How to Find Angle Between Two Vectors?

A quantity that has magnitude, as well as direction, is called a vector. However, scalar quantities have a magnitude and no direction. A vector can be represented in two forms, i.e., in two and three dimensions. The smallest angle between two vectors that are pointing in different directions is called the angle between vectors. Its formula is A⋅B=∣A∣×∣B∣×cos(θ).  The angle between two vectors is only formed at the intersection of the tails of vectors. If the vectors are not joined tail-to-tail, then position will be shifted using parallel shifting. In exams like NEET and JEE Main, questions based on calculation will be asked. Students should, therefore, focus on the calculation part of the vector.

The following are the different types of angles between 2 vectors:

• Acute Angle (θ): An angle between two vectors is acute if it is greater than 0 degrees and less than 90 degrees, i.e. (0° < θ < 90°). The vectors which make an acute angle will be pointing in almost similar directions.

• Right Angle: An angle between vectors will be a right angle if it is exactly 90 degrees (θ = 90°). In this case, both vectors are perpendicular to each other.

• Obtuse Angle: An angle between 2 vectors is obtuse if it is greater than 90 degrees but less than 180 degrees (90° < θ < 180°). The vectors here are almost in the opposite direction.

• Straight Angle: An angle between vectors will be a straight angle if it is exactly 180 degrees (θ = 180°). In this case, both the vectors are pointing exactly in the opposite direction.

• Zero Angle: An angle θ between 2 vectors will be zero if it is exactly 0 degrees (θ = 0°). Here, both vectors are overlapping each other since they are in the same direction

1. Angle Between Two Vectors Using Dot Product

$$\mathbf{A}\cdot \mathbf{B}=A_x\cdot B_x+A_y\cdot B_y+A_z\cdot B_z$$

$$\parallel \mathbf{A}\parallel =\sqrt{{A_x}^2+{A_y}^2+{A_z}^2}$$

$$\parallel \mathbf{B}\parallel =\sqrt{{B_x}^2+{B_y}^2+{B_z}^2}$$

$$\cos \left(\theta \right)=\frac{\mathbf{A}\cdot \mathbf{B}}{\parallel \mathbf{A}\parallel \cdot \parallel \mathbf{B}\parallel }$$

$$\theta ={\cos }^{-1}\left(\frac{\mathbf{A}\cdot \mathbf{B}}{\parallel \mathbf{A}\parallel \cdot \parallel \mathbf{B}\parallel }\right)$$

2. Angle Between 2 Vectors Using Cross Product

This method are important for exams like IISER and IIT JAM. Therefore, do understand both of them carefully: 

Step 1: Let us consider 2 vectors A and B

$$\mathbf{A}=\left(A_x,A_y,A_z\right)$$

$$\mathbf{B}=\left(B_x,B_y,B_z\right)$$

Step 2: Let us first find the cross product of A and B which is as follows:

$$\mathbf{A}\times \mathbf{B}=\left({\mathrm{(A}}_y{B}_z-A_z{B}_y,A_z{B}_x-A_x{B}_z,A_x{B}_y-A_y{B}_{\mathrm{x)}}\right)$$

Step 3: Now, we will be calculating the magnitude of cross product vector:

$$\parallel \mathbf{A}\times \mathbf{B}\parallel =\sqrt{{\left({\mathrm{(A}}_y{B}_z-A_z{B}_{\mathrm{y)}}\right)}^2+{\left({\mathrm{(A}}_z{B}_x-A_x{B}_{\mathrm{z)}}\right)}^2+{\left({\mathrm{(A}}_x{B}_y-A_y{B}_{\mathrm{x)}}\right)}^2}$$

Step 4: The magnitude of the cross product is related to θ through the below-given formula

∣∣A×B∣∣=∣∣A∣∣⋅∣∣B∣∣⋅sin(θ)

Step 5: Now, we will rearrange the formula to solve it and get value for sin θ

$$\sin \left(\theta \right)=\frac{\parallel \mathbf{A}\times \mathbf{B}\parallel }{\parallel \mathbf{A}\parallel \cdot \parallel \mathbf{B}\parallel }$$

Step 6: Let us now take the inverse of Sin (arcsine)

$$\theta ={\sin }^{-1}\left(\frac{\parallel \mathbf{A}\times \mathbf{B}\parallel }{\parallel \mathbf{A}\parallel \cdot \parallel \mathbf{B}\parallel }\right)$$

Some of the examples of angle between two vectors are as below:

1. Calculate the angle between two vectors 3i + 4j and 2i – j + k

Solution.

→                         →

a = 3i + 4j – k and b = 2i – j + k

The dot product is defined as:

→  →

a .  b = (3i + 4j – k).(2i – j + k)

= (3)(2) + (4)(-1) + (-1)(1)

=6-4-1

=1

          →   →

Thus,  a .   b = 1

The magnitude of vectors is given as below:

   →

  │a│= √(32 + 42 + (-1)2) = √26 =   5.09

   →

  │b│= √(22 + (-1)2 + 12) = √6 =     2.45

 The angle between the two vectors is as follows:

             →   →

 = cos^-1 A.   B 

           →    →

          │A││B│

=cos^-1 1 /(5.09)(2.45)

= cos^-1 1/12.47

 = cos^-1 (0.0802)

 = 85.39°

2. Compute the angle between two vectors 5i - j + k and i + j - k

Solution.

→                       →

a = 5i - j + k and b = i + j - k

 The dot product is defined as:

→    →

  a .   b = (5i - j + k)(i + j - k)

= (5)(1) + (-1)(1) + (1)(-1)

=5-1-1

=3

         →  →

Thus, a . b    = 1

The magnitude of vectors is given as below:

    →

  │a│= √(52 + (-1)2 + 12) = √27 =   5.19

   →

 │b│= √(12 + 12 + (-1)2) = √3 =     1.73

 The angle between the two vectors is as follows:

              →   →

 =   cos^-1A.  B

             →       →          

             │A││B│

 = cos^-1 3 /(5.19)(1.73)

  = cos^-1 3 / 8.97

 = cos^-1 (0.334)

 = 70.48°