# Position vector

> Edexcel International GCSE Further Pure Mathematics · 4PM1
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s7-position-vector/

This guide covers 2D position vectors for Edexcel IGCSE Further Pure Math (4PM1) Section 7, including defining vectors relative to an origin and calculating displacement between two points.

**Prerequisites:** [Basic vector notation and arithmetic](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s7-vector-arithmetic/); [Magnitude of vectors](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s7-vector-magnitude/)

## Learning objectives

- Define 2D position vectors relative to a fixed origin O
- Apply the rule $\boldsymbol{\bigtriangleup AB = OB - OA = b - a}$ to calculate displacement between two points
- Solve standard 4PM1 exam problems using position vectors

## 1. Definition of Position Vectors Relative to Origin O

**Position Vector** — A vector that starts at the fixed origin O and ends at point A. For a point with 2D coordinates $(x_A, y_A)$, its position vector is written as a column vector: $\textbf{a} = \begin{pmatrix} x_A \ y_A \end{pmatrix}$.

*Notation:* \textbf{a} or \bigtriangleup OA

*Example:* If point A has coordinates (3, 4) relative to origin O, its position vector is $\textbf{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.

All position vectors in a single exam question are measured from the same fixed origin (stated at the start of the question). 4PM1 only assesses 2D position vectors, so you will never see a third (z) component in official questions.

**Worked example:** Write the position vectors for points P (2, -5) and Q (-1, 0) relative to origin O.

1. For point P, use the x-coordinate as the top component and y-coordinate as the bottom component of the column vector:
2. $$\textbf{p} = \bigtriangleup OP = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$$
3. For point Q, repeat the process using its coordinates:
4. $$\textbf{q} = \bigtriangleup OQ = \begin{pmatrix} -1 \\ 0 \end{pmatrix}$$

> **tip**
>
> Position vectors are always column vectors in 4PM1, so avoid writing them as horizontal row vectors in your exam answers to prevent confusion.

## 2. The $\boldsymbol{\bigtriangleup AB = b - a}$ Displacement Rule

To find the displacement vector from point A to point B, you subtract the position vector of the starting point (A) from the position vector of the end point (B). This is one of the most frequently tested rules in 4PM1 Section 7.

**Displacement between two points** — The vector that travels directly from point A to point B, calculated as: $\bigtriangleup AB = \bigtriangleup OB - \bigtriangleup OA = \textbf{b} - \textbf{a}$, where $\textbf{a}$ and $\textbf{b}$ are the position vectors of A and B respectively.

*Notation:* \bigtriangleup AB

*Example:* If $\textbf{a} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ and $\textbf{b} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}$, then $\bigtriangleup AB = \begin{pmatrix} 4 - 1 \\ 7 - 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.

**Worked example:** Point A has position vector $\textbf{a} = \begin{pmatrix} 2 \\ 6 \end{pmatrix}$ and point B has position vector $\textbf{b} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}$ relative to origin O. Calculate the vector $\bigtriangleup AB$.

1. Recall the displacement rule: $\bigtriangleup AB = \textbf{b} - \textbf{a}$ (end point position vector minus start point position vector).
2. Substitute the given position vectors into the rule:
3. $$\bigtriangleup AB = \begin{pmatrix} 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ 6 \end{pmatrix}$$
4. Subtract corresponding components of the vectors:
5. $$\bigtriangleup AB = \begin{pmatrix} 5 - 2 \\ -1 - 6 \end{pmatrix} = \begin{pmatrix} 3 \\ -7 \end{pmatrix}$$

> **Order matters!**
>
> Swapping the order of subtraction will give you $\bigtriangleup BA = \textbf{a} - \textbf{b}$, which is the negative of the correct vector $\bigtriangleup AB$, so you will lose marks for this mistake.

*Calculator:* allowed

## 3. Exam-style Position Vector Problems

Exam questions for this topic often ask you to find unknown components of position vectors, or relate position vectors to coordinate geometry concepts you already know.

**Worked example:** The vector $\bigtriangleup MN = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$. Point M has position vector $\textbf{m} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ relative to origin O. Find the coordinates of point N.

1. Start with the displacement rule: $\bigtriangleup MN = \textbf{n} - \textbf{m}$
2. Rearrange to solve for the unknown position vector $\textbf{n}$:
3. $$\textbf{n} = \bigtriangleup MN + \textbf{m}$$
4. Substitute the given values and add corresponding components:
5. $$\textbf{n} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} + \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 + 1 \\ -2 + 3 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$$
6. Convert the position vector to coordinates: the x-coordinate is the top component, y-coordinate is the bottom component, so N is at (5, 1).

