Angle Between Two Vector Products

Bookmark added to your notes.
View Notes

What are Vectors?

Before we get to know the angle between two vectors, let us first understand what a vector is. A vector quantity has a magnitude and a direction as well, unlike a scalar quantity which only has a magnitude. It is denoted by an arrow (→). The length of the arrow represents its magnitude and the direction of the arrow represents the direction of the vector. Suppose, ‘a’ is a vector quantity, then it will be written as \[\overrightarrow{a}\]. [|\overrightarrow{a}|\] denotes the magnitude of a vector. Acceleration, velocity, displacement, force, and momentum are all examples of vector quantity as they have both magnitude and direction.

(Image will be uploaded soon)

Angle Between Two Vectors

Two vectors are said to be equal when their magnitude and direction is the same. However, when the direction of the two vectors is unequal, they will form an angle between them. The angle between the two vectors is denoted by θ.

(Image will be uploaded soon)

Angle Between Two Vectors Using Dot Product

The dot product formula of two vectors ‘\[\overrightarrow{a}\]’ and ‘\[\overrightarrow{b}\]’ is:

\[\overrightarrow{a}\] \[\cdot\] \[\overrightarrow{b}\] = \[|\overrightarrow{a}|\] \[|\overrightarrow{b}|\] cos θ, where \[|\overrightarrow{a}|\] and \[|\overrightarrow{b}|\] are the magnitude of \[\vec{a}\] and \[\vec{b}\] and is the angle between ‘a’ and ‘b’.

The angle (θ) between two vectors can be found using this formula:

\[Cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\]

Or, \[\theta = cos^{-1} \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\]

Now, to find out the formula of the magnitude of the vector using Pythagoras theorem:

\[|\vec{a}| = \sqrt{u_{1}^{2} + u_{2}^{2}}\], where u1 and u2 are the points of \[\vec{a}\] on the x-axis and y-axis respectively in a 2-dimensional graph.

\[|\vec{b}| = \sqrt{v_{1}^{2} + v_{2}^{2}}\], where v1and v2 are the points of \[\vec{b}\] on the x-axis and y-axis respectively in a 2-dimensional graph.

Therefore, the formula becomes:

\[\theta = cos^{-1} \frac{[\vec{a} \cdot \vec{b}}{\sqrt{u_{1}^{2} + u_{2}^{2}} \sqrt{v_{1}^{2} + v_{2}^{2}}]}\]

Now, \[\vec{a} \cdot \vec{b} = u_{1} \cdot v_{1} + u_{2} \cdot v_{2}\]

Hence, the final formula is:

\[\theta = cos^{-1} \frac{[u_{1} \cdot v_{1} + u_{2} \cdot v_{2}}{\sqrt{u_{1}^{2} + u_{2}^{2}} \sqrt{v_{1}^{2} + v_{2}}]}\]

Angle Between Two Vectors Using Cross Product

The formula of the angle between two vectors using the cross product is as follows:

\[\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| sin \theta \widehat{n}\], where,

\[\widehat{n}\] denotes the unit vector that shows the direction of the multiplication of two vectors.

Solved Examples

1. Compute the angle between two vectors using dot product:-

\[\vec{a} = 2 \widehat{i} + 2\widehat{j} + 2\widehat{k}\]

\[\vec{b} = 3\widehat{i} + 3\widehat{j} + 3\widehat{k}\]

Answer: \[\widehat{i}, \widehat{j}\] and \[\widehat{k}\] are called unit vectors.

The dot product of unit vectors are:

\[\widehat{i} \cdot \widehat{i}  = 1\]

\[\widehat{j} \cdot \widehat{j}  = 1\]

\[\widehat{k} \cdot \widehat{k}  = 1\]

\[\widehat{i} \cdot \widehat{j}  =0\]

\[\widehat{j} \cdot \widehat{k}  = 0\]

\[\widehat{k} \cdot \widehat{i}  = 0\]

Therefore, \[\vec{a} \cdot \vec{b} = (2 \widehat{i} + 2\widehat{j} + 2\widehat{k}) \cdot 3 \widehat{i} + 3\widehat{j} + 3\widehat{k}\]

= (2)(3) + (2)(3) + (2)(3)

            = 6 + 6 + 6

            = 18

Thus, \[\vec{a} \cdot \vec{b} = 18\].

The magnitude of \[\vec{a}\] and \[\vec{b}\] are:

\[|\vec{a} = \sqrt{(2)^{2} + (2)^{2} + (2)^{2}}\]

\[= \sqrt{4 + 4 + 4}\]

\[= \sqrt{12} = 3.46\]

\[|\vec{b} = \sqrt{(3)^{2} + (3)^{2} + (3)^{2}}\]

\[= \sqrt{9 + 9 + 9}\]

\[= \sqrt{27} = 5.19\]

We know that, \[cos \theta = \vec{a} \cdot \vec{b}/|\vec{a}| |\vec{b}|\]

                  Or, cos θ = 18/ 3.46* 5.19

                   Or, cos θ = 18/ 17.95

                   Or, cos θ = 1.002

                    Or, θ = cos-1 (1.002) where, 1.002 ≈ 1

                    Or, θ = cos-1(1)

                     Or, θ = 0°

2. The value of \[|\vec{a}| = 2, |\vec{b}| = 3\] and \[\vec{a} \cdot \vec{b} = 0\]. Compute the angle between \[\vec{a}\] and \[\vec{b}\] .

Answer: Given,  \[cos \theta = \vec{a} \cdot \vec{b}/|\vec{a}| |\vec{b}|\]

Here, \[|\vec{a}| = 2, |\vec{b}| = 3\] and  \[\vec{a} \cdot \vec{b} = 0\].

Therefore, cos θ = 0/ 2・3

               Or, cos θ = 0/6

               Or, cos θ = 0

                Or, θ = cos-1(0)

                 Or, θ =  90°

Hence, \[\vec{a}\] and \[\vec{b}\] are perpendicular to each other since the angle between them is 90°.

FAQ (Frequently Asked Questions)

1. What are the Types of Vectors?

Answer: The major types of vectors are as follows:-

  • Zero Vector - A zero vector is a vector whose magnitude is zero. Hence, the starting point of the vector coincides with the endpoint of the vector since the length of the vector is zero.

  • Unit Vector - A unit vector has a unit length and its magnitude is one. 

  • Collinear Vectors - Collinear vectors are also known as parallel vectors and they lie in the same line or parallel lines.

  • Equal Vectors - Equal vectors are those vectors whose magnitude and direction is the same.

  • Negative Vectors - The negative vectors are vectors that are of the same magnitude but have different directions.

2. What are the Real-Life Applications of Vectors?

Answer: The real-life applications of vectors are:

  • The application of vectors is found in the domain of Engineering.

  • Vectors can be used to find out the direction and magnitude of the force applied to move an object.

  • Vectors are used in various oscillators.

  • It is used to define the force that is applied to the body in the three dimensions simultaneously.

  • Vectors can be used to obtain the motion of a body in a plane.

  • It can be applied to find out the velocity of water in a pipe.

  • It can be applied to understand the mechanism of gravity that works on an object.