Dot Product Formula

In the vector geometry of Physics, the vectors that are directional in nature and the angle that they are oriented at, are the two most important factors in deriving the vector product formula or as commonly said the dot product formula. So, if it is assumed that there are two directional vectors, say a and b, that are oriented at a specific angle as their ends are joined together making a certain angle between the two vector lines, then the angle of the vector is defined as the shortest angle made by the two vectors upon which any of the two vectors is rotated over the other so that the direction of both the vectors become same. Therefore, geometrically the dot product formula is expressed as follows:

$\vec{a} \times \vec{b} = |a||b| \cos \theta$

Where,

$|a|$ = magnitude length of vector a

$|b|$ = magnitude length of vector b

$\theta$ = angle formed by vector unit a and b

Dot Product Angle Formula

Vector multiplication is the method to find out the angle between the two vectors and the multiplication is of two types:-

1. Scalar Product

2. Cross Product

Scalar Product

The scaler product of two vectors as suggested is the multiplication product of the two vectors that give a scaler quantity whereas the cross product, as the name suggests is a quantity produced by the product of the multiplication of the two vectors.

Thus the angle between the two vectors according to the vector product formula for vectors a and b is as follows:

$\vec{a} \times \vec{b} = |a||b| \cos \theta$

The above equation is clearly a scaler formula. Therefore to find the angle formed between the two vectors the equation is rewritten as follows:

$\cos \theta = \dfrac{a \times b}{|a||b|}$

Or, $\theta = \cos^{-1} \left(\dfrac{a \times b}{|a||b|}\right)$

So, $\theta$ is the angle between the two vectors according to the scaler product.

Solved Example

1. Find out the angle between the two vectors whose magnitudes are equal to each other.

Ans: Let the two equal vector magnitude be,

$|a| = |b|$

Thus by applying the cosine law in the vector product formula, whose resultant vector is assumed as $|R|$ we get:

$|R|^2 =  |a|^2 +  |b|^2 + 2|a||b| \cos \theta$ 

Or,  $|R| = \sqrt{(|a|^2 +  |b|^2 + 2|a||b|\cdot \cos \theta)}$

Or,  $|R| = \sqrt{(|a|^2 +  |b|^2 + 2|a||a| \cdot \cos \theta)}$

Or,  $|R| = \sqrt{(|a|^2 +  |b|^2 +2|a|^2 \cdot \cos \theta)}$

Or,  $|R| = \sqrt{(2|a|^2 ( 1 + \cos \theta)}$

With the half-angle identity 

$|R| = \sqrt{4|a|^2 \cos^2 \left(\dfrac{\theta}{2}\right)}$

Or $|R| = 2a \cos \left(\dfrac{\theta}{2}\right)$

Now calculating the angle of the two vectors,

$\tan \alpha = \dfrac{2a \cos \left(\dfrac{\theta}{2}\right) \cdot \sin \left(\dfrac{\theta}{2}\right)}{2a \cos^2 \left(\dfrac{\theta}{2}\right)}$

$\tan \alpha = \tan \left(\dfrac{\theta}{2}\right)$

Thus, the vector angle is equal to $\alpha = \left(\dfrac{\theta}{2}\right)$

Cross Vectors

The cross vector is calculated for the case when one vector is perpendicular to both the vectors that can be expressed by the right-hand rule. Thus the mathematical expression for the cross vector is given as follows:-

$a \times b = |a||b|\cdot \sin \theta n$

This is also a mathematical representation of the vector product of two vector formulas when a third vector is perpendicular to both the vectors by the right-hand rule.

Where,

$|a|$ = magnitude length of vector a

$|b|$ = magnitude length of vector b

$\theta$ = angle formed by vector unit a and b

$n$ = the unit vector that is perpendicular to both the vectors

Therefore,

$\sin \theta = \dfrac{a \times b}{|a||b|}$

Solved Example

1. What will be the angle between the two vectors such that the units of the two vectors be a and b where the product of two vector units a and b are $a \times b = \dfrac{1}{3i} + \dfrac{1}{4j}$.

Ans: According to the cross vector dot product formula, $\sin \theta = \dfrac{a \times b}{|a||b|}$

And as it is given that $|a| = |b| = 1$

Therefore, $|a \times b| = \sqrt{\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{4}\right)^2} = \dfrac{5}{12}$

Now applying the cross vector dot product formula,

$|a \times b|=|a||b| \sin \theta$

$\dfrac{ 1}{ 5} = (1)(1) \sin \theta$

$\theta=\sin ^{-1}\left(\dfrac{1}{5}\right)$

$\theta=30^{\circ}$

Mathematical form of vector product formula

$a = a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k}$

$b = b_{x} \hat{i} + b_{y} \hat{j} + b_{z} \hat{k}$

$a \times b = \left(a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k}\right) \times \left(b_{x} \hat{i} + b_{y} \hat{j} + b_{z} \hat{k}\right)$

$a \times b=a_{x} \hat{i}\left(b_{x} \hat{i}+b_{y} \hat{j}+b_{z} \hat{k}\right)+a_{y} \hat{j}\left(b_{x} \hat{i}+b_{y} \hat{j}+b_{z} \hat{k}\right)+a_{z} \hat{k}\left(b_{x} \hat{i}+b_{y} \hat{j}+b_{z} \hat{k}\right)$

$a \times b=a_{x} b_{y} \hat{k}-a_{x} b_{z} \hat{j}-a_{y} b_{z} \hat{k}+a_{z} b_{x} \hat{j}-a_{z} b_{y} \hat{i}$

$a \times b=\left(a_{y} b_{z}-a_{z} b_{y}\right) \hat{i}+\left(a_{z} b_{x}-a_{x} b_{z}\right) \hat{j}+\left(a_{x} b_{y}-a_{y} b_{z}\right) \hat{k}$

So the final determination of the product of vector

$a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \end{vmatrix}$

Conclusion

The scalar product, often known as the dot product, is a method of multiplying two vectors.

The dot product is the product of the lengths of the vectors multiplied by the cosine angle between them, $\vec{a} \times \vec{b} = |a||b| \cos \theta$