**Check your understanding**

1. If $\textbf{a} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}$ and $\textbf{b} = \begin{pmatrix} 1 \\ -4 \end{pmatrix}$, what is $\bigtriangleup BA$?

   - $\begin{pmatrix} 4 \\ -6 \end{pmatrix}$
   - $\begin{pmatrix} -4 \\ 6 \end{pmatrix}$
   - $\begin{pmatrix} -2 \\ -2 \end{pmatrix}$

   *Answer:* $\begin{pmatrix} -4 \\ 6 \end{pmatrix}$

   *Why:* $\bigtriangleup BA = \textbf{a} - \textbf{b} = \begin{pmatrix} -3 - 1 \\ 2 - (-4) \end{pmatrix} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}$. Remember that $\bigtriangleup BA$ is the vector from B to A, so you subtract B's position vector from A's position vector.

**Summary**

- Position vectors describe points relative to a fixed origin O, with components matching the point's x (top) and y (bottom) coordinates.
- The displacement vector from A to B is always $\bigtriangleup AB = \textbf{b} - \textbf{a}$, where $\textbf{a}$ and $\textbf{b}$ are the position vectors of A and B.
- You can rearrange the displacement rule to find unknown position vectors if you know one point and the vector between them.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Calculating $\bigtriangleup AB$ as $\textbf{a} - \textbf{b}$ instead of $\textbf{b} - \textbf{a}$.
  - Why it fails: Swapping the order of subtraction gives the vector from B to A, which is the negative of the correct answer, leading to lost marks.
  - Correct: Always subtract the position vector of the starting point from the position vector of the end point: end minus start.
- **Wrong:** Writing position vectors as $\begin{pmatrix} y \\ x \end{pmatrix}$ instead of $\begin{pmatrix} x \\ y \end{pmatrix}$.
  - Why it fails: The top component of a 2D vector corresponds to the horizontal x-coordinate, the bottom to the vertical y-coordinate. Swapping these gives the wrong position.
  - Correct: Copy coordinates directly into the vector: first the x value, then the y value, matching the order of coordinate pairs.
- **Wrong:** Using 3D vectors with a z-component for 4PM1 position vector questions.
  - Why it fails: The 4PM1 specification explicitly limits position vectors to 2D only, no 3D content is assessed in this topic.
  - Correct: Only use two components for all position vectors in the 4PM1 exam, as official questions will never include a z-coordinate.
- **Wrong:** Assuming the origin changes between points in a single question.
  - Why it fails: All position vectors in a single question are measured relative to the same fixed origin stated at the start of the question, unless explicitly told otherwise.
  - Correct: Always reference the stated origin O for all position vectors in a question, unless you are given explicit instructions to use a different reference point.

## Cheatsheet

| Concept | Rule | Example |
| --- | --- | --- |
| Position vector of A (x,y) | $\textbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}$ | A=(2,5) → $\textbf{a} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$ |
| Displacement $\bigtriangleup AB$ | $\bigtriangleup AB = \textbf{b} - \textbf{a}$ | $\textbf{a}=\begin{pmatrix}1\\3\end{pmatrix}, \textbf{b}=\begin{pmatrix}4\\1\end{pmatrix}$ → $\bigtriangleup AB=\begin{pmatrix}3\\-2\end{pmatrix}$ |
| Find unknown position vector of B | $\textbf{b} = \bigtriangleup AB + \textbf{a}$ | $\bigtriangleup AB=\begin{pmatrix}2\\1\end{pmatrix}, \textbf{a}=\begin{pmatrix}3\\0\end{pmatrix}$ → $\textbf{b}=\begin{pmatrix}5\\1\end{pmatrix}$ |

## What's next

Now that you have mastered position vectors, you are ready to progress to more advanced vector topics in Edexcel IGCSE Further Pure Math. Next, you will learn about unit vectors, which build directly on the position vector and magnitude concepts you have already covered. You will also use position vectors to solve ratio division problems, such as finding the position vector of a point that divides a line segment in a given ratio. These topics are all core to Section 7 of the 4PM1 specification, and are frequently tested together in multi-mark exam questions. Make sure you practice applying the $\bigtriangleup AB = \textbf{b} - \textbf{a}$ rule regularly to avoid common order-of-subtraction mistakes in your exam.

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