Magnitude, Direction and Unit Vectors — Edexcel A-level Mathematics (Pure)

SPEC 10.1 • VECTORS

Magnitude, Direction & Unit Vectors

Calculate lengths, find angles, and understand unit vectors in 2D and 3D

Learning Objectives

By the end of this lesson

You will be able to:

Calculate the magnitude (length) of 2D and 3D vectors

Find the direction of a 2D vector as an angle

Understand what a unit vector is

Find the unit vector in the direction of any vector

Recap from 10.01

Remember: A vector has both magnitude (size/length) and direction. Today we learn how to calculate these!

What is Magnitude?

Definition

The magnitude of a vector is its length.

Written as |a| or just a (non-bold)

Notation: | a | means "magnitude of a "

Example:

|a| = 5 means the vector has length 5

Key Fact

Magnitude is always positive (or zero for the zero vector).

Magnitude in 2D

We use Pythagoras' theorem to find the magnitude:

Formula

|a| = √(x² + y²)

Why It Works

The vector forms the hypotenuse of a right-angled triangle with sides x and y.

Magnitude in 3D

We extend Pythagoras to three dimensions:

Formula

|a| = √(x² + y² + z²)

For vector: x y z

Magnitude:

√(x² + y² + z²)

Example

For a = 236: |a| = √(4 + 9 + 36) = √49 = 7

Vector & Distance Between Points

Vector Between Two Points

To find vector from point A to point B:Vector A→B = x_B − x_Ay_B − y_Az_B − z_AEnd point − Start point (for each component)

Distance Between Two Points

Distance AB = |Vector A→B| = √[(x_B−x_A)² + (y_B−y_A)² + (z_B−z_A)²]

Key Connection

The distance between two points equals the magnitude of the vector between them!

Direction in 2D

For 2D vectors, we find the angle from the positive x-axis using trigonometry:

Formula

tan θ = y/x θ = tan⁻¹(y/x)

Watch Out - Quadrants!

Calculator gives angles between −90° and 90°. For vectors in quadrants 2 or 3 (negative x), add 180° to your answer!

Direction - Quadrant Check

Quick Rule

If x is negative:

Add 180° to calculator answer

If x is positive:

Use calculator answer directly

What is a Unit Vector?

Definition

A unit vector is a vector with magnitude exactly 1.

Standard Unit Vectors

i = 10

x-directionj = 01

y-directionk = 001

z-direction

Why Unit Vectors Matter

Unit vectors give us direction only, without any magnitude. They're essential for describing directions in physics and engineering.

Finding a Unit Vector

To find the unit vector in the direction of any vector a:

Formula

â = a|a| = 1|a| × a(Divide the vector by its magnitude)

Example

If a = 34 then |a| = 5

â = 1534 = 3545 ← Leave as fractions!

Notation

A unit vector is often written with a "hat": â (pronounced "a-hat"). Keep answers as exact fractions where possible.

Worked Example01

Find the magnitude of vector a = 34

1

Write down the formula: |a| = √(x² + y²)

2

Substitute the values: |a| = √(3² + 4²)

3

Calculate: |a| = √(9 + 16) = √25 = 5

Answer|a| = 5

Worked Example02

Find the magnitude of vector b = 2−12

1

Write down the 3D formula: |b| = √(x² + y² + z²)

2

Substitute: |b| = √(2² + (−1)² + 2²)

3

Calculate: |b| = √(4 + 1 + 4) = √9 = 3

Answer|b| = 3

Worked Example03

Find the angle that vector a = 55 makes with the positive x-axis.

1

Use the formula: tan θ = y/x

2

Substitute: tan θ = 5/5 = 1

3

Find the angle: θ = tan⁻¹(1) = 45°

4

Check quadrant: x > 0 and y > 0 → Quadrant 1 ✓ (no adjustment needed)

Answerθ = 45°

Worked Example04

Find the angle that vector b = −34 makes with the positive x-axis.

1

Use the formula: tan θ = y/x = 4/(−3) = −4/3

2

Calculator gives: tan⁻¹(−4/3) = −53.13°

3

Check quadrant: x < 0 and y > 0 → Quadrant 2

4

Since x is negative, add 180°: θ = −53.13° + 180° = 126.87°

Answerθ = 126.9° (1 d.p.) or 127° (nearest degree)

Worked Example05

Find the unit vector in the direction of a = 68

1

Find the magnitude: |a| = √(6² + 8²) = √(36 + 64) = √100 = 10

2

Divide each component by 10:

•

x-component: 6 ÷ 10 = 610 = 35

y-component: 8 ÷ 10 = 810 = 45

3

Unit vector: â = 11068 = 3545

Answerâ = 3545

Worked Example06

Find the unit vector in the direction of b = 2i − 2j + k

1

Find the magnitude: |b| = √(2² + (−2)² + 1²) = √(4 + 4 + 1) = √9 = 3

2

Divide each component by 3:

•

i: 2 ÷ 3 = 23

j: −2 ÷ 3 = −23

k: 1 ÷ 3 = 13

Answerb̂ = 13(2i − 2j + k) = 23i − 23j + 13k

Worked Example07

A vector has magnitude 10 and makes an angle of 30° with the positive x-axis. Write this vector in component form.

1

Use trigonometry to find components:

•

x = |a| × cos θ = 10 × cos 30° = 10 × (√3/2) = 5√3

y = |a| × sin θ = 10 × sin 30° = 10 × 0.5 = 5

2

Write as a vector: a = 5√35

Answera = 5√3i + 5j or 5√35

Worked Example08

Points A(1, 2, 3) and B(4, 6, 3). Find the distance AB and the unit vector from A to B.

1

Find vector A→B (end − start for each component):

•

x: 4 − 1 = 3, y: 6 − 2 = 4, z: 3 − 3 = 0

Vector A→B = 340

2

Distance = magnitude: |A→B| = √(9 + 16 + 0) = √25 = 5

3

Unit vector = 15 × A→B = 35450

AnswerDistance = 5, Unit vector = 35450

Practice

Question 1: Find the magnitude of:

(a) a = 512 (b) b = 122

(a) |a| = √(5² + 12²) = √(25 + 144) = √169 = 13

(b) |b| = √(1² + 2² + 2²) = √(1 + 4 + 4) = √9 = 3

Practice

Question 2: For vector p = 43, find:

(a) The magnitude |p|

(b) The angle it makes with the positive x-axis

(c) The unit vector in the direction of p

(a) |p| = √(16 + 9) = √25 = 5

(b) tan θ = 34, so θ = tan⁻¹(0.75) = 36.9° (1 d.p.)

(c)p̂ = 15p = 4535

Practice

Question 3: A drone flies from point A(2, 1, 0) to point B(5, 5, 12).

Find the distance travelled and the unit vector in the direction of travel.

Step 1: Find vector A→B (end − start):

x: 5 − 2 = 3, y: 5 − 1 = 4, z: 12 − 0 = 12

A→B = 3412

Step 2: Distance = |A→B| = √(9 + 16 + 144) = √169 = 13 units

Step 3: Unit vector = 113 × A→B = 3134131213

Summary

Concept2D3D

Magnitude| a | = √(x² + y²)| a | = √(x² + y² + z²)

Directionθ = tan⁻¹(y/x)Needs 2 angles

Unit Vectorâ = a / | a |

Key Exam Skills

Use Pythagoras correctly (don't forget to square root!)

Check your quadrant when finding direction angles

To verify a unit vector: its magnitude should equal 1

Distance between points = magnitude of the vector between them