Vectors

Pure Mathematics (YPM01) · Pearson Edexcel International A Level · 10 min read · Updated: 6 Sept 2026

Welcome to the World of Vectors! (P4 Mathematics)

Hello future mathematician! This chapter, Vectors, is one of the most powerful tools you will learn in Pure Mathematics 4. Don't worry if words like "magnitude" and "scalar product" sound complicated right now—we are going to break everything down into simple, manageable steps.

Vectors are essentially instructions for movement in space. They allow us to describe geometry and movement not just on a flat piece of paper (2D) but in the real, three-dimensional world (3D). Mastering this chapter is crucial for understanding physics, engineering, and advanced mathematics.
Let’s get started!

Key Takeaway from the Introduction

Vectors describe position and movement in 3D space, which is essential for applied mathematics.

1. Fundamentals: 3D Space and Vector Notation

1.1. Coordinates in Three Dimensions

Before we deal with movement, we need a map. In 3D space, we use three axes: the $x$-axis (width), the $y$-axis (depth), and the $z$-axis (height). A point in space is defined by three coordinates: $(x, y, z)$.

Analogy: Imagine standing in the corner of a room. The $x$-axis goes along the wall to your left/right, the $y$-axis goes into the room away from the wall, and the $z$-axis goes straight up to the ceiling.

1.2. Defining a Vector

A Vector is a quantity that has both magnitude (size or length) and direction. This is different from a Scalar (like temperature or time), which only has magnitude.

We represent vectors using bold letters ($\mathbf{a}$ or $\mathbf{r}$) or by drawing an arrow over the letters representing the starting and ending points (e.g., $\vec{AB}$).

Vector Notation: Column Vectors and Basis Vectors

A vector $\mathbf{a}$ can be written in two main ways:

Column Vector Notation: $$\mathbf{a} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$ Example: A vector moving 2 units in the $x$ direction, -1 in the $y$ direction, and 5 in the $z$ direction is $\begin{pmatrix} 2 \\ -1 \\ 5 \end{pmatrix}$.
Basis Vector Notation (i, j, k):
The basis vectors $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ represent movement of 1 unit along the $x, y,$ and $z$ axes respectively. $$\mathbf{a} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$$
Example: $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 5\mathbf{k}$

1.3. Magnitude (Length) of a Vector

The magnitude (or modulus) of a vector $\mathbf{a}$ is its length. We use the notation $|\mathbf{a}|$ or $|\vec{AB}|$. It’s calculated using the 3D version of Pythagoras' Theorem.

For a vector $\mathbf{a} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$:

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

Example: If $\mathbf{v} = 3\mathbf{i} - 4\mathbf{j} + 0\mathbf{k}$, then $$|\mathbf{v}| = \sqrt{3^2 + (-4)^2 + 0^2} = \sqrt{9 + 16 + 0} = \sqrt{25} = 5$$

Unit Vectors

A unit vector is any vector with a magnitude of exactly 1. To find the unit vector $\hat{\mathbf{a}}$ in the same direction as $\mathbf{a}$:

$$\hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|} \mathbf{a}$$

You are essentially dividing the vector by its own length, shrinking it down to size 1.

Quick Review: Section 1

Vectors have magnitude and direction. We use column vectors or $\mathbf{i}, \mathbf{j}, \mathbf{k}$ notation. Magnitude is found using $|\mathbf{a}| = \sqrt{x^2 + y^2 + z^2}$.

2. Vector Operations and Position Vectors

2.1. Addition and Subtraction

Adding or subtracting vectors is easy: you simply add or subtract the corresponding components.

If $\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$, then:

$$\mathbf{a} + \mathbf{b} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix}$$

Analogy: Adding vectors is like following one instruction ($\mathbf{a}$) and then following the next instruction ($\mathbf{b}$). The result is the single instruction that gets you from the start of $\mathbf{a}$ to the end of $\mathbf{b}$.

2.2. Position Vectors and Displacement Vectors

A Position Vector ($\mathbf{r}$ or $\mathbf{a}$) tells you the location of a point relative to the origin $(0, 0, 0)$. If point A has coordinates $(x_A, y_A, z_A)$, its position vector is $\vec{OA} = \mathbf{a} = x_A\mathbf{i} + y_A\mathbf{j} + z_A\mathbf{k}$.

A Displacement Vector connects two specific points, A and B. To find the vector $\vec{AB}$ (the path from A to B), we use a detour via the origin:

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

Memory Aid: To get from A to B, you take the position of B and subtract the position of A. (The letter you end at minus the letter you start at.)

3. The Scalar Product (Dot Product)

The scalar product is a key concept in P4. It tells us how much two vectors are pointing in the same direction. When you 'multiply' two vectors using the scalar product, the result is always a scalar (a single number), not another vector.

3.1. Calculating the Scalar Product

There are two essential ways to calculate the scalar product of two vectors, $\mathbf{a}$ and $\mathbf{b}$.

