# Unit Vectors – Examples and Practice Problems

Unit vectors are vectors that have a magnitude of 1 and have no units. These vectors are used to describe a direction in space. To find the unit vector of a vector, we divide each component by its magnitude.

In this article, we will learn how to calculate unit vectors of vectors. We will learn about the formulas that we can use, and we will apply them to solve some practice problems.

##### PHYSICS

Relevant for

Learning about the unit vectors of a vector.

See examples

##### PHYSICS

Relevant for

Learning about the unit vectors of a vector.

See examples

## How to find the unit vector of a vector?

To find the unit vector of a given vector, we have to normalize the original vector. A unit vector is a vector with a magnitude (length) of 1, which points in the same direction as the original vector.

Then, we can follow two simple steps to find the unit vector:

Step 1: Calculate the magnitude (length) of the original vector: For a vector $latex \vec{V} = \langle x, y, z\rangle$, the magnitude is given by:

$latex |V| = \sqrt{x^2 + y^2 + z^2}$

Step 2: Divide each component of the original vector by its magnitude:

$$\hat{V} = \frac{\vec{V}}{|V|}$$

$$= \langle \frac{x}{|V|}, ~\frac{y}{|V|}, ~\frac{z}{|V|}\rangle$$

Note: We use the notation (^) or “hat” on the letter that represents unit vectors. This allows us to distinguish them from normal vectors.

## 10 Solved examples of unit vectors

### EXAMPLE 1

Find the unit vector of the vector $latex \vec{A} = \langle 3, ~4 \rangle$.

We start by calculating the magnitude of the vector:

$$|A| = \sqrt{3^2 + 4^2} =$$

$$\sqrt{9 + 16} = \sqrt{25} = 5$$

Now, we normalize the original vector:

$$\hat{A} = \left\langle \frac{3}{5}, ~\frac{4}{5} \right\rangle$$

$$= \langle 0.6, ~0.8 \rangle$$

The unit vector of $latex \vec{A} = \langle 3, ~4 \rangle$ is $latex \hat{A} = \langle 0.6, ~0.8 \rangle$.

### EXAMPLE 2

What is the unit vector of $latex \vec{B} = \langle -6, ~8 \rangle$?

By calculating the magnitude of the vector, we have:

$$|B| = \sqrt{(-6)^2 + 8^2}$$

$$= \sqrt{36 + 64} = \sqrt{100} = 10$$

Normalizing the vector, we have:

$$\hat{B} = \left\langle \frac{-6}{10}, ~\frac{8}{10} \right\rangle$$

$$= \langle -0.6, ~0.8 \rangle$$

The unit vector of $latex \vec{B} = \langle -6, ~8 \rangle$ is $latex \hat{B} = \langle -0.6, ~0.8 \rangle$.

### EXAMPLE 3

Find the unit vector of the vector $latex \vec{C} = \langle 5, ~12 \rangle$.

We start by calculating the magnitude of the given vector:

$$|C| = \sqrt{5^2 + 12^2}$$

$$= \sqrt{25 + 144} = \sqrt{169} = 13$$

Now, we divide each component of the vector by the magnitude:

$$\hat{C} = \left\langle \frac{5}{13},~ \frac{12}{13} \right\rangle$$

$$= \langle \frac{5}{13}, ~\frac{12}{13} \rangle$$

$$\approx \langle 0.385, ~0.923 \rangle$$

The unit vector of $latex \vec{C} = \langle 5,~ 12 \rangle$ is $latex \hat{C} \approx \langle 0.385, ~0.923 \rangle$.

### EXAMPLE 4

What is the unit vector of the vector $latex \vec{A} = \langle 2,~ -4, ~1 \rangle$?

The magnitude of vector $latex \vec{A}$ is:

$$|A| = \sqrt{2^2 + (-4)^2 + 1^2}$$

$$= \sqrt{4 + 16 + 1} = \sqrt{21}$$

Normalizing the vector, we have:

$$\hat{A} = \left\langle \frac{2}{\sqrt{21}}, ~\frac{-4}{\sqrt{21}}, ~\frac{1}{\sqrt{21}} \right\rangle$$

$$\approx \langle 0.44, -0.88, 0.22 \rangle$$

The unit vector of $latex \vec{A} = \langle 2, ~-4, ~1 \rangle$ is $latex \hat{A} \approx \langle 0.44, ~-0.88,~ 0.22 \rangle$.

### EXAMPLE 5

If we have the vector $latex \vec{B} = \langle -3, ~6,~2 \rangle$, find its unit vector.

The magnitude of vector $latex \vec{B}$ is:

$$|B| = \sqrt{(-3)^2 + 6^2 + 2^2}$$

$$= \sqrt{9 + 36 + 4} = \sqrt{49} = 7$$

When we normalize the vector, we have:

$$\hat{B} = \left\langle \frac{-3}{7}, ~\frac{6}{7}, ~\frac{2}{7} \right\rangle$$

$$\approx \langle -0.43, ~0.86,~ 0.29 \rangle$$

The unit vector of $latex \vec{B} = \langle -3, ~6, ~2 \rangle$ is $latex \hat{B} \approx \langle -0.43,~ 0.86,~ 0.29 \rangle$.

