# Dot Product of Vectors – Examples and Practice

There are two main methods we can use to calculate the scalar product or dot product of two vectors. We can use their magnitudes and the angle between the vectors, or we can use their components.

In this article, we will look at some examples with answers of the dot product of two vectors, where we will apply the methods mentioned above.

##### PHYSICS

Relevante para

Solving examples of the dot product of vectors.

See examples

##### PHYSICS

Relevante para

Solving examples of the dot product of vectors.

See examples

## 10 Solved examples on the dot product of vectors

### EXAMPLE 1

Vectors A and B have magnitudes of 5 and 3, respectively. The angle between them is 45°. Find the dot product of these vectors.

We have the following information:

• $latex |A|=5$
• $latex |B|=3$
• $latex \theta=45^{\circ}$

To find the dot product using magnitudes and the angle, we can use the following formula:

$latex \vec{A} \cdot \vec{B} = |A| ~|B| \cos(\theta)$

$latex = 5 \times 3 \cos(45^{\circ})$

$$= 15 \left(\frac{\sqrt{2}}{2}\right)$$

$$\approx 10.61$$

### EXAMPLE 2

Vectors C and D have magnitudes of 4 and 6, respectively, and the angle between them is 120°. Find the dot product of these vectors.

We can observe the following:

• $latex |C|=4$
• $latex |D|=6$
• $latex \theta=120^{\circ}$

Use the same formula as in the previous exercise to find the dot product:

$latex \vec{C} \cdot \vec{D} = |C| ~|D| \cos(\theta)$

$latex = 4 \times 6 \cos(120^{\circ})$

$$= 24 \left(-\frac{1}{2}\right)$$

$$= -12$$

### EXAMPLE 3

Vector E has magnitude 7 and F has magnitude 2. If the angle between them is 90°, find the dot product of these vectors.

We can observe the following:

• $latex |E|=7$
• $latex |F|=2$
• $latex \theta=90^{\circ}$

Since the angle between vectors E and F is 90°, cos(90°) = 0, so the dot product will be zero:

$latex \vec{E} \cdot \vec{F} = |E| ~|F| \cos(\theta)$

$latex = 7 \times 2 \cos(90^{\circ})$

$$= 14 \left(0\right)$$

$$= 0$$

### EXAMPLE 4

Find the dot product of the vectors $latex \vec{A} = \langle 2, 3, 1\rangle$ and $latex \vec{B} = \langle -1, 4, 2\rangle$.

The scalar product (dot product) of two vectors A and B can be found with the following formula:

$latex \vec{A} \cdot \vec{B} = A_{x} \times B_{x} + A_{y} \times B_{y} + A_z \times B_{z}$

where $latex A_{x},~ A_{y},~ A_{z}$ are the components of vector $latex \vec{A}$, and $latex B_{x},~ B_{y},~ B_{z}$ are the components of vector $latex \vec{B}$.

Then, the dot product of $latex \vec{A}$ and $latex \vec{B}$ is:

$latex \vec{A} \cdot \vec{B} = (2 \times -1) + (3 \times 4) + (1 \times 2)$

$latex = -2 + 12 + 2 = 12$

### EXAMPLE 5

Find the dot product of vectors $latex \vec{A} = \langle 5, ~4, ~2\rangle$ and $latex \vec{B} = \langle -3,~ 0,~ 3\rangle$.

We apply the same formula as in the previous example. Then, we have:

$latex \vec{A} \cdot \vec{B} = A_{x} \times B_{x} + A_{y} \times B_{y} + A_z \times B_{z}$

$latex \vec{A} \cdot \vec{B} = (5 \times -3) + (4 \times 0) + (2 \times 3)$

$latex = -15 + 0 + 6$

$latex = -9$

### EXAMPLE 6

We have the vectors $latex \vec{C} = \langle 4,~ -5,~ 6\rangle$ and $latex \vec{D} = \langle 1,~ 2, ~3\rangle$.

We apply the dot product formula of two vectors using the given components:

$latex \vec{C} \cdot \vec{D} = C_{x} \times D_{x} + C_{y} \times D_{y} + C_z \times D_{z}$

$latex C \cdot D = (4 \times 1) + (-5 \times 2) + (6 \times 3)$

$latex = 4 – 10 + 18$

$latex = 12$

### EXAMPLE 7

Find the dot product of vectors $latex \vec{E} = \langle -3,~ 7,~ 1\rangle$ and $latex \vec{F} = \langle 5, ~2,~ 4\rangle$.

To find the scalar product of two vectors $latex \vec{E}$ and $latex \vec{F}$ using their components, we can use the following formula:

$latex \vec{E} \cdot \vec{F} = E_{x} \times F_{x} + E_{y} \times F_{y} + E_z \times F_{z}$

$latex \vec{E} \cdot \vec{F} = (-3 \times 5) + (7 \times 2) + (1 \times 4)$

$latex = -15 + 14 + 4$

$latex = 3$

### EXAMPLE 8

If we have the vectors $latex \vec{G} = \langle 3,~ -2,~ 4\rangle$ and $latex \vec{H} = \langle 1,~ 2,~ -2\rangle$, find their scalar product.

We can find the scalar product of the vectors as follows:

$latex \vec{G} \cdot \vec{H} = G_{x} \times H_{x} + G_{y} \times H_{y} + G_z \times H_{z}$

$latex \vec{G} \cdot \vec{H} = (3 \times 1) + (-2 \times 2) + (4 \times -2)$

$latex = 3 – 4 – 8$

$latex = -9$

### EXAMPLE 9

Find the dot product of vectors $latex \vec{I} = \langle 7, ~1,~ -3\rangle$ and $latex \vec{J} =\langle -2,~ 6,~ 5\rangle$.

We apply the scalar product formula of two vectors using the components:

$latex \vec{I} \cdot \vec{J} = I_{x} \times J_{x} + I_{y} \times J_{y} + I_z \times J_{z}$

$latex \vec{I} \cdot \vec{J} = (7 \times -2) + (1 \times 6) + (-3 \times 5)$

$latex = -14 + 6 – 15$

$latex = -23$

### EXAMPLE 10

Determine the angle between the vectors $latex \vec{C} = \langle1,~ 1,~ 0\rangle$ and $latex \vec{D} = \langle 0, ~-1,~ 1\rangle$.

To find the angle θ between two vectors C and D, we can use the following formula:

$$\cos(\theta) = \frac{\vec{C} \cdot \vec{D}}{ |C|~|D|}$$

where $latex \vec{C} \cdot \vec{D}$ is the dot product of $latex \vec{C}$ and $latex \vec{D}$, and |C| and |D| are the magnitudes of the vectors.

First, we find the dot product:

$latex \vec{C} \cdot \vec{D} = (1 \times 0) + (1 \times -1) + (0 \times 1) = -1$

Now, we find the magnitudes:

$latex |C| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$

$latex |D| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$

Then, we calculate cos(θ):

$$\cos(\theta) = \frac{-1}{\sqrt{2} \times \sqrt{2}}$$

$$= -\frac{1}{2}$$

Finally, we find the angle θ:

$latex \theta = \text{arccos}(-\frac{1}{2}) \approx 120^{\circ}$

## Dot product of vectors – Practice problems

Dot product of vectors quiz  You have completed the quiz!

Interested in learning more about vectors? You can look at these pages: ### Jefferson Huera Guzman

Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.  