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Here we will learn about the magnitude of a vector, including what the magnitude of a vector is and how to calculate it.

There are also vector worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

The **magnitude of a vector** is the length of a vector. It is also known as the **modulus** or the **absolute value** of the vector.

The magnitude of the vector \textbf{a} is written as \lvert \textbf{a} \rvert

To work out the magnitude of a vector we use the Pythagorean theorem.

In general, we can develop a formula:

\textbf{a}= \begin{pmatrix} \; x \;\\ \; y \; \end{pmatrix} \lvert \textbf{a} \rvert=\sqrt{x^2+y^2}If a vector has a magnitude of 1 , it is a unit vector

E.g.

\textbf{a}= \begin{pmatrix} \; 4 \;\\ \; 3 \; \end{pmatrix}Vector \textbf{a} has two components.

There is a horizontal component and a vertical component. Using the initial point and the terminal point of the vector, the components of the vector make a right-angled triangle. The length of the vector is the hypotenuse of the right-angled triangle.

The magnitude of vector \textbf{a} is

\sqrt{4^2+3^2}=\sqrt{25}=5The length of the vector is 5 .

In order to calculate the magnitude of a vector:

**Note the components of the vector.****Use Pythagoras’ theorem.****Write down the answer.**

Get your free magnitude of a vector worksheet of 20+ vectors questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free magnitude of a vector worksheet of 20+ vectors questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Magnitude of a vector** is part of our series of lessons to support revision on **vectors**. You may find it helpful to start with the main vectors lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

Find the magnitude of vector \textbf{b} , giving your answer to 1 decimal place:

\textbf{b}= \begin{pmatrix} \; 5 \;\\ \; 2 \; \end{pmatrix}**Note the components of the vector.**

The horizontal component is x=5

The vertical component is y=2

2**Use Pythagoras’ theorem.**

Use the components

\sqrt{x^2+y^2}=\sqrt{5^2+2^2}=\sqrt{29}3**Write down the answer.**

The magnitude of the vector is

\sqrt{29}=5.3851…=5.4 \ \text{(to } 1 \text{ decimal place)}Find the magnitude of vector \textbf{v} , giving your answer to 1 decimal place:

\textbf{v}= \begin{pmatrix} \; -1 \;\\ \; 3 \; \end{pmatrix}**Note the components of the vector.**

The horizontal component is x=-1

The vertical component is y=3

**Use Pythagoras’ theorem.**

Use the components

\sqrt{x^2+y^2}=\sqrt{(-1)^2+3^2}=\sqrt{10}

**Write down the answer.**

The magnitude of the vector is

\sqrt{10}=3.1622…=3.2 \ \text{(to } 1 \text{ decimal place)}

Find the magnitude of vector \textbf{a} , giving your answer to 1 decimal place:

\textbf{a}= \begin{pmatrix} \; -4 \;\\ \; 5 \; \end{pmatrix}**Note the components of the vector.**

The horizontal component is x=-4

The vertical component is y=5

**Use Pythagoras’ theorem.**

Use the components

\sqrt{x^2+y^2}=\sqrt{(-4)^2+5^2}=\sqrt{41}

**Write down the answer.**

The magnitude of the vector is

\sqrt{41}=6.4031…=6.4 \ \text{(to } 1 \text{ decimal place)}

Find the magnitude of vector \textbf{b} , giving your answer to 1 decimal place:

\textbf{b}= \begin{pmatrix} \; -2 \;\\ \; -3 \; \end{pmatrix}**Note the components of the vector.**

The horizontal component is x=-2

The vertical component is y=-3

**Use Pythagoras’ theorem.**

Use the components

\sqrt{x^2+y^2}=\sqrt{(-2)^2+(-3)^2}=\sqrt{13}

**Write down the answer.**

The magnitude of the vector is

\sqrt{13}=3.6055…=3.6 \ \text{(to } 1 \text{ decimal place)}

Find the magnitude of vector \textbf{c} , giving your answer to 1 decimal place:

\textbf{c}= \begin{pmatrix} \; -4 \;\\ \; -6 \; \end{pmatrix}**Note the components of the vector.**

The horizontal component is x=-4

The vertical component is y=-6

**Use Pythagoras’ theorem.**

Use the components

\sqrt{x^2+y^2}=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}

**Write down the answer.**

The magnitude of the vector is

\sqrt{52}=7.2111…=7.2 \ \text{(to } 1 \text{ decimal place)}

Find the magnitude of vector \textbf{v} , giving your answer to 1 decimal place:

\textbf{v}= \begin{pmatrix} \; 2 \;\\ \; -7 \; \end{pmatrix}**Note the components of the vector.**

The horizontal component is x=2

The vertical component is y=-7

**Use Pythagoras’ theorem.**

Use the components

\sqrt{x^2+y^2}=\sqrt{2^2+(-7)^2}=\sqrt{53}

**Write down the answer.**

The magnitude of the vector is

\sqrt{53}=7.2801…=7.3 \ \text{(to } 1 \text{ decimal place)}

**Be careful when squaring negative values**

A negative number squared gives a positive answer. If you need to square a negative number using a calculator, it is a good idea to use brackets to make sure that you get the correct answer.

