- Home
- >
- Angle between the two Vectors – Explanation and Examples

**Angle between the two Vectors **– Explanation and Examples

Vectors, specifically the direction of vectors and the angles they are oriented at, have significant importance in vector geometry and physics. If there are two vectors, let’s say **a **and **b** in a plane such that the tails of both vectors are joined, then there exists some angle between them, and that **angle between the two vectors** is defined as:

* “**Angle between two vectors is the shortest angle at which any of the two vectors is rotated about the other vector such that both of the vectors have the same direction.”*

Furthermore, this discussion focuses on finding the angle between two standard vectors, which means their origin is at (0, 0) in the x-y plane.

In this topic, we shall briefly discuss the following points:

- What is the angle between two vectors?
- How to find out the angle between two vectors?
- The angle between two 2-D vectors.
- The angle between two 3-D vectors.
- Examples.
- Problems.

**Angle Between Two Vectors**

*Vectors are oriented in different directions while forming different angles. This angle exists between two vectors and is responsible for specifying the erection of vectors. *

The angle between two vectors can be found using vector multiplication. There are two types of vector multiplication, i.e., scalar product and cross product**. **

The scalar product is the product or the multiplication of two vectors such that they yield a scalar quantity. As the name suggests, vector product or cross product produces a vector quantity due to the two vectors’ product or multiplication.

For instance, if we talk about the tennis ball’s motion, its position is described by a position vector and movement by a velocity vector whose length indicates the ball’s speed. The direction of the vector explains the direction of motion. Similarly, the ball’s momentum is also an example of a vector quantity that is mass times the velocity.

Sometimes we have to deal with two vectors acting on some object, so the vectors’ angle is critical. In the real world, any working system combines several vectors linked to each other and makes some angles with each other in the given plane. Vectors can be two-dimensional or 3-dimensional. Therefore, it’s necessary to calculate the angle between the vectors.

Let’s first discuss scalar products.

**Angle Between Two Vectors Using Dot Product**

Consider two vectors **a** and **b **separated by some angle θ. Then according to the formula of the dot product is:

**a.b** = |a| |b|.cosθ

where **a.b** is the dot product of two vectors. |a| and |b| is the magnitude of vectors **a **and **b,** and θ is the angle between them.

To find the angle between two vectors, we will start with the formula of the dot product that gives the cosine of angle θ.

According to the formula of the scalar product,

**a.b** = |a| |b|.cosθ

This states that the dot product of two vectors a and b is equal to the magnitude of two vectors a and b multiplied by the cosine of the angle. To find the angle between two vectors, a and b, we will solve the angle θ,

cosθ = **a.b **/ |a|. |b|

θ = arccos ( **a.b** / |a| . |b| )

So, θ is the angle between two vectors.

If vector **a **= < ax , ay > and **b** = < bx, by >,

Then the dot product between two vectors **a **and **b** is given as,

**a.b ** = <ax, ay > . < bx, by >

**a.b** = ax.bx + ay.by

Here, we can have an example of work done as the work done is defined as the force applied to move an object at some distance. Both force and displacement are vectors, and their dot product yield a scalar quantity, i.e**.,** work. Work done is the dot product of force and displacement, which can be defined as,

**F . d ** = |F| |d| cos (θ)

Where** θ** is the angle between force and displacement. For instance, if we consider a car moving on the road, covering some distance in a certain direction, a force is acting on the car, whereas force is making some angle θ with displacement.

Following are some properties of the dot product:

- The dot product is commutative in nature.
- It is distributive in nature over vector addition:

a . ( b + c) = ( a . b ) + ( a . c )

- It is non associative in nature.
- 4. A scalar quantity can be multiplied with the dot product of two vectors.

c . ( a . b ) = ( c a ) . b = a . ( c b)

- The dot product is maximum when two non-zero vectors are parallel to each other.
- 6. Two vectors are perpendicular to each other if and only if a . b = 0 as dot product is the cosine of the angle between two vectors a and b and cos ( 90 ) = 0.
- For unit vectors

i . i = 1

j . j = 1

k . k = 1

- Dot multiplication does not follow the cancellation law

a . b = a . c

a . ( b – c ) = 0

Similarly, we can also use cross products for this purpose.

The formula for the cross product is as follows:

a x b = |a|.|b|.sinθ. **n**

Let’s first evaluate the angle between the two vectors by using the dot product.

*Example 1*

Find out the angle between two vectors having equal magnitude, and the magnitude of their resultant vector is equivalent to the magnitude of any one of the given vectors.

Solution

Let’s consider two vectors, **A ** and ** B, **and the resultant of two vectors is **R**.

