Vectors

### Introduction to Vectors

A vector is a mathematical quantity that possesses both magnitude (or size) and direction. This distinguishes it from a scalar, which has only magnitude. Examples of vectors include displacement, velocity, and force, while examples of scalars include distance, speed, and mass.

Vectors can be represented in several ways:

* As a directed line segment, e.g., from point A to point B, denoted as $\vec{AB}$.

* Using bold lowercase letters, e.g., a, b.

* As a column vector, showing its components in each dimension. In 2D, $\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}$. In 3D, $\mathbf{a} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$.

* Using unit vectors i, j, and k. These are vectors of magnitude 1 in the direction of the positive x, y, and z axes, respectively. A 3D vector can be written as $\mathbf{a} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$.

### Position Vectors and Displacement

A position vector is a vector that gives the position of a point in space relative to a fixed origin, O. The position vector of a point A is denoted by $\vec{OA}$ or simply a.

To find the vector that describes the displacement from a point A to a point B, we use their position vectors a and b. The journey from A to B can be thought of as going from A to the origin O (which is -a), and then from the origin O to B (which is b).

Therefore, the displacement vector $\vec{AB}$ is given by the fundamental formula:

Process: Finding the vector between two points

$\mathbf{\vec{AB} = b - a}$

For example, if A has position vector $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}$ and B has position vector $\mathbf{b} = 5\mathbf{i} + 7\mathbf{j}$, then $\vec{AB} = (5\mathbf{i} + 7\mathbf{j}) - (2\mathbf{i} + 3\mathbf{j}) = 3\mathbf{i} + 4\mathbf{j}$.

### Magnitude of a Vector

The magnitude (or modulus) of a vector is its length. It is denoted by vertical bars, e.g., $|\mathbf{a}|$. The magnitude is calculated using Pythagoras' theorem extended to three dimensions.

Formula: Magnitude of a vector

For a vector $\mathbf{a} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, its magnitude is:

$|\mathbf{a}| = \sqrt{x^2 + y^2 + z^2}$

A unit vector is a vector with a magnitude of exactly 1. To find the unit vector in the direction of a given vector a, you divide the vector by its own magnitude.

Formula: Unit vector

$\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}$

### The Scalar (Dot) Product

The scalar product, also known as the dot product, is one way of multiplying two vectors. The result of a scalar product is a scalar number, not another vector.

There are two formulas to calculate the scalar product of vectors a and b:

Formula: Scalar product (components)

For $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$:

$\mathbf{a \cdot b} = a_1b_1 + a_2b_2 + a_3b_3$

Formula: Scalar product (geometric)

$\mathbf{a \cdot b} = |\mathbf{a}| |\mathbf{b}| \cos\theta$

where $\theta$ is the angle between the two vectors when they are placed tail-to-tail ($0^\circ \le \theta \le 180^\circ$).

An important property of the scalar product arises when two vectors are perpendicular (or orthogonal). In this case, the angle $\theta = 90^\circ$, and since $\cos(90^\circ) = 0$, their scalar product is zero.

Condition for Perpendicular Vectors: $\mathbf{a \cdot b} = 0$

### Finding the Angle Between Two Vectors

By equating the two formulas for the scalar product, we can derive a method to find the angle between any two non-zero vectors.

Process: Finding the angle between vectors

For example, to find the angle between $\mathbf{a} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and $\mathbf{b} = 3\mathbf{i} - 4\mathbf{k}$:

* $\mathbf{a \cdot b} = (2)(3) + (1)(0) + (-2)(-4) = 6 + 0 + 8 = 14$.

* $|\mathbf{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.

* $|\mathbf{b}| = \sqrt{3^2 + 0^2 + (-4)^2} = \sqrt{9+0+16} = \sqrt{25} = 5$.

* $\cos\theta = \frac{14}{3 \times 5} = \frac{14}{15}$.

* $\theta = \arccos(\frac{14}{15}) \approx 21.0^\circ$.