Vectors (Edexcel IGCSE Maths): Revision notes

Section 1

What is a column vector?

A vector has both magnitude (size) and direction. In component form, a vector is written as a column vector:

$\vec{a} = \begin{pmatrix} x \\ y \end{pmatrix}$

where $x$ is the horizontal displacement and $y$ is the vertical displacement. Vectors can be labelled with a bold letter (e.g. a) or an arrow (e.g. $\vec{AB}$ ), where $\vec{AB}$ means the vector from $A$ to $B$ .

Two vectors are equal if they have the same magnitude and direction — they do not need to start at the same point. Vectors with the same direction (one is a scalar multiple of the other) are parallel.

Key termsvectormagnitudedirectioncolumn vectorparallel vectors

Common mistake

Don't confuse a vector $\vec{AB}$ with a coordinate. $\vec{AB} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ is a movement of 3 right, 2 up — it says nothing about where $A$ and $B$ actually are unless you're also told a starting point.

Exam tip

Always check the direction of the arrow. $\vec{AB} = -\vec{BA}$ : reversing the letters reverses the sign of every component.

Section 2

How do you add, subtract and scale vectors?

Addition/subtraction: combine component-wise.

$\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a+c \\ b+d \end{pmatrix}, \qquad \begin{pmatrix} a \\ b \end{pmatrix} - \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a-c \\ b-d \end{pmatrix}$

Scalar multiplication: multiply every component by the scalar $k$ .

$k\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}$

Geometrically, adding vectors means following one displacement then the other (triangle/parallelogram law). Scaling by $k$ stretches (or shrinks) the vector by a factor of $k$ ; a negative $k$ also reverses its direction. If $\vec{b} = k\vec{a}$ for some scalar $k$ , then $\vec{a}$ and $\vec{b}$ are parallel.

Key termsscalar multiplicationresultant vector

Example

If $\vec{a} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$ , then $2\vec{a} - \vec{b} = \begin{pmatrix} 8 \\ -2 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 10 \\ -5 \end{pmatrix}$ .

Think of it like this

Think of scalar multiplication like adjusting a recipe: doubling every ingredient (component) keeps the dish the same 'flavour' (direction) just bigger — unless you use a negative amount, which flips it into its opposite.

Section 3

How do you find the magnitude of a vector?

The magnitude (length) of $\vec{a} = \begin{pmatrix} x \\ y \end{pmatrix}$ is found using Pythagoras' theorem, since $x$ and $y$ form the two shorter sides of a right-angled triangle:

$|\vec{a}| = \sqrt{x^2 + y^2}$

Magnitude is always given as a positive value (or zero for the zero vector) with no direction attached — it's just a number, often left as a surd unless a decimal is requested.

Key termsmagnitude of a vector

Example

If $\vec{p} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}$ , then $|\vec{p}| = \sqrt{6^2 + (-8)^2} = \sqrt{36+64} = \sqrt{100} = 10$ .

Common mistake

Don't forget to square the components even if they're negative — $(-8)^2 = 64$ , not $-64$ . A negative squared is always positive.

Section 4

What is a position vector and how do you use it?

A position vector describes the location of a point relative to a fixed origin $O$ . The position vector of point $A$ is written $\vec{OA}$ , often shortened to $\vec{a}$ .

To find the vector between two points whose position vectors you know, subtract the position vector of the start point from the position vector of the end point:

$\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$

This is the single most useful rule in exam vector-geometry questions: any route can be broken into a chain of position vectors, and $\vec{AB} = \vec{b} - \vec{a}$ lets you convert between 'position relative to $O$ ' and 'displacement between two named points'.

For a point $M$ that divides a line $AB$ in the ratio $m:n$ from $A$ to $B$ :

$\vec{OM} = \vec{a} + \frac{m}{m+n}(\vec{b}-\vec{a})$

For the midpoint specifically ( $m=n=1$ ): $\vec{OM} = \vec{a} + \tfrac{1}{2}(\vec{b}-\vec{a}) = \tfrac{1}{2}(\vec{a}+\vec{b})$ .

Key termsposition vectororiginratio division of a line

Exam tip

"End minus start" — to go from point $A$ to point $B$ , always compute (position vector of $B$ ) minus (position vector of $A$ ). Getting this backwards is the most common vector exam error.

Example

If $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$ , and $M$ is the midpoint of $AB$ , then $\vec{OM} = \tfrac{1}{2}(\vec{a}+\vec{b})$ .

Section 5

How do you prove points are collinear or lines are parallel?

Parallel lines: two vectors are parallel if and only if one is a scalar multiple of the other, i.e. $\vec{PQ} = k\vec{RS}$ for some scalar $k$ . Express both vectors in terms of the same base vectors (usually $\vec{a}$ and $\vec{b}$ ), then compare coefficients to find $k$ .

Collinear points: three points $A$ , $B$ , $C$ are collinear (lie on a single straight line) if $\vec{AB}$ and $\vec{BC}$ (or any two vectors formed from the three points) are parallel and share a common point (e.g. $B$ ). Being parallel alone only proves the lines are parallel — sharing a point as well proves they lie on the same straight line.

Exam method:

Write each required vector in terms of the given base vectors, simplifying fully.
Show one vector equals a scalar multiple of the other: $\vec{AB} = k\vec{BC}$ .
State the shared point (e.g. $B$ ) to conclude $A$ , $B$ , $C$ are collinear.

Key termscollinear pointsbase vectors

Example

If $\vec{AB} = 2\vec{a} - \vec{b}$ and $\vec{BC} = 4\vec{a} - 2\vec{b}$ , then $\vec{BC} = 2\vec{AB}$ , so $AB$ and $BC$ are parallel. Since they share point $B$ , $A$ , $B$ , $C$ are collinear.

Common mistake

Showing two vectors are parallel is NOT enough to prove collinearity on its own — you must also state that they share a common point, otherwise they could be parallel but on two separate, non-intersecting lines.

Must Know

$\vec{AB} = \vec{b} - \vec{a}$ : always end position vector minus start position vector.
Magnitude: $|\vec{a}| = \sqrt{x^2+y^2}$ , using Pythagoras' theorem, always positive.
$k\vec{a}$ scales the length by $|k|$ and reverses direction if $k$ is negative.
Midpoint of $AB$ : $\tfrac{1}{2}(\vec{a}+\vec{b})$ ; point dividing $AB$ in ratio $m:n$ : $\vec{a} + \frac{m}{m+n}(\vec{b}-\vec{a})$ .
Vectors are parallel only if one is a scalar multiple of the other.
Collinearity needs BOTH parallel vectors AND a shared point.

That's the notes covered.

Carry on to the next subtopic.