Method 1: Component Form (The Calculation)

If $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$, the scalar product $\mathbf{a} \cdot \mathbf{b}$ is:

$$\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$

Example: If $\mathbf{a} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k}$ and $\mathbf{b} = 4\mathbf{i} + 0\mathbf{j} + \mathbf{k}$. $$\mathbf{a} \cdot \mathbf{b} = (2)(4) + (1)(0) + (-3)(1) = 8 + 0 - 3 = 5$$

Method 2: Angle Form (The Result)

The scalar product is also defined in terms of the angle $\theta$ between the two vectors:

$$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta$$

Did you know? In physics, if $\mathbf{a}$ is a Force and $\mathbf{b}$ is a Displacement, the scalar product represents the Work Done!

3.2. Finding the Angle Between Two Vectors

The main use of the scalar product in P4 is to find the angle $\theta$ between two vectors. By rearranging the angle formula:

$$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$$

Step-by-Step Guide to Finding $\theta$:

Calculate the Scalar Product ($\mathbf{a} \cdot \mathbf{b}$) using Method 1.
Calculate the Magnitude of each vector ($|\mathbf{a}|$ and $|\mathbf{b}|$).
Substitute these values into the $\cos \theta$ formula.
Use $\theta = \arccos(...) $ to find the angle. (Ensure your calculator is in Degree or Radian mode as required by the question!)

3.3. The Perpendicularity Condition

This is one of the most important concepts in P4 Vectors.

If two vectors are perpendicular (meet at a right angle, $\theta = 90^\circ$), then $\cos 90^\circ = 0$.

Therefore, if $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos 90^\circ$:

Two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ are perpendicular if and only if: $$\mathbf{a} \cdot \mathbf{b} = 0$$

Common Mistake to Avoid: If a question asks if two vectors are perpendicular, DO NOT try to prove $\theta = 90^\circ$. Simply calculate the scalar product. If it equals zero, they are perpendicular. If it doesn't, they are not.

Key Takeaway: Section 3

The scalar product ($\mathbf{a} \cdot \mathbf{b}$) is found by multiplying corresponding components and summing them. It is used to find angles, and if $\mathbf{a} \cdot \mathbf{b} = 0$, the vectors are perpendicular.

4. Equations of Straight Lines in 3D

In 2D, we use $y = mx + c$. In 3D space, this doesn't work well. Instead, we define a line using a starting point and a direction.

4.1. The Vector Form of a Line

A line $L$ passing through a fixed point A (with position vector $\mathbf{a}$) and parallel to a direction vector $\mathbf{d}$ is given by:

$$\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$$

$\mathbf{r}$ is the position vector of any point on the line.
$\mathbf{a}$ is the position vector of a known point on the line (the starting point).
$\mathbf{d}$ is the direction vector (the slope/path).
$\lambda$ (lambda) is the scalar parameter. It tells us how far we travel along the direction $\mathbf{d}$.

Analogy: Think of $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ as a treasure map: "Start at location $\mathbf{a}$, then move in direction $\mathbf{d}$ for some amount of time $\lambda$."

Example of Vector Form

If a line passes through the point $(1, 5, -2)$ and is parallel to the vector $\mathbf{d} = 3\mathbf{i} - \mathbf{j} + 4\mathbf{k}$:

$$\mathbf{r} = \begin{pmatrix} 1 \\ 5 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}$$

Any point on this line can be written in parametric form: $$x = 1 + 3\lambda$$ $$y = 5 - \lambda$$ $$z = -2 + 4\lambda$$

4.2. The Cartesian (Symmetric) Form of a Line

You can eliminate the parameter $\lambda$ from the parametric equations to find the Cartesian form. This is often required in exams.

From the example above: $$\lambda = \frac{x-1}{3} \quad | \quad \lambda = \frac{y-5}{-1} \quad | \quad \lambda = \frac{z-(-2)}{4}$$

Since all these expressions equal $\lambda$, they must equal each other:

$$\frac{x-x_1}{d_1} = \frac{y-y_1}{d_2} = \frac{z-z_1}{d_3}$$

Where $(x_1, y_1, z_1)$ is the point A, and $\mathbf{d} = (d_1, d_2, d_3)$ is the direction vector.

4.3. Intersection of Two Lines

If two lines, $L_1$ and $L_2$, intersect, there must be a point $(x, y, z)$ that satisfies both equations.

Line 1: $\mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{d}_1$
Line 2: $\mathbf{r} = \mathbf{a}_2 + \mu \mathbf{d}_2$
(Crucial: Always use a different parameter, like $\mu$ (mu), for the second line!)

