Chapter 7: Vectors
A vector is a quantity that has both a magnitude (size) and a direction. Vectors are used to describe displacements, forces, and velocities. In IGCSE mathematics the focus is on geometric vectors: how to represent and calculate them, and how to use them to prove geometric facts about shapes.
Exam Board Coverage
Vector arithmetic (Sections 7.1–7.5) is examined at both Foundation and Extended tier. Section 7.6 (Parallel Vectors and Collinearity) and Section 7.7 (Geometric Proofs) are primarily Extended tier topics and involve the most complex exam questions.
7.1 What is a Vector?
Vector Notation
- Column vector: $\mathbf{a} = \dbinom{x}{y}$ where $x$ is the horizontal component and $y$ is the vertical component.
- Bold letter: $\mathbf{a}$ (in printed text); underlined in handwriting: $\underline{a}$.
- Arrow notation: $\overrightarrow{AB}$ denotes the vector from point $A$ to point $B$.
- Negative vector: $-\mathbf{a} = \overrightarrow{BA}$ has the same magnitude as $\mathbf{a}$ but the opposite direction.
Triangle law of vector addition: $\overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OB}$
7.2 Adding and Subtracting Vectors
Vector Addition — Triangle Law
To add vectors geometrically, place the tail of the second vector at the head of the first. The resultant goes from the tail of the first to the head of the second.
$$\mathbf{a} + \mathbf{b} = \dbinom{a_1+b_1}{a_2+b_2}$$
Vector Subtraction
$$\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b}) = \dbinom{a_1-b_1}{a_2-b_2}$$
Example 7.1 — Adding column vectors
$\mathbf{a} = \dbinom{3}{-1}$ and $\mathbf{b} = \dbinom{-2}{4}$. Find $\mathbf{a} + \mathbf{b}$ and $\mathbf{a} - \mathbf{b}$.
$\mathbf{a} + \mathbf{b} = \dbinom{3+(-2)}{-1+4} = \dbinom{1}{3}$
$\mathbf{a} - \mathbf{b} = \dbinom{3-(-2)}{-1-4} = \dbinom{5}{-5}$
7.3 Scalar Multiplication
Multiplying a vector by a scalar (a number) changes its magnitude but not its direction (unless the scalar is negative, which reverses the direction).
Scalar Multiple
$$k\mathbf{a} = k\dbinom{a_1}{a_2} = \dbinom{ka_1}{ka_2}$$
- $k > 0$: same direction, magnitude multiplied by $k$.
- $k < 0$: opposite direction, magnitude multiplied by $|k|$.
- $k = 0$: the zero vector $\mathbf{0} = \dbinom{0}{0}$.
Example 7.2 — Scalar multiplication
If $\mathbf{p} = \dbinom{2}{-3}$, find $3\mathbf{p}$ and $-2\mathbf{p}$.
$3\mathbf{p} = \dbinom{6}{-9}$ $-2\mathbf{p} = \dbinom{-4}{6}$
7.4 Magnitude and Unit Vectors
Magnitude of a Vector
The magnitude (length) of $\mathbf{a} = \dbinom{a_1}{a_2}$ is:
$$|\mathbf{a}| = \sqrt{a_1^2 + a_2^2}$$
This is simply Pythagoras' Theorem applied to the components.
Unit Vector
A unit vector has magnitude $1$. The unit vector in the direction of $\mathbf{a}$ is:
$$\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}$$
Example 7.3 — Magnitude and unit vector
Find the magnitude of $\mathbf{v} = \dbinom{3}{4}$, and find the unit vector in its direction.
$|\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$
$\hat{\mathbf{v}} = \dfrac{1}{5}\dbinom{3}{4} = \dbinom{0.6}{0.8}$
7.5 Position Vectors
The position vector of a point $A$ is the vector $\overrightarrow{OA}$ from the origin $O$ to $A$. If $A = (a_1, a_2)$ then $\overrightarrow{OA} = \dbinom{a_1}{a_2}$.
Vector Between Two Points
$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$$
where $\mathbf{a} = \overrightarrow{OA}$ and $\mathbf{b} = \overrightarrow{OB}$.
