Dot Product Of Two Vectors

Dot Product of two vectors:

There are various mathematical operations possible on vectors. We have already seen addition and subtraction of vectors. In the upcoming discussion, we will see how to achieve multiplication by using dot product of two vectors

Vectors are multiplied using two methods

• scalar product of vectors or dot product
• vector product of vectors or cross product

The difference between both the methods is just that, using the first method we get a scalar value as resultant and using the second technique the resultant obtained is also a vector in nature.

Let us discuss scalar product of two vectors.

Dot product of two vectors:

The scalar product of two vectors a and b of magnitude |a|

and |b| is given as |a||b| cosθ, where θ represents the angle between the vectors a and b taken in the direction of the vectors.
We can express the scalar product as:

a.b=|a||b| cosθ

where |a| and |b| represent the magnitude of the vectors a and b, cosθ denotes the cosine of the angle between both the vectors and a.b indicate the dot product of the two vectors.

In case where any of the vectors is zero the angle θ is not defined and in such a scenario a.b is given as zero.

Projection of Vectors:

BP is known to be the projection of vector a on vector b in the director of vector b given by |a| cosθ.

Similarly, the projection of vector b on vector a in the director of vector a is given by |b| cosθ.

Projection of vector a in direction of vector b is expressed as

$BP = \frac{a.b}{|b|}$

$\Rightarrow \overrightarrow{BP} = \frac{a.b}{|b|} × \hat{b}$

$\Rightarrow \overrightarrow{BP}$ =$\frac{a.b}{|b|}.\frac{b}{|b|}$

$\Rightarrow \overrightarrow{BP}$ =$\frac{a.b}{|b|^2}b$

Similarly projection of vector b in direction of vector a is expressed as

$BQ$ = $\frac{a.b}{|a|}$

$\Rightarrow \overrightarrow{BQ}$ = $\frac{a.b}{|a|} × \hat{a}$

$\Rightarrow \overrightarrow{BQ}$ = $\frac{a.b}{|a|} \frac{a}{|a|}$

$\Rightarrow \overrightarrow{BQ}$ = $\frac{a.b}{|a|^2}a$

Thus, we see that the dot product of two vectors is the product of magnitude of one vector with the resolved component of the other in the direction of the first vector.

Properties of dot product of vector:

i.   Dot product of two vectors is commutative i.e. a.b = b.a = ab cosθ.

ii.  If a.b = 0 then it can be clearly seen that either

b or a is zero or $cos\theta=0$$\Rightarrow \theta$ = $\frac{\pi}{2}$. It suggests that either of the vectors is zero or they are perpendicular to each other.

iii. Also we know that using scalar product of vectors
(pa).(qb)=(pb).(qa)=pqa.b

iv. The dot product of a vector to itself is the magnitude squared of the vector i.e. a.a=a.a  cos0 = $a^2$

v. The dot product follows the distributive law also i.e. a.(b+c) = a.b + a.c

vi. In terms of orthogonal co-ordinates for mutually perpendicular vectors it is seen that $\hat{i}.\hat{i}$ = $\hat{j}.\hat{j}$= $\hat{k}.\hat{k}$ =$1$

vii. In terms of unit vectors if $a$= $a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $b = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$then

$a.b$=$(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}).(b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k})$

$\Rightarrow a_1b_1 + a_2b_2 + a_3b_3$ = $ab cos\theta$<

Example: Let there be two vectors [6,2,-1] and [5,-8,2]. Find the dot product of the vectors.

Solution: Given vectors: [6,2,-1] and [5,-8,2] be a and b respectively.

a.b=(6)(5)+(2)(-8)+(-1)(2)

a.b=30-16-2

a.b=12

Example: Let there be two vectors |a|=4 and |b|=2 and ϴ=60. Find their dot product.

Solution: a.b=|a||b|cos 60

a.b=4.2cos60

a.b=4

Now we are thorough with the concept of dot product of two vectors. To know more about vectors and their vector product visit our page.

Practise This Question

If f(a) = 2, f’(a) = 1, g(a) = –1, g’(a) = 2, then the value of limxag(x)f(a)g(a)f(x)xa is