A scalar is a numeric quantity. A vector is a quantity that has a magnitude (the length) and the direction associated with it. A vector is represented by an arrow over a variable. For ease we will simply follow writing vector as a normal variable in here. In general we have a multi-dimensional plane ($x_1$, $x_2$, …, $x_n$). When we have two dimensional plane we write (x, y) and for three dimensional plane we refer (x, y, z). In general we discuss vectors on two or three dimensional planes.

Both scalar (like distance, speed) and vector (like displacement, velocity) fields are part of vector calculus. A scalar when zero does not have any significance while a zero vector will have significance. A zero vector may have magnitude zero but it will have some direction that will define something about that vector quantity.

Let us say we have u = (x, y).

Then the magnitude of u which is denoted by ||u|| or |u| =sqrt ($x^2$ + $y^2$)

Also the direction of the vector is given by $tan^{-1}$ ($\frac{y}{x}$). in case of three dimensional space the direction is associated with a unit vector in the direction of the third coordinate with n cap.

Consider u and v are two vectors, then the following operations holds true.

If m is a scalar, then m . u = mu which is a vector a again.

v + u or v – u which results in a vector again.

v . u which results in a scalar number.

v x u which yields a complete vector quantity again.

All these operations have their methods and procedures to be followed for evaluation. In addition to them we also have divergence, curl and gradient. All these are full topics in themselves which require separate space to be discussed.

Then u + v = (3 + 4, 5 - 2, -4 + 0) = (7, 3, -4)

And u – v = (3 - 4, 5 + 2, -4 - 0) = (-1, 7, -4)

Similarly we can follow various steps associated with every operations on vectors to obtain the result.

Let vector x = (-6, 8) and vector y = (5, 12)

This implies x$_1$ = -6, x$_2$ = 5 and y$_1$ = 8, y$_2$ = 12.

The dot product of two vectors is given by the sum of the product of the respective coordinates.

So let c represent the resultant product.

Then c = a . b = -6 * 5 + 8 * 12 = -30 + 96 = 66

$\vec{A}$ $\times$ $\vec{B}$ = $\begin{bmatrix}

i& j & k\\

2& -3 &1 \\

4& 5 &-1

\end{bmatrix}$

= i(3 - 5) - j(-2 - 4) + k(10 + 12)

= -2i + 6j + 22k

Given $\vec{A}$ = 3i - 6j + 4k and $\vec{B}$ = 10i − 13k

Dot product of both the vectors is

$\vec{A}$ . $\vec{B}$ = (3i - 6j + 4k).(10i − 13k)

= (3 $\times$ 10) + (-6 $\times$ 0) + (4 $\times$ -13)

= 30 + 0 - 52

= -22