Hence, according to the condition given in the question:

|A| = |B| = |R|

Now, according to the law of cosines,

|R|^2 = |A|^2 + |B|^2 + 2|A||B| . cos (θ)

Since, |A| = |B| = |R|

|A|^2 = |A|^2 + |A|^2 + 2|A||A| . cos (θ)

|A|^2 = |A|^2 + |A|^2 + |A|^2 . cos (θ)

|A|^2 = 2|A|^2 + |A|^2 . cos (θ)

|A|^2 = 2|A|^2 ( 1 + cos (θ) )

|A|^2 / 2|A|^2 = ( 1 + cos (θ) )

1 / 2 = 1 + cos (θ)

1 / 2 – 1 = cos (θ)

-1 / 2 = cos (θ)

θ = cos-1 ( -1 / 2 )

θ = 120º

So, the angle between two vectors having equal magnitude is equal to 120º.

*Example 2*

Find the angle between two vectors having equal magnitude. Also, calculate the magnitude of the resultant vector.

Solution

It is given that,

|A| = |B|

Using the law of cosine to calculate the magnitude of the resultant vector **R**.

|R|^2 = |A|^2 + |B|^2 + 2|A||B| . cos (θ)

|R| = √( |A|^2 + |B|^2 + 2|A||B| . cos (θ))

|R| = √|A|^2 + |A|^2 + 2|A||A| . cos (θ)

|R| = √ ( 2|A|^2 + 2|A|^2 . cos (θ) )

|R| = √ ( 2|A|^2 ( 1 + cos (θ)) )

Applying half angle identity,

|R| = √ (4A^2 cos^2 ( θ / 2))

|R| = 2 A cos ( θ / 2 )

Now, for calculating the resultant angle α that it will make with the first vector,

tan α = ( A sin θ ) / ( A + A cos θ )

tan α = (2 A cos (θ / 2) . sin (θ / 2) / ( 2 A cos2 (θ / 2))

tan α = tan (θ / 2)

α = θ / 2

Hence, this shows that the resultant will bisect the angle between the two vectors having equal magnitude.

*Example 3*

Find out the angle between the given two vectors.

**A** = 6**i **+ 5**j** + 7**k**

**B **= 3**i **+ 8**j** + 2**k**

Solution

Use the formula of the dot product,

**A . B **= |A| |B| . cos (θ)

Find out the magnitude of **A** and **B.**

So, the magnitude of **A **is given as,

|A| = √ ( (6)^2 + (5)^2 + (7)^2 )

|A| = √ ( 36 + 25 + 49 )

|A| = √ ( 110 )

The magnitude of **B** is given as,

|B| = √ ( (3)^2 + (8)^2 + (2)^2 )

|B| = √ ( 9 + 64 + 4 )

|B| = √ ( 77 )

Now, finding the dot product,

**A.B** = ( 6**i **+ 5**j** +7**k **) . ( 3**i **+ 8**j** + 2**k** )

**A.B **= 18 + 40 + 14

**A.B** = 72

Putting in the formula of dot product,

72 = (√(110)) . (√(77)) . cos (θ)

72 / (√ ( 110 x 77 )) = cos (θ)

cos (θ) = 0.78

θ = cos-1 (0.78)

θ = 51.26º

*Example 4*

Find out the angle between the given two vectors

**A** = < 4, 3, 2 >

**B **= < 1, 2, 5 >

Solution

Use the formula of the dot product,

**A . B **= |A| |B| . cos (θ)

Find out the magnitude of **A** and **B.**

So, the magnitude of **A **is given as,

|A| = √ ( (4)^2 + (3)^2 + (2)^2 )

|A| = √ ( 16 + 9 + 4 )

|A| = √ ( 29 )

The magnitude of **B** is given as,

|B| = √ ( (1)^2 + (2)^2 + (5)^2 )

|B| = √ ( 1 + 4 + 25 )

|B| = √ ( 30 )

Now, finding the dot product,

**A.B** = <4, 3, 2> . <1, 2, 5>

**A.B **= 4 + 6 + 10

**A.B** = 20

Putting in the formula of the dot product,

20 = (√(29)) . (√(30)) . cos (θ)

20 / (√ (29 x 30)) = cos (θ)

cos (θ) = 0.677

θ = cos-1 (0.677)

θ = 42.60º

**Angle Between Two Vectors Using Cross Product**

**Another method of finding the angle between two vectors is the cross product. Cross product is defined as:**

*“The vector that is perpendicular to both the vectors and direction is given by the right-hand rule. *

So, the cross product is represented mathematically as,

**a x b** = |a| |b| . sin (θ) **n**

Where **θ** is the angle between two vectors, |a| and |b| are the magnitudes of two vectors **a **and **b,** and **n** is the unit vector perpendicular to the plane containing two vectors **a ** and **b** in the direction that is given by the right-hand rule.

