Angle Between Two Vector Products in Vector Algebra

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.

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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.

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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°.