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Module 6 · L11 of 20 ~35 min ⚡ +95 XP available

Unit Vectors

A GPS navigator always knows the magnitude and direction of your journey, but for pure direction it strips away the distance and works with a unit vector — length exactly 1. In this lesson you'll define $\hat{\mathbf{u}} = \dfrac{\mathbf{u}}{|\mathbf{u}|}$, derive it from any non-zero vector, and use the standard basis vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ to decompose any vector in 2D or 3D.

Today's hook — The vector $\mathbf{v} = 3\mathbf{i} + 4\mathbf{j}$ has magnitude 5. Before reading the formula, what do you think the unit vector in the direction of $\mathbf{v}$ looks like? Jot a guess — you'll verify it after card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

The vector $\mathbf{v} = 3\mathbf{i} + 4\mathbf{j}$ has magnitude $|\mathbf{v}| = 5$. Without using a formula — what do you think a "unit vector" in the same direction looks like? Write your reasoning below.

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The core idea

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A unit vector is any vector with magnitude exactly 1. To find one in the direction of $\mathbf{u}$, divide by its magnitude:

Given a non-zero vector $\mathbf{u}$, the unit vector in its direction is:

$\hat{\mathbf{u}} = \dfrac{\mathbf{u}}{|\mathbf{u}|}$

This scales $\mathbf{u}$ so its length becomes 1 while preserving its direction. The notation $\hat{\mathbf{u}}$ (read "u-hat") signals a unit vector.

$\hat{\mathbf{u}} = \dfrac{\mathbf{u}}{|\mathbf{u}|}$

Magnitude check

Always verify: $|\hat{\mathbf{u}}| = \left|\dfrac{\mathbf{u}}{|\mathbf{u}|}\right| = \dfrac{|\mathbf{u}|}{|\mathbf{u}|} = 1$. If your answer is not length 1, recheck your calculation of $|\mathbf{u}|$.

Standard basis vectors

$\mathbf{i}$, $\mathbf{j}$ (and $\mathbf{k}$ in 3D) are already unit vectors: $|\mathbf{i}| = |\mathbf{j}| = |\mathbf{k}| = 1$.

Zero vector excluded

The zero vector $\mathbf{0}$ has no direction and cannot be normalised. Always check $\mathbf{u} \neq \mathbf{0}$ before applying the formula.

What you'll master

Know

Key facts

A unit vector has magnitude exactly 1
$\hat{\mathbf{u}} = \dfrac{\mathbf{u}}{|\mathbf{u}|}$ for any non-zero $\mathbf{u}$
$\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ are the standard basis unit vectors

Understand

Concepts

Why dividing by $|\mathbf{u}|$ preserves direction but sets length to 1
How any vector $\mathbf{u}$ can be written as $|\mathbf{u}|\,\hat{\mathbf{u}}$ (magnitude times direction)
The geometric meaning of normalisation

Can do

Skills

Calculate the unit vector in the direction of any 2D or 3D vector
Express a vector as a scalar multiple of a unit vector
Verify that a given vector is a unit vector

Key terms

Unit vectorA vector with magnitude exactly 1. Written $\hat{\mathbf{u}}$ ("u-hat"). Used to represent pure direction.

NormaliseThe process of dividing a vector by its magnitude to produce a unit vector: $\hat{\mathbf{u}} = \mathbf{u}/|\mathbf{u}|$.

$\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$Standard basis unit vectors along the positive $x$-, $y$-, and $z$-axes respectively.

Magnitude $|\mathbf{u}|$The length (Euclidean norm) of a vector: $|\mathbf{u}| = \sqrt{u_1^2 + u_2^2}$ in 2D, $\sqrt{u_1^2 + u_2^2 + u_3^2}$ in 3D.

Scalar multiple of $\hat{\mathbf{u}}$Any vector can be written as $\mathbf{u} = |\mathbf{u}|\,\hat{\mathbf{u}}$ — magnitude times direction.

DirectionThe orientation of a vector, independent of its magnitude. Two parallel vectors with equal direction have the same unit vector.

Finding the unit vector

core concept

To find the unit vector in the direction of $\mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j}$ (or in 3D with a $u_3\mathbf{k}$ component):

$$\hat{\mathbf{u}} = \frac{\mathbf{u}}{|\mathbf{u}|} = \frac{u_1\mathbf{i} + u_2\mathbf{j}}{\sqrt{u_1^2 + u_2^2}}$$

Step 1: Find $|\mathbf{u}| = \sqrt{u_1^2 + u_2^2}$. Step 2: Divide each component by $|\mathbf{u}|$.

Example: Find the unit vector in the direction of $\mathbf{v} = 3\mathbf{i} + 4\mathbf{j}$.

