# Components and resolved parts of a vector

> Edexcel International GCSE Further Pure Mathematics · 4PM1 2016 Spec (Higher)
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s7-components-and-resolved-parts-of/

This guide teaches you to write 2D vectors using i and j base units, calculate horizontal and vertical resolved parts of vectors, and combine component values, aligned exactly to the Edexcel 4PM1 Further Pure IGCSE specification.

**Prerequisites:** [Understanding of basic vector definitions and scalar vs vector quantities](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s7-vector-algebra/); Fluency with right triangle trigonometry (sine, cosine of angles)

## Learning objectives

- Express 2D vectors correctly in i-j component form
- Calculate horizontal and vertical resolved parts of vectors given magnitude and direction
- Combine vector components to find resultant vectors accurately

## Component Form of Vectors Using i and j Base Units

**Component form** — A vector is written in component form as $a\mathbf{i} + b\mathbf{j}$, where $a$ is the horizontal displacement and $b$ is the vertical displacement from the start to end point of the vector. $\mathbf{i}$ is the unit vector (magnitude 1) in the positive x-direction, $\mathbf{j}$ is the unit vector in the positive y-direction.

*Example:* A vector that moves 3 units right and 2 units up is written as $3\mathbf{i} + 2\mathbf{j}$

If you are given the coordinates of the start and end points of a vector, subtract the start coordinates from the end coordinates to find the i and j components. For a vector from point $(x_1, y_1)$ to $(x_2, y_2)$: the i component is $x_2 - x_1$, the j component is $y_2 - y_1$.

**Worked example:** Find the component form of the vector from point $P(2, 5)$ to point $Q(-1, 7)$.

1. Step 1: Calculate the horizontal i component: subtract the x-coordinate of P from the x-coordinate of Q.
2. $$i\text{ component} = x_Q - x_P = -1 - 2 = -3$$
3. Step 2: Calculate the vertical j component: subtract the y-coordinate of P from the y-coordinate of Q.
4. $$j\text{ component} = y_Q - y_P = 7 - 5 = 2$$
5. Step 3: Combine the components to write the final vector.
6. $$\overrightarrow{PQ} = -3\mathbf{i} + 2\mathbf{j}$$

> **Exam tip:** Always double check the direction of the vector: if you are calculating the vector from P to Q, you subtract P's coordinates from Q's, not the other way around, which is a common sign error.

*Calculator:* allowed

## Calculating Resolved Parts of a Vector

**Resolved parts of a vector** — The resolved parts of a vector of magnitude $V$ at an angle $\theta$ to the positive x-axis are the horizontal (x-axis) and vertical (y-axis) scalar components of the vector.

*Notation:* $V_x = V\cos\theta$, $V_y = V\sin\theta$

The angle $\theta$ is always measured anticlockwise from the positive horizontal x-axis unless stated otherwise. If the vector points below the x-axis, $\theta$ will be negative, or you can add a negative sign to the vertical component directly. If the angle is given relative to the y-axis, swap cosine and sine for the respective components.

**Worked example:** A force vector has magnitude 12 N and acts at an angle of 40° above the positive x-axis. Find its horizontal and vertical resolved parts, and write the vector in i-j form. Give your answers to 2 decimal places.

1. Step 1: Identify the given values: magnitude $V=12$, angle $\theta=40^\circ$.
2. Step 2: Calculate the horizontal resolved part using $V_x = V\cos\theta$.
3. $$V_x = 12 \times \cos(40^\circ) = 12 \times 0.7660 = 9.19 \text{ (2 d.p.)}$$
4. Step 3: Calculate the vertical resolved part using $V_y = V\sin\theta$.
5. $$V_y = 12 \times \sin(40^\circ) = 12 \times 0.6428 = 7.71 \text{ (2 d.p.)}$$
6. Step 4: Write the vector in component form.
7. $$9.19\mathbf{i} + 7.71\mathbf{j} \text{ N}$$

> **tip**
>
> If the angle is given as 40° above the negative x-axis, either use $\theta = 180^\circ - 40^\circ = 140^\circ$ in your calculation, or add a negative sign to the horizontal component directly to avoid mistakes.

