Vectors: Addition, subtraction and scalar multiplication of vectors

What you'll learn

This topic covers the fundamental operations with vectors that form the basis of coordinate geometry and transformation questions in CIE IGCSE Mathematics. You'll master how to add and subtract vectors both geometrically and algebraically, multiply vectors by scalars, and apply these skills to solve problems involving displacement, position vectors and geometric proofs. These concepts appear regularly in Paper 2 and Paper 4 examinations.

Key terms and definitions

Vector — a quantity with both magnitude (size) and direction, represented by a directed line segment or column notation.

Scalar — a quantity with magnitude only (a real number), used to multiply vectors.

Magnitude — the length or size of a vector, written as |a| or |AB|, calculated using Pythagoras' theorem in 2D or 3D.

Position vector — a vector that describes the position of a point relative to the origin O, usually written as r or $\overrightarrow{OA}$.

Displacement vector — a vector representing movement from one point to another, written as $\overrightarrow{AB}$ or AB.

Column vector — a vector written in the form $\begin{pmatrix} x \ y \end{pmatrix}$ for 2D or $\begin{pmatrix} x \ y \ z \end{pmatrix}$ for 3D, where x, y, and z are scalar components.

Resultant vector — the single vector obtained by combining two or more vectors through addition.

Unit vector — a vector with magnitude 1, often used to indicate direction only.

Core concepts

Representing vectors

Vectors can be represented in multiple ways, and CIE IGCSE Mathematics examinations expect fluency in converting between them:

Geometric representation:

• Draw a directed line segment with an arrow showing direction
• Label using capital letters (AB means from A to B)
• Equal vectors have the same length and direction, regardless of position

Algebraic representation:

• Column notation: $\begin{pmatrix} 3 \ -2 \end{pmatrix}$ means 3 units in the x-direction, 2 units down in the y-direction
• Component form: a = 3i + 2j (less common at IGCSE but may appear)
• Bold lowercase letters (a, b) or letters with arrows above ($\vec{a}$)

Key relationships:

• $\overrightarrow{AB}$ = position vector of B − position vector of A
• $\overrightarrow{BA}$ = −$\overrightarrow{AB}$
• If $\overrightarrow{OA}$ = a, then $\overrightarrow{AO}$ = −a

Vector addition

Addition of vectors follows specific rules that apply both geometrically and algebraically.

Geometric method (triangle law):

1. Place the tail of the second vector at the head of the first vector
2. The resultant is drawn from the tail of the first to the head of the second
3. This represents the combined displacement: a + b = c

Geometric method (parallelogram law):

1. Place both vectors tail-to-tail
2. Complete the parallelogram
3. The diagonal from the common tail gives the resultant

Algebraic method: Add corresponding components: $$\begin{pmatrix} a_1 \ a_2 \end{pmatrix} + \begin{pmatrix} b_1 \ b_2 \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \ a_2 + b_2 \end{pmatrix}$$

Properties of vector addition:

• Commutative: a + b = b + a
• Associative: (a + b) + c = a + (b + c)
• Zero vector: a + 0 = a

Vector subtraction

Subtracting vectors involves adding the negative of the second vector.

Conceptual understanding:

• a − b = a + (−b)
• The negative vector −b has the same magnitude as b but opposite direction

Geometric method:

1. Reverse the direction of the vector being subtracted
2. Apply the triangle law for addition
3. Alternatively: a − b is the vector from the head of b to the head of a when both start from the same point

Algebraic method: Subtract corresponding components: $$\begin{pmatrix} a_1 \ a_2 \end{pmatrix} - \begin{pmatrix} b_1 \ b_2 \end{pmatrix} = \begin{pmatrix} a_1 - b_1 \ a_2 - b_2 \end{pmatrix}$$

Common examination context: Finding a displacement vector when given two position vectors: $$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$$

This formula appears frequently in CIE papers when working with coordinate geometry problems.

Scalar multiplication of vectors

Multiplying a vector by a scalar changes its magnitude and possibly its direction.

Effect of scalar multiplication:

• Multiply each component by the scalar
• If k > 1: the vector becomes longer
• If 0 < k < 1: the vector becomes shorter
• If k < 0: the vector reverses direction
• If k = 0: the result is the zero vector

Algebraic method: $$k\begin{pmatrix} a_1 \ a_2 \end{pmatrix} = \begin{pmatrix} ka_1 \ ka_2 \end{pmatrix}$$

Geometric interpretation:

• The scalar multiple ka is parallel to a
• Vectors are parallel if one is a scalar multiple of the other
• This property is crucial for proving collinearity in examination questions

Key applications in CIE papers:

Finding parallel vectors: If a = $\begin{pmatrix} 2 \ 3 \end{pmatrix}$, then 3a = $\begin{pmatrix} 6 \ 9 \end{pmatrix}$ is parallel to a

Dividing a line segment: The point dividing AB in ratio m:n has position vector: $$\frac{n \cdot \overrightarrow{OA} + m \cdot \overrightarrow{OB}}{m+n}$$

Proving collinearity: Points A, B, C are collinear if $\overrightarrow{AB}$ = k$\overrightarrow{AC}$ for some scalar k

Magnitude of vectors

Calculating the magnitude (length) of a vector uses Pythagoras' theorem.