Step-by-Step Guide to Finding Intersection:

Set them equal: Set $\mathbf{a}_1 + \lambda \mathbf{d}_1 = \mathbf{a}_2 + \mu \mathbf{d}_2$.
Form simultaneous equations: This equality generates three scalar equations (one for $x$, one for $y$, one for $z$).
Example: If $L_1: x = 1+\lambda$ and $L_2: x = 5+2\mu$, then $1+\lambda = 5+2\mu$.
Solve for the parameters: Use the first two equations to solve simultaneously for $\lambda$ and $\mu$.
Check Consistency: Substitute the found values of $\lambda$ and $\mu$ into the third equation.
- If the third equation holds true (LHS = RHS), the lines intersect.
- If the third equation is false, the lines do not intersect (they are skew or parallel).
Find the Intersection Point: If they intersect, substitute the value of $\lambda$ back into $L_1$ (or $\mu$ into $L_2$) to find the coordinates $(x, y, z)$.

4.4. The Angle Between Two Lines

The angle between two lines is defined as the angle between their direction vectors, $\mathbf{d}_1$ and $\mathbf{d}_2$.

You use the scalar product formula:

$$\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|}$$

(Note: We often take the absolute value of the numerator, $|\mathbf{d}_1 \cdot \mathbf{d}_2|$, to ensure we find the acute angle, $0^\circ \le \theta \le 90^\circ$, between the lines, unless the question specifically asks for the obtuse angle.)

Key Takeaway: Section 4

The equation of a line is $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$. Intersection requires solving for $\lambda$ and $\mu$ and then checking consistency using the third coordinate. The angle between lines is the angle between their direction vectors.

5. Encouragement and Final Tips

A Note to Struggling Students

The most common difficulty in Vectors isn't the calculus—it's managing the geometry and the notation. If you feel lost, go back to the analogies:
1. Position Vector $\mathbf{a}$: Your starting base at the origin.
2. Displacement $\vec{AB}$: The shortcut path between two points.
3. Scalar Product ($\mathbf{a} \cdot \mathbf{b}$): A quick check for perpendicularity.
4. Line Equation $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$: A set of GPS instructions.

Practice your algebraic skills when solving simultaneous equations for intersection, as this is where most arithmetic errors occur. You can do this!

Common Mistakes Checklist

Forgetting the Minus Sign: When finding $\vec{AB}$, students often write $\mathbf{a} + \mathbf{b}$ instead of $\mathbf{b} - \mathbf{a}$. Remember: END minus START.
Mixing Parameters: Using $\lambda$ for both lines when finding an intersection. Use $\lambda$ and $\mu$.
Solving vs. Checking: Always use the third equation only for the consistency check in intersection problems.
Using Position Vectors for Angles: When finding the angle between two lines, only use the direction vectors ($\mathbf{d}_1$ and $\mathbf{d}_2$).

Good luck with your revision. You have successfully navigated the complexities of 3D space!

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the difference between a vector and a scalar?

A vector is a quantity with both magnitude (size) and direction, whereas a scalar has only magnitude.

How is a vector represented in basis vector notation (̂i, ̂j, ̂k)?

A vector is written as $\mathbf{a} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, where $\mathbf{i}, \mathbf{j},$ and $\mathbf{k}$ are unit vectors along the $x, y,$ and $z$ axes respectively.

How do you calculate the magnitude (length) of a vector in three dimensions?

The magnitude of a vector $\mathbf{a} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ is calculated using $|\mathbf{a}| = \sqrt{x^2 + y^2 + z^2}$.

What is a unit vector and how is it calculated?

A unit vector has a magnitude of 1 and is calculated by dividing a vector by its own magnitude: $\hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|} \mathbf{a}$.

What is the formula to find the displacement vector $\vec{AB}$ between two points A and B?

The displacement vector is found by subtracting the position vector of the start from the position vector of the end: $\vec{AB} = \mathbf{b} - \mathbf{a}$.

How do you calculate the scalar product (dot product) of two vectors using components?

The scalar product $\mathbf{a} \cdot \mathbf{b}$ is the sum of the products of corresponding components: $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$.

How do you find the angle $\theta$ between two vectors using the scalar product?

The angle can be found using the formula: $\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$.

What is the condition for two non-zero vectors to be perpendicular?

Two vectors are perpendicular if and only if their scalar product is zero ($\mathbf{a} \cdot \mathbf{b} = 0$).

What is the general vector equation of a straight line in 3D?

The equation is $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$, where $\mathbf{a}$ is a point on the line and $\mathbf{d}$ is the direction vector.

What is the Cartesian form of a 3D line equation?

The Cartesian form is $\frac{x-x_1}{d_1} = \frac{y-y_1}{d_2} = \frac{z-z_1}{d_3}$, where $(x_1, y_1, z_1)$ is a point on the line and $(d_1, d_2, d_3)$ is the direction vector.

How do you verify if two lines intersect in 3D space?

Solve the simultaneous equations for the first two coordinates to find the parameters, then check if these values satisfy the third coordinate equation (the consistency check).

How is the angle between two 3D lines calculated?

The angle between two lines is the angle between their direction vectors ($\mathbf{d}_1$ and $\mathbf{d}_2$), found using the scalar product formula.

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