Example 7.4 — Vector between two points
$A = (2, 5)$ and $B = (7, 1)$. Find $\overrightarrow{AB}$ and $\overrightarrow{BA}$.
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \dbinom{7}{1} - \dbinom{2}{5} = \dbinom{5}{-4}$
$\overrightarrow{BA} = -\overrightarrow{AB} = \dbinom{-5}{4}$
Example 7.5 — Midpoint using position vectors
$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $AB$. Find $\overrightarrow{OM}$.
$\overrightarrow{OM} = \overrightarrow{OA} + \tfrac{1}{2}\overrightarrow{AB} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$
7.6 Parallel Vectors and Collinear Points Extended
Parallel Vectors
Two vectors are parallel if and only if one is a scalar multiple of the other: $\mathbf{a} \parallel \mathbf{b}$ iff $\mathbf{a} = k\mathbf{b}$ for some non-zero scalar $k$.
Collinear Points
Three points $A$, $B$, $C$ are collinear (lie on the same straight line) if and only if $\overrightarrow{AB}$ is parallel to $\overrightarrow{AC}$ (they share the point $A$ and their directions are parallel, so all three points lie on one line).
Method: show $\overrightarrow{AC} = k\,\overrightarrow{AB}$ for some scalar $k$. The shared point $A$ confirms collinearity.
Example 7.6 — Proving collinearity
$A$, $B$, $C$ have position vectors $\mathbf{a}=\dbinom{1}{2}$, $\mathbf{b}=\dbinom{3}{5}$, $\mathbf{c}=\dbinom{7}{11}$. Show that $A$, $B$, $C$ are collinear.
Step 1 $\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \dbinom{2}{3}$
Step 2 $\overrightarrow{AC} = \mathbf{c}-\mathbf{a} = \dbinom{6}{9} = 3\dbinom{2}{3} = 3\,\overrightarrow{AB}$
Conclusion $\overrightarrow{AC} = 3\,\overrightarrow{AB}$, so $\overrightarrow{AC}$ is parallel to $\overrightarrow{AB}$. Since $A$ is common to both, $A$, $B$, $C$ are collinear. $\square$
7.7 Geometric Proofs with Vectors Extended
Vector proofs use vector algebra to establish geometric properties — such as that a point lies on a line, or that a quadrilateral is a parallelogram. The key skill is expressing every vector in terms of a small set of base vectors.
Strategy for Vector Proofs
- Define two base vectors (usually $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$).
- Express every other vector in terms of $\mathbf{a}$ and $\mathbf{b}$ by following paths around the diagram.
- To prove lines are parallel: show one vector is a scalar multiple of another.
- To prove a point divides a segment in a given ratio: substitute the ratio into the position vector formula.
Example 7.7 — Midpoint proof
$OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $M$ is the midpoint of $AB$. Show that $\overrightarrow{OM} = \mathbf{a} + \tfrac{1}{2}\mathbf{c}$.
Step 1 In parallelogram: $\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}$ (opposite sides).
Step 2 $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \tfrac{1}{2}\overrightarrow{AB} = \mathbf{a} + \tfrac{1}{2}\mathbf{c}$ $\square$
Exercises
Exercise 7.1
$\mathbf{a} = \dbinom{5}{-2}$ and $\mathbf{b} = \dbinom{-1}{4}$. Find: (i) $\mathbf{a} + \mathbf{b}$, (ii) $\mathbf{a} - \mathbf{b}$, (iii) $3\mathbf{a}$, (iv) $2\mathbf{a} - 3\mathbf{b}$.
Show Solution
(i) $\dbinom{4}{2}$ (ii) $\dbinom{6}{-6}$ (iii) $\dbinom{15}{-6}$ (iv) $\dbinom{13}{-16}$
Exercise 7.2
Find the magnitude of $\mathbf{v} = \dbinom{5}{12}$.
Show Solution
$|\mathbf{v}| = \sqrt{25+144} = \sqrt{169} = \mathbf{13}$
Exercise 7.3
Find the unit vector in the direction of $\mathbf{w} = \dbinom{1}{2}$, giving components in surd form.