$|\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

$\hat{\mathbf{v}} = \dfrac{3\mathbf{i} + 4\mathbf{j}}{5} = \dfrac{3}{5}\mathbf{i} + \dfrac{4}{5}\mathbf{j}$

Verify: $|\hat{\mathbf{v}}| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25}+\frac{16}{25}} = \sqrt{1} = 1$ ✓

Real-world connection. When a computer game engine moves a character, it calculates the unit vector toward the target, then multiplies by the speed: $\mathbf{v}_\text{move} = \text{speed} \times \hat{\mathbf{u}}_\text{target}$. Unit vectors separate the "where" from the "how fast".

To find the unit vector in the direction of $\mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j}$ (or in 3D with a $u_3\mathbf{k}$ component):

Pause — copy the unit vector formula $\hat{u}=\frac{\vec{u}}{|\vec{u}|}$ with the verification $|\hat{u}|=1$ and a worked example into your book.

Quick check: The unit vector in the direction of $\mathbf{u} = 3\mathbf{i} + 4\mathbf{j}$ is:

Standard basis vectors and component form

core concept

We just saw that the unit vector in the direction of $\vec{u}$ is $\hat{u}=\frac{\vec{u}}{|\vec{u}|}$, always with magnitude 1. That raises a question: the standard basis vectors $\mathbf{i}=(1,0)^T$ and $\mathbf{j}=(0,1)^T$ (and in 3D $\mathbf{k}=(0,0,1)^T$) are themselves unit vectors — how does any vector $\vec{a}=(a_1,a_2)^T$ decompose into a linear combination of them? This card answers it → $\vec{a}=a_1\mathbf{i}+a_2\mathbf{j}$, so the components are exactly the coefficients of the basis vectors.

The standard basis vectors in 2D are:

$$\mathbf{i} = \begin{pmatrix}1\\0\end{pmatrix}, \quad \mathbf{j} = \begin{pmatrix}0\\1\end{pmatrix}$$

In 3D, $\mathbf{k} = (0, 0, 1)^T$ is added. These are unit vectors — check: $|\mathbf{i}| = \sqrt{1^2+0^2} = 1$.

Any vector $\mathbf{u} = \begin{pmatrix}a\\b\end{pmatrix}$ can be decomposed as $\mathbf{u} = a\mathbf{i} + b\mathbf{j}$, and written as a scalar multiple of its unit vector:

$$\mathbf{u} = |\mathbf{u}|\,\hat{\mathbf{u}}$$

3D example. $\mathbf{w} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$: magnitude $= \sqrt{4+1+4} = 3$; unit vector $\hat{\mathbf{w}} = \tfrac{2}{3}\mathbf{i} - \tfrac{1}{3}\mathbf{j} + \tfrac{2}{3}\mathbf{k}$. So $\mathbf{w} = 3\,\hat{\mathbf{w}}$.

In 3D, $\mathbf{k} = (0, 0, 1)^T$ is added. These are unit vectors — check: $|\mathbf{i}| = \sqrt{1^2+0^2} = 1$.

Pause — copy the basis decomposition: $\vec{a}=a_1\mathbf{i}+a_2\mathbf{j}$ in 2D, $\vec{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ in 3D, with $\mathbf{i},\mathbf{j},\mathbf{k}$ defined into your book.

Did you get this? True or false: the vector $\mathbf{k} = (0, 0, 1)$ is a unit vector.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · 2D UNIT VECTOR

Find the unit vector in the direction of $\mathbf{a} = -5\mathbf{i} + 12\mathbf{j}$.

$|\mathbf{a}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

Compute the magnitude using $|\mathbf{a}| = \sqrt{a_1^2+a_2^2}$. Note signs disappear under the square.

PROBLEM 2 · 3D UNIT VECTOR

Find the unit vector in the direction of $\mathbf{b} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$.

$|\mathbf{b}| = \sqrt{1^2+(-2)^2+2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$

Magnitude in 3D uses all three components.

PROBLEM 3 · SCALAR MULTIPLE FORM

Express $\mathbf{p} = 6\mathbf{i} - 8\mathbf{j}$ in the form $|\mathbf{p}|\,\hat{\mathbf{p}}$.

$|\mathbf{p}| = \sqrt{36+64} = \sqrt{100} = 10$

Find the magnitude first.

Fill the gap: The magnitude of $\mathbf{v} = -5\mathbf{i}+12\mathbf{j}$ is , so the unit vector has components $-\tfrac{5}{13}\mathbf{i}+\tfrac{12}{13}\mathbf{j}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting to take the square root

$|\mathbf{u}| = \sqrt{u_1^2+u_2^2}$, not $u_1^2+u_2^2$. A common error is dividing components by $u_1^2+u_2^2$ instead of $\sqrt{u_1^2+u_2^2}$. Always take the positive square root.

