# Introduction to Vectors

## Introduction Figure 1. Vector in 2 dimensions. A vector is a mathematical concept to quantify, or describes mathematically, quantities that have a magnitude , and a direction . Forces , velocity and displacement are examples of quantities that have magnitude and direction and may therefore be described using vectors.
Vectors in one dimension describe quantities along one direction and its opposite. Vectors in two dimensions describe quantities in a plane, and vectors in three dimensions describe quantities in 3 dimensional space.

## 2-dimensional Vectors

   A two dimensional vector is used to represent a quantity in a plane. On the right, is shown vector A in two dimensions with components $$A_x$$ and $$A_y$$ that may be written as
$\vec{A} = A_x \vec{i} + A_y \vec{j}$ or $\vec{A} = < A_x \; , \; A_y >$
The magnitude of vector $$\vec{A}$$ written as $$|\vec{A}|$$ is given by
$|\vec{A}| = \sqrt{A_x^2 + A_y^2}$

## 3-dimensional Vectors

A 3 dimensional vector is used to represent a quantity in a 3 dimensional space. Below, is shown vector $$\vec{A}$$ in three dimensions with components $$A_x$$, $$A_y$$ and $$A_z$$ that may written as $\vec{A} = A_x \vec{i} + A_y \vec{j} + A_z \vec{k}$ or $\vec{A} = < A_x \; , \; A_y ; , \; A_z >$

The magnitude of vector $$\vec{A}$$ written as $$|\vec{A}|$$ is given by
$|\vec{A}| = \sqrt{A_x^2 + A_y^2+A_z^2}$

## Magnitude and Direction of 2-dimensional Vector

Referring to figure 1 above, $$\theta$$ , the angle between the vector and the positive x-axis direction, in counterclockwise direction, is called the direction of the vector. The relationships between $$\theta$$, $$A_x = |\vec{A}| \cos \theta$$ and the components $$A_x$$ and $$A_y$$ of vector $$\vec{A}$$ are:
$$A_x = |\vec{A}| \cos \theta$$
$$A_y = |\vec{A}| \sin \theta$$
$$\tan \theta = \dfrac {A_y}{A_x}$$

Example 1
The magnitude of a 2-dimensional vector is 10 and its direction $$\theta = 135^{\circ}$$. Find its components $$A_x$$ and $$A_y$$.
Solution
$$A_x = |\vec{A}| \cos \theta = 10 \cos 135^{\circ} = -5 \sqrt{2}$$
$$A_y = |\vec{A}| \sin \theta = 10 \sin 135^{\circ} = 5 \sqrt{2}$$

Example 2
Find the magnitude and direction of the vector $$\vec{B} = 2 \vec{i} - 2 \sqrt{3} \vec{j}$$
Solution
Magnitude
$$|\vec{B}| = \sqrt {2^2 + (- 2 \sqrt{3})^2 = 4 }$$
Direction $$\theta$$

$$\tan(\theta)= \dfrac{A_y}{A_x} = \dfrac{- 2 \sqrt{3}}{2} = - \sqrt{3}$$
Ignoring the sign, the reference angle to angle $$\theta$$ is equal to $$\arctan(\sqrt{3}) = 60^{\circ}$$
Since the x-component of the given vector is positive and its y-component is negative, the terminal side of angle $$\theta$$ is in quadrant IV, hence
$$\theta = 360^{\circ} - 60^{\circ} = 300^{\circ}$$
Vector B makes an angle of $$300^{\circ}$$ with the positive x-axis in counterclockwise direction.