1. Calculate the magnitude of the vector. Give your answer to 1 decimal place.

\begin{pmatrix} \; 3 \;\\ \; 8 \; \end{pmatrix}

8.6

4.8

4.9

8.5

The magnitude of the vector is

\sqrt{x^2+y^2}=\sqrt{3^2+8^2}=\sqrt{73}=8.544…=8.5 \ \text{to } 1 \text{ decimal place}

2. Calculate the magnitude of the vector. Give your answer to 1 decimal place.

\begin{pmatrix} \; 4 \;\\ \; 7 \; \end{pmatrix}

8.0

5.3

8.1

5.2

The magnitude of the vector is

\sqrt{x^2+y^2}=\sqrt{4^2+7^2}=\sqrt{65}=8.062…=8.1 \ \text{to } 1 \text{ decimal place}

3. Calculate the magnitude of the vector. Give your answer to 1 decimal place.

\begin{pmatrix} \; 1 \;\\ \; -3 \; \end{pmatrix}

3.1

1.7

3.2

1.8

The magnitude of the vector is

\sqrt{x^2+y^2}=\sqrt{1^2+(-3)^2}=\sqrt{10}=3.162…=3.2 \ \text{to } 1 \text{ decimal place}

4. Calculate the magnitude of the vector. Give your answer to 1 decimal place.

\begin{pmatrix} \; 2 \;\\ \; -4 \; \end{pmatrix}

4.5

4.4

5.1

5.2

The magnitude of the vector is

\sqrt{x^2+y^2}=\sqrt{2^2+(-4)^2}=\sqrt{20}=4.472…=4.5 \ \text{to } 1 \text{ decimal place}

5. Calculate the magnitude of the vector. Give your answer to 1 decimal place.

\begin{pmatrix} \; -3 \;\\ \; 2 \; \end{pmatrix}

3.7

3.6

2.4

2.5

The magnitude of the vector is

\sqrt{x^2+y^2}=\sqrt{(-3)^2+2^2}=\sqrt{13}=3.605…=3.6 \ \text{to } 1 \text{ decimal place}

6. Calculate the magnitude of the vector. Give your answer to 1 decimal place.

\begin{pmatrix} \; -1 \;\\ \; -4 \; \end{pmatrix}

4.2

5.1

4.1

5.2

The magnitude of the vector is

\sqrt{x^2+y^2}=\sqrt{(-1)^2+(-4)^2}=\sqrt{17}=4.123…=4.1 \ \text{to } 1 \text{ decimal place}

1. What is the magnitude of this vector?

\begin{pmatrix} \; -4 \;\\ \; 0 \; \end{pmatrix}

**(2 marks)**

Show answer

\sqrt{(-4)^{2}+0^{2}}

(For using Pythagoras’ theorem)

**(1)**

=\sqrt{16}=4

(For the correct answer)

**(1)**

2. Here is a column vector.

\textbf{a}= \begin{pmatrix} \; 7 \;\\ \; 3 \; \end{pmatrix}

Calculate the magnitude of the vector \textbf{a} .

Give your answer to 3 significant figures.

**(2 marks)**

Show answer

\sqrt{7^2+3^2}

(For using Pythagoras’ theorem)

**(1)**

=\sqrt{58}=7.61577…=7.62 \ \text{to } 3sf

(For the correct answer)

**(1)**

3. Here is a column vector.

\textbf{b}= \begin{pmatrix} \; -4 \;\\ \; 9 \; \end{pmatrix}

Calculate the magnitude of the vector \textbf{b} .

Give your answer to 3 significant figures.

**(2 marks)**

Show answer

\sqrt{(-4)^2+9^2}

(For using Pythagoras’ theorem)

**(1)**

=\sqrt{97}=9.84885…=9.85 \ \text{to } 3sf

(For the correct answer)

**(1)**

You have now learned how to:

- Calculate the magnitude of a vector using Pythagoras’ theorem

Vectors are studied more beyond GCSE. If a vector has a magnitude of 1 , it is a unit vector. The magnitude of a vector can be used to calculate the dot product of two vectors. It is also possible to calculate the cross product of two vectors.

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