$$\newcommand\mycolv[1]{\begin{pmatrix}#1\end{pmatrix}}$$

# Magnitude and Direction

## Calculate the magnitude and direction of a vector and convert between forms.

 1 Find the length of the vector: $$\mycolv{3\\4}$$ ☐ ☐ ✓ ✗ 2 Find the magnitude of: $$\mycolv{12\\5}$$ ☐ ☐ ✓ ✗ 3 Find the length of the vector: $$\mycolv{8\\6}$$ ☐ ☐ ✓ ✗ 4 Find the magnitude of $$\mycolv{9\\12}$$ ☐ ☐ ✓ ✗ (to 3 significant figures) 5 Find the magnitude of $$\mycolv{-10\\19}$$ ☐ ☐ ✓ ✗ (to 3 significant figures) 6 Find the magnitude of $$\mycolv{19\\-9}$$ ☐ ☐ ✓ ✗ (to 3 significant figures) 7 Find the magnitude of $$\mycolv{-17\\19}$$ ☐ ☐ ✓ ✗ (to 3 significant figures) 8 Find the magnitude of $$\mycolv{6\\12}$$ ☐ ☐ ✓ ✗ (to 3 significant figures) 9 Find the magnitude of $$\mycolv{-9\\-15}$$ ☐ ☐ ✓ ✗ (to 3 significant figures)
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This is Vectors - Magnitude and Direction level 1. You can also try:
Vector Connectors Level 2 Level 3

## Instructions

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## Description of Levels

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Vector Connectors - A basic set of exercises on vectors which could be done before attempting the following.

Vectors - Addition and subtraction of vectors and multiplication of a vector by a scalar.

Level 1 - Magnitude of a vector from a column vector

Level 2 - Direction of a vector from a column vector

Level 3 - Finding the column vector from magnitude-direction form

Exam Style Questions - A collection of problems in the style of GCSE or IB/A-level exam paper questions (worked solutions are available for Transum subscribers).

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## Curriculum Reference

See the National Curriculum page for links to related online activities and resources.

## Example

This set of exercises is to provide you with practice calculating the magnitude and direction of a vector and converting between component form and magnitude/direction form.

The magnitude of a vector is the length of the vector. A length cannot be negative. If the horizontal and vertical components are known it can be calculated using Pythagoras' theorem.

For example consider the vector:

$$\mycolv{9\\12}$$

The magnitude is:

$$\sqrt{9^2 + 12^2} = 15$$

The direction of the vector can be calculated using trigonometry. The angle the vector makes with the horizontal is the inverse tan of the vertical component divide by the horizontal component.

$$\tan^{-1} \frac{12}{9} = 53.1^o$$

If the magnitude and direction are given, as in the diagram below, trigonometry can be used to calculate the components for the column vector.

$$x = 18 \cos 61^o = 8.73$$ $$y = 18 \sin 61^o = 15.7$$

So the column vector, with components expressed to 3 significant figures, is:

$$\mycolv{8.73\\15.7}$$

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