For 2D vectors: $$\left|\begin{pmatrix} x \ y \end{pmatrix}\right| = \sqrt{x^2 + y^2}$$

For 3D vectors: $$\left|\begin{pmatrix} x \ y \ z \end{pmatrix}\right| = \sqrt{x^2 + y^2 + z^2}$$

Unit vectors: A unit vector in the direction of a is found by dividing a by its magnitude: $$\hat{a} = \frac{\mathbf{a}}{|\mathbf{a}|}$$

This ensures the resulting vector has magnitude 1 while maintaining the same direction.

Combined operations and problem-solving

CIE IGCSE examinations frequently test multiple operations within a single question.

Standard problem types:

Express one vector in terms of others: Given $\overrightarrow{OA}$ = a and $\overrightarrow{OB}$ = b, find $\overrightarrow{AB}$: $$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$$

Find the position of a point: If M is the midpoint of AB, then: $$\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}) = \frac{1}{2}(\mathbf{a} + \mathbf{b})$$

Prove geometric properties: Show that ABCD is a parallelogram by proving $\overrightarrow{AB}$ = $\overrightarrow{DC}$

Worked examples

Example 1: Vector addition and subtraction

Given a = $\begin{pmatrix} 4 \ -1 \end{pmatrix}$ and b = $\begin{pmatrix} -2 \ 5 \end{pmatrix}$, find:

(a) a + b [1 mark] (b) 2a − b [2 marks] (c) |a + b| [2 marks]

Solution:

(a) a + b = $\begin{pmatrix} 4 \ -1 \end{pmatrix}$ + $\begin{pmatrix} -2 \ 5 \end{pmatrix}$ = $\begin{pmatrix} 4+(-2) \ -1+5 \end{pmatrix}$ = $\begin{pmatrix} 2 \ 4 \end{pmatrix}$ ✓

(b) 2a = $\begin{pmatrix} 8 \ -2 \end{pmatrix}$

2a − b = $\begin{pmatrix} 8 \ -2 \end{pmatrix}$ − $\begin{pmatrix} -2 \ 5 \end{pmatrix}$ = $\begin{pmatrix} 8-(-2) \ -2-5 \end{pmatrix}$ = $\begin{pmatrix} 10 \ -7 \end{pmatrix}$ ✓✓

(c) From part (a), a + b = $\begin{pmatrix} 2 \ 4 \end{pmatrix}$

|a + b| = $\sqrt{2^2 + 4^2}$ = $\sqrt{4 + 16}$ = $\sqrt{20}$ = $2\sqrt{5}$ ✓✓

Example 2: Position vectors and midpoints

The points A and B have position vectors $\overrightarrow{OA}$ = $\begin{pmatrix} 6 \ 2 \end{pmatrix}$ and $\overrightarrow{OB}$ = $\begin{pmatrix} -2 \ 8 \end{pmatrix}$ respectively.

(a) Find $\overrightarrow{AB}$ [2 marks] (b) M is the midpoint of AB. Find the position vector of M. [2 marks]

Solution:

(a) $\overrightarrow{AB}$ = $\overrightarrow{OB}$ − $\overrightarrow{OA}$ ✓

$\overrightarrow{AB}$ = $\begin{pmatrix} -2 \ 8 \end{pmatrix}$ − $\begin{pmatrix} 6 \ 2 \end{pmatrix}$ = $\begin{pmatrix} -8 \ 6 \end{pmatrix}$ ✓

(b) $\overrightarrow{OM}$ = $\frac{1}{2}$($\overrightarrow{OA}$ + $\overrightarrow{OB}$) ✓

$\overrightarrow{OM}$ = $\frac{1}{2}$($\begin{pmatrix} 6 \ 2 \end{pmatrix}$ + $\begin{pmatrix} -2 \ 8 \end{pmatrix}$) = $\frac{1}{2}\begin{pmatrix} 4 \ 10 \end{pmatrix}$ = $\begin{pmatrix} 2 \ 5 \end{pmatrix}$ ✓

Example 3: Proving collinearity

Points P, Q and R have position vectors p = $\begin{pmatrix} 3 \ 1 \end{pmatrix}$, q = $\begin{pmatrix} 7 \ 5 \end{pmatrix}$ and r = $\begin{pmatrix} 9 \ 7 \end{pmatrix}$ respectively.

