There are two way of * multiplying vectors* – and scalar product or dot product is one of them. Scalar product of two vectors $\vec{A}$ and $\vec{B}$ refers to to as length of $\vec{A}$ times the component of $\vec{B}$ on $\vec{A},$ (or to the direction of $\vec{A}$). Sometimes we also say that the component of $\vec{B}$ is parallel to $\vec{A}$. There is another way of multiplying vectors, which is vector product, or cross product. So it refers to as length of $\vec{A}$ times component of $\vec{B}$ perpendicular to $\vec{A}$. Later this article we will individually discuss about that.

## Scalar Product or Dot Product

If a force $\vec{B}$ on a particle cause its displacement vector, $\vec{A},$ then the Work-Done by the force $\vec{B}$ represented by the * dot product* of two vectors. Such as, work-done by force $\vec{B}$ is equal to $\vec{B}.\vec{A}$.

So, the scalar product, $\vec{B}.\vec{A} = |\vec{B}|.|\vec{A}|cos θ,$ also refers to as $\vec{B}.\vec{A} =|\vec{B}|cos θ.|\vec{A}|$.

$|\vec{B}|cos θ$ is the projection of vector $\vec{B}$ on $\vec{A},$ or the component of vector $\vec{B}$. Later in a separate article we will discuss about vector projection, * here>>*. We also can say that $|\vec{B}|cos θ$ is the part of vector $\vec{B}$ is going to the same direction of $\vec{A}$. $|\vec{A}|$ is the magnitude of displacement vector $\vec{A}$. $θ$ is the angle between two vectors. We find the component of vector $\vec{B}$ on $\vec{A}$ using

*on right angle. Remember that, in vector multiplication we always have to consider angle between the vectors.*

**pythagorean theorem**Now, if we have two vectors with their components, $\vec{A}=8\hat{i}+2\hat{j},$ and $\vec{B}=2\hat{i}+(-3)\hat{j},$ the dot product of two vectors, $\vec{A}.\vec{B} = 16-6=10$. Because, $\hat{i}.\hat{i}=1,$ for $\hat{i}$ ‘s projection or component on $\hat{i}$ is equal to 1 (both are parallel, therefore $cos 0 =1$). But has no component on other coordinates (bcoz, coordinates are perpendicular on each other, therefore $cos 90 =0$), so $\hat{i}.\hat{j},$ or $\hat{i}.\hat{k}$ always equal to zero. This properties are similar for other * unit vectors* also.

As we can see that dot product omit the *unit vectors*, therefore the product is no longer a vector. It is now a number, or scalar. This is because, the dot product we also refer to as scalar product.