What you'll learn
This revision guide covers vector notation, magnitude and direction—topics that regularly appear in CIE IGCSE Mathematics Paper 2 and Paper 4 (Extended). You'll master how to represent vectors using different notations, calculate their magnitude using Pythagoras' theorem, and determine direction using trigonometry. These skills form the foundation for vector arithmetic and geometry questions worth 4-8 marks in typical examination papers.
Key terms and definitions
Vector — a quantity that has both magnitude (size) and direction, represented by a directed line segment
Scalar — a quantity that has only magnitude with no direction, such as distance, speed or temperature
Magnitude — the size or length of a vector, denoted |a| or |AB→|, always expressed as a positive value
Position vector — a vector that describes the position of a point relative to the origin O, written as r or OP where P is the point
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 1, often used to indicate direction only
Direction — the angle a vector makes with a reference line (typically the positive x-axis), measured anticlockwise in degrees
Displacement vector — a vector representing the straight-line distance and direction from one point to another, written as AB→ or AB
Core concepts
Vector notation methods
Vectors can be represented in multiple ways, and CIE IGCSE Mathematics examiners expect you to recognize and use all three standard notations:
Directed line segment notation: AB→ or BA→ represents the vector from point A to point B (or B to A). The order matters—AB→ = -BA→ because they point in opposite directions.
Bold letter notation: Vectors are written as lowercase bold letters such as a, b or v in printed text. In handwritten work, use underlining: a, b or v.
Column vector notation: The most common form in exam questions, written as: $$\begin{pmatrix} x \ y \end{pmatrix}$$
where x represents the horizontal component (positive right, negative left) and y represents the vertical component (positive up, negative down).
For three-dimensional vectors at IGCSE level (Extended tier only): $$\begin{pmatrix} x \ y \ z \end{pmatrix}$$
where z represents vertical displacement in 3D space.
Calculating magnitude in two dimensions
The magnitude of a vector represents its length and is always calculated as a positive value. For a two-dimensional vector a = (x/y), the magnitude is found using Pythagoras' theorem:
$$|\mathbf{a}| = \sqrt{x^2 + y^2}$$
This formula emerges because any vector forms the hypotenuse of a right-angled triangle, where x and y are the two shorter sides.
Step-by-step process:
- Identify the x-component and y-component from the column vector
- Square each component separately
- Add the squared values together
- Take the positive square root of the sum
- Express your answer to appropriate accuracy (usually 3 significant figures unless stated otherwise)
For example, the magnitude of vector p = (6/8) is: $$|\mathbf{p}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
When dealing with vectors between two points A(x₁, y₁) and B(x₂, y₂), first calculate the displacement vector: $$\overrightarrow{AB} = \begin{pmatrix} x_2 - x_1 \ y_2 - y_1 \end{pmatrix}$$
Then apply the magnitude formula to this result.
Calculating magnitude in three dimensions
For three-dimensional vectors v = (x/y/z), extend Pythagoras' theorem to three dimensions:
$$|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$$
This represents the direct distance from the origin to the point (x, y, z) in 3D space.
Example: Find the magnitude of q = (2/-3/6) $$|\mathbf{q}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$
Determining direction using bearings and angles
The direction of a vector is typically expressed as an angle measured from a reference direction. In CIE IGCSE Mathematics, you'll encounter two main methods:
Angle from the positive x-axis: Measured anticlockwise from the positive horizontal direction, giving angles from 0° to 360°. For a vector a = (x/y), use:
$$\tan(\theta) = \frac{y}{x}$$
Then calculate θ = tan⁻¹(y/x).
Important considerations:
- If x > 0 and y > 0 (first quadrant): θ = tan⁻¹(y/x)
- If x < 0 and y > 0 (second quadrant): θ = 180° - tan⁻¹(|y/x|)
- If x < 0 and y < 0 (third quadrant): θ = 180° + tan⁻¹(|y/x|)
- If x > 0 and y < 0 (fourth quadrant): θ = 360° - tan⁻¹(|y/x|)
Bearing notation: A three-figure bearing measured clockwise from North (000° to 360°). To convert from a standard angle θ measured anticlockwise from the positive x-axis (East):
Bearing = 090° - θ (adjusting as necessary to keep within 000° to 360°)
Unit vectors and direction vectors
A unit vector has magnitude 1 and indicates pure direction. To find the unit vector in the direction of any vector a:
$$\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}$$
Divide each component of the vector by its magnitude.
Process:
- Calculate the magnitude |a|
- Divide the x-component by |a|
- Divide the y-component by |a|
- Write as a column vector
For instance, if a = (3/4), then |a| = 5, so: $$\hat{\mathbf{a}} = \begin{pmatrix} 3/5 \ 4/5 \end{pmatrix} = \begin{pmatrix} 0.6 \ 0.8 \end{pmatrix}$$
You can verify this is correct by checking that the magnitude equals 1.
Position vectors and displacement
Position vectors describe locations relative to a fixed origin O. If point P has coordinates (x, y), its position vector is: $$\overrightarrow{OP} = \mathbf{p} = \begin{pmatrix} x \ y \end{pmatrix}$$
To find the displacement vector from point A with position vector a to point B with position vector b: $$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$
This "destination minus starting point" rule appears frequently in exam questions and must be applied correctly to avoid sign errors.