### EXAMPLE 6

Find the unit vector of vector $latex \vec{D} = \langle 4,~ -8, ~4 \rangle$.

We begin by finding the magnitude of vector D:

$$|D| = \sqrt{4^2 + (-8)^2 + 4^2}$$

$$= \sqrt{16 + 64 + 16} = \sqrt{96} = 4\sqrt{6}$$

Now, we divide each component by the magnitude of the vector:

$$\hat{D} = \left\langle \frac{4}{4\sqrt{6}}, ~\frac{-8}{4\sqrt{6}}, ~\frac{4}{4\sqrt{6}} \right\rangle$$

$$= \langle \frac{1}{\sqrt{6}},~ \frac{-2}{\sqrt{6}},~ \frac{1}{\sqrt{6}} \rangle$$

$$\approx \langle 0.41, ~-0.82,~ 0.41 \rangle$$

The unit vector of $latex \vec{D} = \langle 4,~ -8,~ 4 \rangle$ is $latex \hat{D} \approx \langle 0.41,~ -0.82,~ 0.41 \rangle$.

### EXAMPLE 7

What is the unit vector of the vector $latex \vec{E} = \langle -6, ~3, ~9 \rangle$?

We find the magnitude of the vector:

$$|E| = \sqrt{(-6)^2 + 3^2 + 9^2}$$

$$= \sqrt{36 + 9 + 81} = \sqrt{126} = 3\sqrt{14}$$

When we divide each component by the magnitude, we have:

$$\hat{E} = \left\langle \frac{-6}{3\sqrt{14}}, ~\frac{3}{3\sqrt{14}},~ \frac{9}{3\sqrt{14}} \right\rangle$$

$$= \langle \frac{-2}{\sqrt{14}}, ~\frac{1}{\sqrt{14}}, ~\frac{3}{\sqrt{14}} \rangle$$

$$\approx \langle -0.53,~ 0.27, ~0.80 \rangle$$

The unit vector of $latex \vec{E} = \langle -6, ~3, ~9 \rangle$ is $latex \hat{E} \approx \langle -0.53, ~0.27,~ 0.80 \rangle$.

### EXAMPLE 8

Find the unit vector of the vector $latex \vec{G} = \langle 5, ~10, ~-5 \rangle$.

The magnitude of the given vector is:

$$|G| = \sqrt{5^2 + 10^2 + (-5)^2}$$

$$= \sqrt{25 + 100 + 25} = \sqrt{150} = 5\sqrt{6}$$

By normalizing the vector, we have:

$$\hat{G} = \left\langle \frac{5}{5\sqrt{6}},~ \frac{10}{5\sqrt{6}},~ \frac{-5}{5\sqrt{6}} \right\rangle$$

$$= \langle \frac{1}{\sqrt{6}}, ~\frac{2}{\sqrt{6}},~ \frac{-1}{\sqrt{6}} \rangle$$

$$\approx \langle 0.41,~ 0.82, ~-0.41 \rangle$$

The unit vector of $latex \vec{G} = \langle 5,~ 10, ~-5 \rangle$ is $latex \hat{G} \approx \langle 0.41,~ 0.82,~ -0.41 \rangle$.

### EXAMPLE 9

If we have the vector $latex \vec{A} = \langle 2,~ 4,~ -4 \rangle$, find its unit vector.

We find the magnitude of vector A:

$$|A| = \sqrt{2^2 + 4^2 + (-4)^2}$$

$$= \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$

We divide each component of the vector by its magnitude:

$$\hat{A} = \left\langle \frac{2}{6},~ \frac{4}{6},~ \frac{-4}{6} \right\rangle$$

$$= \langle \frac{1}{3}, ~\frac{2}{3}, ~-\frac{2}{3} \rangle$$

$$\approx \langle 0.333, ~0.667, ~-0.667 \rangle$$

The unit vector of $latex \vec{A} = \langle 2, ~4,~ -4 \rangle$ is $latex \hat{A} \approx \langle 0.333, ~0.667,~ -0.667 \rangle$.

### EXAMPLE 10

Find the unit vector of the vector $latex \vec{B} = \langle -3, ~6, ~9 \rangle$.

We calculate the magnitude of the given vector:

$$|B| = \sqrt{(-3)^2 + 6^2 + 9^2}$$

$$= \sqrt{9 + 36 + 81} = \sqrt{126} = 3\sqrt{14}$$

Normalizing the vector, we have:

$$\hat{B} = \left\langle \frac{-3}{3\sqrt{14}}, ~\frac{6}{3\sqrt{14}},~ \frac{9}{3\sqrt{14}} \right\rangle$$

$$= \langle \frac{-1}{\sqrt{14}}, ~\frac{2}{\sqrt{14}},~ \frac{3}{\sqrt{14}} \rangle$$

$$\approx \langle -0.27,~ 0.53, ~0.80 \rangle$$

The unit vector of $latex \vec{B} = \langle -3, ~6, ~9 \rangle$ is $latex \hat{B} \approx \langle -0.27,~ 0.53, ~0.80 \rangle$.

## Unit vectors – Practice problems

Unit vectors quiz
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