*Calculator:* allowed

## Combining Vector Components

When you are given two or more vectors, you can find the resultant vector by adding their respective i and j components separately. This skill is the foundation for all more complex vector calculations you will encounter in the 4PM1 exam.

**Worked example:** Two vectors are given: $\mathbf{a} = 4\mathbf{i} - 2\mathbf{j}$ and $\mathbf{b} = -1\mathbf{i} + 5\mathbf{j}$. Find the resultant vector $\mathbf{a} + \mathbf{b}$ in component form.

1. Step 1: Add the i components of both vectors.
2. $$\text{Total i component} = 4 + (-1) = 3$$
3. Step 2: Add the j components of both vectors.
4. $$\text{Total j component} = -2 + 5 = 3$$
5. Step 3: Write the resultant vector.
6. $$\mathbf{a} + \mathbf{b} = 3\mathbf{i} + 3\mathbf{j}$$

**Check your understanding**

1. A vector of magnitude 10 acts at 30° below the positive x-axis. What is its j component?

   - $5\mathbf{j}$
   - $-5\mathbf{j}$
   - $5\sqrt{3}\mathbf{j}$
   - $-5\sqrt{3}\mathbf{j}$

   *Answer:* $-5\mathbf{j}$

   *Why:* The angle below the x-axis means the vertical component is negative: $10 \times \sin(-30^\circ) = -10 \times 0.5 = -5$, so the j component is $-5\mathbf{j}$.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Subtracting end coordinates from start coordinates when calculating a vector between two points
  - Why it fails: This reverses the direction of the vector, leading to incorrect sign values for both i and j components
  - Correct: Always subtract the coordinates of the starting point from the coordinates of the end point of the vector
- **Wrong:** Using sine for the horizontal component and cosine for the vertical component when angle is measured from the x-axis
  - Why it fails: The horizontal component is adjacent to the angle with the x-axis so uses cosine, while the vertical is opposite so uses sine
  - Correct: Remember: adjacent = hypotenuse × cos(theta) for horizontal x-component, opposite = hypotenuse × sin(theta) for vertical y-component when angle is measured from x-axis
- **Wrong:** Forgetting to add a negative sign to components when the vector points left or down relative to the positive axes
  - Why it fails: This leads to overestimating or misplacing the direction of the resultant vector in calculations
  - Correct: Assign negative values to i components for leftward displacements, and negative j components for downward displacements
- **Wrong:** Using radian mode on the calculator when calculating cosine and sine of angles given in degrees
  - Why it fails: All Edexcel 4PM1 vector questions use angles in degrees, so radian mode will give incorrect component values
  - Correct: Always check your calculator is set to degree mode before starting any vector component calculation in the exam

## Cheatsheet

| Task | Formula / Rule | Example |
| --- | --- | --- |
| Write vector between (x1,y1) and (x2,y2) in i-j form | $(x2-x1)\mathbf{i} + (y2-y1)\mathbf{j}$ | From (1,2) to (4,6): $3\mathbf{i} + 4\mathbf{j}$ |
| Find horizontal resolved part of magnitude V at angle θ to +x axis | $V_x = V\cos\theta$ | V=20, θ=30°: $V_x = 20\cos30° = 10\sqrt{3}$ |
| Find vertical resolved part of magnitude V at angle θ to +x axis | $V_y = V\sin\theta$ | V=20, θ=30°: $V_y = 20\sin30° = 10$ |
| Add two vectors in component form | Add i components, add j components separately | $(2\mathbf{i} + 3\mathbf{j}) + (4\mathbf{i} - 1\mathbf{j}) = 6\mathbf{i} + 2\mathbf{j}$ |

## What's next

Now that you can express vectors in i-j component form and calculate their resolved parts, you are ready to move on to calculating the magnitude of vectors, the next core skill in the vectors unit for Edexcel IGCSE Further Pure Maths. You will also use component form extensively when working with position vectors, and later when solving problems involving forces and kinematics using vectors. Mastering component calculations now will save you time and reduce errors in more complex vector problems that appear frequently in both Paper 1 and Paper 2 of the 4PM1 exam.

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