Worked examples
Example 1: Finding magnitude and direction
Question: Vector v = (5/12). Calculate: (a) the magnitude of v [2 marks] (b) the angle θ that v makes with the positive x-axis [2 marks]
Solution:
(a) Using the magnitude formula: $$|\mathbf{v}| = \sqrt{5^2 + 12^2}$$ $$= \sqrt{25 + 144}$$ $$= \sqrt{169}$$ $$= 13$$ [2 marks: 1 for correct formula, 1 for correct answer]
(b) Both components are positive (first quadrant), so: $$\tan(\theta) = \frac{12}{5}$$ $$\theta = \tan^{-1}\left(\frac{12}{5}\right)$$ $$\theta = \tan^{-1}(2.4)$$ $$\theta = 67.4°$$ (to 1 d.p.) [2 marks: 1 for correct method, 1 for correct answer]
Example 2: Displacement vectors and magnitude
Question: Points A and B have position vectors a = (2/-1) and b = (-3/5) respectively. (a) Find the vector AB→ [2 marks] (b) Calculate the distance from A to B [2 marks]
Solution:
(a) Using the displacement formula: $$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$ $$= \begin{pmatrix} -3 \ 5 \end{pmatrix} - \begin{pmatrix} 2 \ -1 \end{pmatrix}$$ $$= \begin{pmatrix} -3-2 \ 5-(-1) \end{pmatrix}$$ $$= \begin{pmatrix} -5 \ 6 \end{pmatrix}$$ [2 marks: 1 for correct method, 1 for correct answer]
(b) Distance = magnitude of AB→: $$|\overrightarrow{AB}| = \sqrt{(-5)^2 + 6^2}$$ $$= \sqrt{25 + 36}$$ $$= \sqrt{61}$$ $$= 7.81$$ (to 3 s.f.) [2 marks: 1 for correct substitution, 1 for correct answer]
Example 3: Unit vectors
Question: Find the unit vector in the direction of w = (8/-15). [3 marks]
Solution:
First, find the magnitude: $$|\mathbf{w}| = \sqrt{8^2 + (-15)^2} = \sqrt{64 + 225} = \sqrt{289} = 17$$ [1 mark]
Divide each component by the magnitude: $$\hat{\mathbf{w}} = \frac{1}{17}\begin{pmatrix} 8 \ -15 \end{pmatrix}$$ $$= \begin{pmatrix} 8/17 \ -15/17 \end{pmatrix}$$ or $$\begin{pmatrix} 0.471 \ -0.882 \end{pmatrix}$$ (to 3 s.f.) [2 marks: 1 for method, 1 for correct answer]
Common mistakes and how to avoid them
• Confusing AB→ with BA→: These vectors have equal magnitude but opposite directions. Always write displacement vectors in the order "destination minus start". AB→ = b - a, not a - b. Check: the arrow points from the first letter to the second.
• Incorrect angle quadrant: When using tan⁻¹ to find direction, calculators only give angles between -90° and 90°. You must adjust based on the signs of x and y components. Draw a quick sketch to identify which quadrant the vector lies in, then calculate accordingly.
• Forgetting to square root: Students often calculate x² + y² but forget the final square root step when finding magnitude. The formula is √(x² + y²), not just x² + y². Always complete the calculation.
• Negative magnitude values: Magnitude represents length and must always be positive. If your calculation gives a negative result, you've made an error—check your squaring of negative components. For example, (-3)² = +9, not -9.
• Subtracting components incorrectly: When finding displacement vectors, take extra care with negative numbers. For (2) - (-5) = 2 + 5 = 7, not -3. Write out the subtraction explicitly before simplifying.
• Rounding too early: In multi-step problems, store full calculator values for intermediate steps and only round the final answer. Rounding |v| = 7.2111... to 7.2 before using it to find a unit vector will introduce cumulative errors.
Exam technique for Vectors: Vector notation, magnitude and direction
• Command word "Find": Requires a complete numerical answer with working shown. For magnitude questions, write out the full formula √(x² + y²) = √(...) = ... before giving your answer. Examiners award method marks even if arithmetic errors occur.
• Accuracy requirements: Unless stated otherwise, give answers to 3 significant figures for magnitude calculations and 1 decimal place for angles. Three-figure bearings must be written with leading zeros (e.g., 037° not 37°). Read the question carefully for specific instructions.
• Mark allocation patterns: Magnitude questions typically award 2 marks (1 for method, 1 for answer). Direction questions also carry 2 marks. Combined "find magnitude and direction" questions are worth 4 marks total. Displacement vectors between points usually earn 2 marks. Plan your working accordingly.
• Show clear vector notation: In handwritten solutions, underline vectors consistently (a not a) and use arrows for directed segments (AB→). Poor notation loses clarity marks. Draw diagrams for bearing questions—this helps avoid quadrant errors and demonstrates your understanding.
Quick revision summary
Vectors have magnitude and direction. Write them as column vectors (x/y), bold letters a, or directed segments AB→. Calculate magnitude using |a| = √(x² + y²) for 2D or √(x² + y² + z²) for 3D—always positive. Find direction using tan(θ) = y/x, adjusting for quadrant. Displacement from A to B equals b - a (destination minus start). Unit vectors have magnitude 1, found by dividing each component by the vector's magnitude. Always show full working and round only final answers.