Introduction to Vectors
Introduction

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 \( \theta \)
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.
More References and links
- Formulas for Vectors
- Addition and Subtraction of Vectors (with examples)
- Scalar Product of Vectors with questions (some of which may be challenging) that explains the application of the scalar product.
- Parallel and Perpendicular Vectors with questions (some of which may be challenging) and detailed solutions.
- Vector Direction and Bearing With examples, applications and questions with solutions.
- Free SAT II Physics Practice Questions Vectors with detailed solutions and explanations