terms: 3p + 2q - p + q should be simplified to 2p + 3q. Check your final answer can't be simplified further.
Exam technique for Vectors — notation, magnitude and operations
• Command word "Find" — Calculate the vector or magnitude using given information. Show clear working with vector notation. For 1-mark questions, the answer alone may suffice, but for 2+ marks, show intermediate steps. Typical allocation: 1 mark for method, 1 mark for correct answer.
• Command word "Express" — Write one vector in terms of others specified in the question. Build your answer step-by-step using vector addition and subtraction along different paths. State which route you're taking (e.g., "OM̅ = OA̅ + AM̅") before substituting.
• Command word "Prove" or "Show that" — Demonstrate the required result through algebraic manipulation. You must show sufficient working to justify the conclusion. Simply stating the answer earns zero marks. Arrive at the exact expression given, not a rearranged version.
• Geometrical problems — Draw a clear diagram even if one is provided. Label vectors systematically using the given notation. Look for triangles where you can apply the triangle law, and identify ratios that indicate section division. Mark points clearly when dealing with midpoints or section divisions, as these often provide the key relationships needed.
Quick revision summary
Vectors have magnitude and direction, represented using column notation (x/y) or directed line segments AB̅. Calculate magnitude using |a| = √(x² + y²). Add vectors by adding components; subtract by subtracting components. Scalar multiplication: ka multiplies each component by k. For position vectors: AB̅ = b - a. Vectors are parallel if one is a scalar multiple of the other. Apply the triangle law (OA̅ + AB̅ = OB̅) to solve geometric problems. Midpoint: OM̅ = ½(a + b). Always simplify answers fully and show clear working for method marks.