Prove that P, Q and R are collinear. [3 marks]

Solution:

Find $\overrightarrow{PQ}$: $\overrightarrow{PQ}$ = q − p = $\begin{pmatrix} 7 \ 5 \end{pmatrix}$ − $\begin{pmatrix} 3 \ 1 \end{pmatrix}$ = $\begin{pmatrix} 4 \ 4 \end{pmatrix}$ ✓

Find $\overrightarrow{PR}$: $\overrightarrow{PR}$ = r − p = $\begin{pmatrix} 9 \ 7 \end{pmatrix}$ − $\begin{pmatrix} 3 \ 1 \end{pmatrix}$ = $\begin{pmatrix} 6 \ 6 \end{pmatrix}$ ✓

$\overrightarrow{PR}$ = $\begin{pmatrix} 6 \ 6 \end{pmatrix}$ = $\frac{3}{2}\begin{pmatrix} 4 \ 4 \end{pmatrix}$ = $\frac{3}{2}\overrightarrow{PQ}$

Since $\overrightarrow{PR}$ is a scalar multiple of $\overrightarrow{PQ}$, the vectors are parallel and share point P, therefore P, Q and R are collinear. ✓

Common mistakes and how to avoid them

• Mistake: Adding vectors by adding magnitudes rather than components (e.g., claiming |a + b| = |a| + |b|). Correction: Always add vectors component-by-component, then calculate the magnitude of the resultant if required. The magnitude of a sum is generally not equal to the sum of magnitudes.

• Mistake: Reversing the order in subtraction, writing $\overrightarrow{AB}$ = $\overrightarrow{OA}$ − $\overrightarrow{OB}$ instead of $\overrightarrow{OB}$ − $\overrightarrow{OA}$. Correction: Remember that $\overrightarrow{AB}$ means "from A to B", so you need "position of B minus position of A". The destination comes first in the subtraction.

• Mistake: Forgetting to apply the scalar to all components when performing scalar multiplication, or applying it to only the x-component. Correction: When calculating ka, multiply every component by k. For 3a where a = $\begin{pmatrix} 2 \ -5 \end{pmatrix}$, the answer is $\begin{pmatrix} 6 \ -15 \end{pmatrix}$, not $\begin{pmatrix} 6 \ -5 \end{pmatrix}$.

• Mistake: Claiming vectors are equal because they have the same magnitude, ignoring direction. Correction: For vectors to be equal, they must have both the same magnitude and the same direction. Vectors $\begin{pmatrix} 3 \ 4 \end{pmatrix}$ and $\begin{pmatrix} -3 \ -4 \end{pmatrix}$ have the same magnitude but are not equal.

• Mistake: When proving collinearity, only showing that two vectors are parallel without checking they share a common point. Correction: To prove three points A, B, C are collinear, show that $\overrightarrow{AB}$ = k$\overrightarrow{AC}$ (both vectors start from the same point A) or that $\overrightarrow{AB}$ = k$\overrightarrow{BC}$ (they connect in sequence).

• Mistake: Calculating magnitude in 3D using only two components, forgetting the z-component. Correction: For a 3D vector $\begin{pmatrix} x \ y \ z \end{pmatrix}$, the magnitude is $\sqrt{x^2 + y^2 + z^2}$. All three components must be squared and summed.

Exam technique for vectors operations

• Command word "Find": For vector calculations, show your working clearly by writing out the component arithmetic. Examiners award method marks even if the final answer is incorrect, so displaying $\begin{pmatrix} 4 \ -2 \end{pmatrix}$ + $\begin{pmatrix} 3 \ 7 \end{pmatrix}$ = $\begin{pmatrix} 7 \ 5 \end{pmatrix}$ earns more marks than just writing the answer.

• Command word "Prove" or "Show that": You must demonstrate clear logical steps, particularly for collinearity or parallelogram proofs. Calculate the relevant vectors, show one is a scalar multiple of another, and state your conclusion explicitly. Partial working without a concluding statement typically loses the final mark.

• Multi-part questions: Vector questions often build up, with part (b) using the answer from part (a). If you cannot complete part (a), use a pronumeral or assume a value to continue with later parts—you can still earn subsequent method marks. Write "using the result from (a)" to show your reasoning.

• Accuracy with negatives: Take extra care with negative signs, especially in subtraction. Write out intermediate steps like $4 - (-3) = 4 + 3 = 7$ rather than doing the arithmetic mentally. Sign errors are the most common reason for lost marks in vector calculations on CIE papers.

Quick revision summary

Vectors are quantities with magnitude and direction, written as column vectors $\begin{pmatrix} x \ y \end{pmatrix}$. Add or subtract vectors by combining corresponding components. Scalar multiplication multiplies each component by the scalar; negative scalars reverse direction. The magnitude of a 2D vector is $\sqrt{x^2 + y^2}$. Key relationships: $\overrightarrow{AB}$ = position of B minus position of A; parallel vectors are scalar multiples; midpoint M of AB has position vector $\frac{1}{2}$(a + b). Prove collinearity by showing vectors are scalar multiples sharing a common point.