Consider two vectors **a **and **b **whose tails are joined together and hence make some angle θ. In order to find the angle between two vectors, we will manipulate the above-mentioned formula of the cross product.

(** a x b** ) / ( |a| . |b| ) = sin (θ)

If the given vectors **a **and **b** are parallel to each other then according to the above-mentioned formula the cross product will be zero as sin (0) = 0. While dealing with the cross product we have to be careful with the directions.

Following are some properties of the cross product:

- Cross product is anticommutative in nature.
- The self cross product of the vectors is equal to zero.

**A** x **A** = 0

- Cross product is distributive over vector addition

** a **x ( **b + c) **= ( **a **x** b **) + ( **a **x **c **)

- It is non associative in nature.
- A scalar quantity can be multiplied with the dot product of two vectors.

c . ( **a **x** b **) = ( c **a** ) x **b = a** x ( c **b** )** **

- Dot product is maximum when two non-zero vectors are perpendicular to each other.
- Two vectors are parallel ( i.e. if angle between two vectors is 0 or 180 ) to each other if and only if
**a x b**= 1 as cross product is the sine of angle between two vectors**a**and**b**and sine ( 0 ) = 0 or sine (180) = 0. - For unit vectors

**i x i **= 0

**j x j **= 0

**k x k** = 0

**i x j **= **k**

**j x k** = **i**

**k x i **= **j**

- Cross multiplication does not follow the cancellation law

**a x b **= **a x c**

**a x **( **b – c** ) = 0

These are some of the properties of cross product.

Let’s solve some examples to comprehend this concept.

*Example 5*

Calculate the angle between two vectors such that they are unit vectors **a **and **b ** where **a **x **b** = 1 / 3**i** + 1 / 4**j**.

Solution

Since, its given,

|a| = |b| = 1

Where as,

| a x b | = √ ( (1 / 3)^2 + ( 1 / 4)^2) = 1 / 5

Now, putting into the formula,

| a x b | = |a| |b| sin θ

1 / 5 = (1) (1) sin θ

θ = sin-1 (1/ 5)

θ = 30º

*Example 6*

Calculate the angle between two vectors such that **a **= 3**i **– 2**j** – 5**k** ** **and **b **= **i** + 4**j** – 4**k ** where **a **x **b** = 28**i** + 7**j **+ 14**k**.

Solution

So, the **magnitude** of vector** a **is given as,

|a| = √( (3)^2 + (-2)^2 + (-5)^2)

|a| = √( 9 + 4 + 25)

|a| = √(38)

Magnitude of vector **b** is given as,

|b| = √( (1)^2 + (4)^2 + (-4)^2)

|b| = √( 1 + 16 + 16)

|b| = √(33)

Whereas, magnitude of **a x b **is given as,

| a x b | = √ ( (28)2 + (7)2 + (14) )

| a x b | = √(1029)

| a x b | = 32.08

Now, putting into the formula,

| a x b | = |a| |b| sin θ

32.08 = (√ (38)) (√(33)) sin θ

sin θ = 32.08 / (√ (38)) (√(33))

θ = 64.94º

So, the **angle** between two vectors **a ** and **b ** is θ = 64.94º .

Vectors can be both two dimensional as well as three dimensional. The method of finding the angle is the same in both cases. The only difference is that the 2-D vector has two coordinates x and y whereas the 3-D vector has three coordinates x, y, and z. The examples solved above use both 2-D as well as 3-D vectors.

**Practice Problems**

**Practice Problems**

- Given that |A| = 3 and |B| = 5 where as
**a . b**= 7.5, find out the angle between two vectors. - Compute the angle between two vectors 3i + 4j – k and 2i – j + k.
- Calculate the angle between two vectors such that
**a**= 2**i**– 3**j**+ 1**k****b**= -1**i**+ 0**j**+ 5**k**where**a**x**b**= -15**i**– 11**j**– 3**k**. - Calculate the angle between two vectors such that
**a**= 2**i**+ 3**j**+ 5**k****b**=**i**+ 6**j**– 4**k**where**a**.**b**= 0. - Find the angle between given vectors
**t**= (3, 4) and**r**= (−1, 6). - What will be the resultant vector
**R**of the two vectors**A**and**B**having the same magnitude if the angle between them is 90o.

**Answers**

**Answers**

- 60°
- 85.40°
- 81.36°
- 90°
- 36.30°
- 90°

**Previous Lesson | Main Page | Next Lesson**

**5**/

**5**(

**11**votes )