Show Solution
$|\mathbf{w}| = \sqrt{1+4} = \sqrt{5}$
$\hat{\mathbf{w}} = \dfrac{1}{\sqrt{5}}\dbinom{1}{2} = \dbinom{1/\sqrt{5}}{2/\sqrt{5}} = \dbinom{\sqrt{5}/5}{2\sqrt{5}/5}$
Exercise 7.4
$A = (2, 7)$ and $B = (6, -1)$. Find $\overrightarrow{AB}$ and $|\overrightarrow{AB}|$.
Show Solution
$\overrightarrow{AB} = \dbinom{6-2}{-1-7} = \dbinom{4}{-8}$
$|\overrightarrow{AB}| = \sqrt{16+64} = \sqrt{80} = 4\sqrt{5}$
Exercise 7.5
$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $OA$ and $N$ is the midpoint of $OB$. Express $\overrightarrow{MN}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
Show Solution
$\overrightarrow{OM} = \tfrac{1}{2}\mathbf{a}$, $\overrightarrow{ON} = \tfrac{1}{2}\mathbf{b}$
$\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \tfrac{1}{2}\mathbf{b} - \tfrac{1}{2}\mathbf{a} = \tfrac{1}{2}(\mathbf{b}-\mathbf{a})$
Exercise 7.6
Are the vectors $\dbinom{3}{-2}$ and $\dbinom{-9}{6}$ parallel? Justify your answer.
Show Solution
$\dbinom{-9}{6} = -3\dbinom{3}{-2}$. Since one is a scalar multiple of the other, yes, they are parallel.
Exercise 7.7 Extended
Points $P$, $Q$, $R$ have position vectors $\dbinom{1}{3}$, $\dbinom{3}{7}$, $\dbinom{9}{19}$ respectively. Show that $P$, $Q$, $R$ are collinear.
Show Solution
$\overrightarrow{PQ} = \dbinom{2}{4}$; $\overrightarrow{PR} = \dbinom{8}{16} = 4\dbinom{2}{4} = 4\,\overrightarrow{PQ}$
$\overrightarrow{PR}$ is parallel to $\overrightarrow{PQ}$ with common point $P$, so $P$, $Q$, $R$ are collinear. $\square$
Exercise 7.8 Extended
$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $P$ divides $AB$ in the ratio $1:2$. Express $\overrightarrow{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
Show Solution
$\overrightarrow{AP} = \tfrac{1}{3}\overrightarrow{AB} = \tfrac{1}{3}(\mathbf{b}-\mathbf{a})$
$\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \tfrac{1}{3}(\mathbf{b}-\mathbf{a}) = \tfrac{2}{3}\mathbf{a} + \tfrac{1}{3}\mathbf{b}$
Exercise 7.9 Extended
$ABCD$ is a quadrilateral. $\overrightarrow{AB} = \mathbf{p}$ and $\overrightarrow{DC} = \mathbf{p}$. What can you deduce about $ABCD$?
Show Solution
$\overrightarrow{AB} = \overrightarrow{DC}$: equal vectors means $AB \parallel DC$ and $AB = DC$ (same length, same direction). Therefore $ABCD$ is a parallelogram.
Exercise 7.10 Extended
$O$ is the origin. $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $AB$. Show that $\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a}+\mathbf{b})$.
Show Solution
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$. $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b}-\mathbf{a}) = \tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b} = \tfrac{1}{2}(\mathbf{a}+\mathbf{b})$ $\square$
Exam Tips
Tip 1 — Write $\overrightarrow{OA}$ not just $A$ for position vectors
Always use arrow notation or bold notation consistently. In handwriting, underline bold vectors: $\underline{a}$. Confusing position vectors with coordinates costs marks.
Tip 2 — Use a path when finding a vector
To find $\overrightarrow{AC}$, if you cannot go directly, go via another point: $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$. Choose the path that uses known vectors.
Tip 3 — Proving collinearity: two conditions needed
You must show (1) the two vectors are parallel (one is a scalar multiple of the other) AND (2) they share a common point. Without the shared point, the lines could be parallel but different.
Tip 4 — Check your final vector makes geometric sense
If you are asked for a vector in a triangle and your answer has a fraction, check it: midpoints give halves, thirds-of-the-way give thirds. A common error is sign errors when reversing a vector direction.