Trap 02

Losing the negative signs

When computing $|\mathbf{u}|$, negative components are squared so signs vanish. But when dividing to get $\hat{\mathbf{u}}$, negatives must be preserved: $\hat{\mathbf{u}} = \frac{-5\mathbf{i}+12\mathbf{j}}{13} = -\frac{5}{13}\mathbf{i}+\frac{12}{13}\mathbf{j}$.

Trap 03

Assuming the zero vector has a unit vector

The zero vector $\mathbf{0}$ has magnitude 0, so $\mathbf{0}/|\mathbf{0}|$ is undefined. Unit vectors only exist for non-zero vectors. If a question asks for a unit vector, check first that $|\mathbf{u}| \neq 0$.

Did you get this? True or false: the zero vector $\mathbf{0}$ has a unit vector in its direction.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find the unit vector in the direction of $\mathbf{u} = 8\mathbf{i} + 6\mathbf{j}$. Show all steps.

Find the unit vector in the direction of $\mathbf{v} = 2\mathbf{i} + 2\mathbf{j} - \mathbf{k}$.

A vector has unit vector $\hat{\mathbf{w}} = \tfrac{1}{2}\mathbf{i} - \tfrac{\sqrt{3}}{2}\mathbf{j}$ and magnitude 6. Find $\mathbf{w}$.

Show that $\mathbf{p} = \tfrac{3}{5}\mathbf{i} - \tfrac{4}{5}\mathbf{j}$ is a unit vector.

Find the unit vector in the direction of $\mathbf{q} = -3\mathbf{i} - 4\mathbf{j} + 0\mathbf{k}$ and verify it has magnitude 1.

Odd one out: Three of these are unit vectors. Which one is NOT?

Revisit your thinking

Earlier you guessed what the unit vector of $\mathbf{v} = 3\mathbf{i} + 4\mathbf{j}$ looks like.

The exact answer is $\hat{\mathbf{v}} = \tfrac{3}{5}\mathbf{i} + \tfrac{4}{5}\mathbf{j}$. Each component is divided by the magnitude 5. The direction is preserved — $\hat{\mathbf{v}}$ still points the same way as $\mathbf{v}$ — but the length is now exactly 1.

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Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Find the unit vector in the direction of $\mathbf{u} = 5\mathbf{i} - 12\mathbf{j}$. (2 marks)

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ApplyBand 43 marks

Q2. A force has magnitude 15 N and acts in the direction of $\mathbf{d} = 3\mathbf{i} + 4\mathbf{j}$. Write the force as a vector. (3 marks)

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AnalyseBand 53 marks

Q3. Find the unit vector in the direction of $\mathbf{w} = 4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}$. Verify your answer has magnitude 1. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $|\mathbf{u}|=\sqrt{64+36}=10$; $\hat{\mathbf{u}}=\tfrac{4}{5}\mathbf{i}+\tfrac{3}{5}\mathbf{j}$ · 2. $|\mathbf{v}|=\sqrt{4+4+1}=3$; $\hat{\mathbf{v}}=\tfrac{2}{3}\mathbf{i}+\tfrac{2}{3}\mathbf{j}-\tfrac{1}{3}\mathbf{k}$ · 3. $\mathbf{w}=6\hat{\mathbf{w}}=3\mathbf{i}-3\sqrt{3}\,\mathbf{j}$ · 4. $|\mathbf{p}|=\sqrt{\tfrac{9}{25}+\tfrac{16}{25}}=1$ ✓ · 5. $|\mathbf{q}|=5$; $\hat{\mathbf{q}}=-\tfrac{3}{5}\mathbf{i}-\tfrac{4}{5}\mathbf{j}$

Q1 (2 marks): $|\mathbf{u}|=\sqrt{25+144}=\sqrt{169}=13$ [1]. $\hat{\mathbf{u}}=\dfrac{5}{13}\mathbf{i}-\dfrac{12}{13}\mathbf{j}$ [1].

Q2 (3 marks): $|\mathbf{d}|=\sqrt{9+16}=5$ [1]. $\hat{\mathbf{d}}=\tfrac{3}{5}\mathbf{i}+\tfrac{4}{5}\mathbf{j}$ [1]. Force $=15\hat{\mathbf{d}}=9\mathbf{i}+12\mathbf{j}$ N [1].

Q3 (3 marks): $|\mathbf{w}|=\sqrt{16+16+4}=\sqrt{36}=6$ [1]. $\hat{\mathbf{w}}=\tfrac{2}{3}\mathbf{i}-\tfrac{2}{3}\mathbf{j}+\tfrac{1}{3}\mathbf{k}$ [1]. Verify: $\sqrt{\tfrac{4}{9}+\tfrac{4}{9}+\tfrac{1}{9}}=\sqrt{1}=1$ ✓ [1].

Boss battle · The Vector Normaliser

earn bronze · silver · gold

Five timed questions on unit vectors. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering unit vector questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 10 · Direction and Bearing Lesson 12 · The Scalar (Dot) Product →

Module overview · Extension 1 · Checkpoint 3