notes/uni/mmme/1026_maths_for_engineering/vectors.md

---
author: Akbar Rahman
date: \today
title: MMME1026 // Vectors
tags:
- uni
- nottingham
- mechanical
- engineering
- mmme1026
- maths
- vectors
uuid: ef1669d9-b8c3-4813-8b33-5764a2f57a55
---

Vectors have a *magnitude* (size) and *direction*.
Examples of vectors include force, velocity, and acceleration.

In type, vectors are notated in **bold**: $\pmb{a}$.
In handwriting is is \underline{underlined}.

# Vector Algebra

## Equality

Two vectors are said to be equal if their magnitudes and directions are equal.

You can also do this by checking if their vertical and horizontal components are equal.

## Addition

Two vectors, $\pmb{a}$ and $\pmb{b}$, can be summed together by summing their components.

You can also do this graphically by drawing $\pmb{a}$ and then $\pmb{b}$ by putting its tail on
the tip of $\pmb{a}$.
The sum of the vectors is from the tail of $\pmb{a}$ to the tip of $\pmb{b}$:

![](./images/vimscrot-2022-02-18T16:34:17,290235454+00:00.png)

Vector addition is associative[^d_associative] and commutative[^d_commutative].

## Zero Vector

The *zero vector* is denoted by $\pmb{0}$ and has zero magnitude and arbitrary direction.

$$\pmb{a} + \pmb 0 = \pmb a$$

If $\pmb a + \pmb b = 0$ then it is normal to write


$$\pmb b = -\pmb a$$

$-\pmb a$ is a vector with the same magnitude to $\pmb a$ but opposite direction.

## Multiplication

### Multiplication by a Scalar

Let $k$, an arbitrary scalar and $\pmb a$, an arbitrary vector.

- $k\pmb a$ is a vector of magnitude $|k|$ times that of $\pmb a$ and is parralel to it
- $0\pmb a = 0$
- $1\pmb a = a$
- $(-k)\pmb a = -(k\pmb a)$
- $(-1)\pmb a = -\pmb a$
- $k(\pmb a + \pmb b) = k\pmb a + k\pmb b$
- $(k_1 + k_2)\pmb a = k_1\pmb a + k_2\pmb a$
- $(k_1k_2)\pmb a = k_1(k_2\pmb a)$

### The Scalar Product (Inner Product, Dot Product)

The scalar product of two vectors $\pmb a$ and $\pmb b$ is a scalar defined by

$$\pmb a \cdot \pmb b = |\pmb a||\pmb b|\cos\theta =  a_1b_1 + a_2b_2 + a_3b_3$$

where $\pmb a = (a_1, a_2, a_3)$ and $\pmb b = (b_1, b_2, b_3)$

where $\theta$ is the angle between the two vectors (note that $\cos\theta = \cos(2\pi - \theta)$).
This definition does not depend on a coordinate system.

- The dot product is commutative[^d_commutative]
- The dot product is distributive[^d_distributive]
- If $\pmb a$ is perpendicular to $\pmb b$, then $\pmb a \cdot \pmb b = 0$ and they are said to be
  orthogonal
- If $\pmb a \cdot \pmb b = 0$ then either

    i. The vectors are orthogonal
    ii. One or both of the vectorse are zero vectors

- $\pmb a \cdot \pmb a = |\pmb a|^2 = a^2$

The base vectors are said to be *orthonormal* when $\pmb i^2 = \pmb j^2 = \pmb k^2 = 1$ and
$i\cdot j = i\cdot k = j\cdot k = 0$.

### The Vector Product (Cross Product)

The vector product between two vectors is defined by:

$$\pmb a \times \pmb b = |\pmb a||\pmb b|\sin\theta \pmb n$$

where $0 \le \theta \le \pi$ is the angle between $\pmb a$ and $\pmb b$ and $\pmb n$ is a unit
vector such that the three vectors from a right handed system:

![](./images/vimscrot-2022-02-18T20:11:12,072203286+00:00.png)

- $\pmb a \times \pmb b = -\pmb b \times \pmb a$ (the vector product is non-commutative[^d_commutative])
- If $\pmb a \times \pmb b = 0$ then either

    i. The vectors are parralel
    ii. One or both of the vectors are a zero vector
- $(k_1\pmb a)\times(k_2\pmb b) = (k_1k_2)(\pmb a \times \pmb b)$ where $k_1$, $k_2$ are scalars
- If $\pmb a = (a_1, a_2, a_3)$ and $\pmb b = (b_1, b_2, b_3)$ then

    $$\pmb a \times \pmb b = (a_2b_3 - a_3b_2, a_1b_3-a_3b_1, a_1b_2-a_2b_1)$$

- In the notation of determinants, provided we **expand by row 1**:

    $$\pmb a \times \pmb b = \begin{vmatrix} \pmb i & \pmb j & \pmb k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$

    This is technically not a determinant because not all the elements are numbers but shhhhhh...


### Scalar Triple Product

\begin{align*}
        [ \pmb a, \pmb b, \pmb c ] &= \pmb a \cdot (\pmb b \times \pmb c) \\
                                   &= \pmb b \cdot (\pmb c \times \pmb a) \\
                                   &= \pmb c \cdot (\pmb a \times \pmb b) \\
                                   &= (\pmb b \times \pmb c) \cdot \pmb a \\
                                   &= (\pmb c \times \pmb a) \cdot \pmb b \\
                                   &= (\pmb a \times \pmb b) \cdot \pmb c
\end{align*}

In terms of determinants:

$$
[ \pmb a, \pmb b, \pmb c ] = \begin{vmatrix} 
                a_1 & a_2 & a_3 \\ 
                b_1 & b_2 & b_3 \\ 
                c_1 & c_2 & c_3 
        \end{vmatrix}
$$

If $[\pmb a, \pmb b, \pmb c] = 0$ then the vectors are coplanar.

The absolute value of the scalar triple product reperesents the volume of the parallelepiped defined
by those vectors:

![](./images/vimscrot-2022-02-18T20:27:53,322534334+00:00.png)


## The Unit Vector

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

## Components of a Vector

The component of a vector $\pmb a$ in the direction of the unit vector $\pmb n$ is

$$\pmb a \cdot \pmb n$$

![](./images/vimscrot-2022-02-18T20:34:50,128465689+00:00.png)

Vectors are often written in terms of base vectors, such as the Cartesian system's $\pmb i$,
$\pmb j$, and $\pmb k$ in three dimensions.

![](./images/vimscrot-2022-02-18T19:31:50,686289623+00:00.png)

These vectors have unit magnitude, are perpendicular to each other, and are right handed.

If $\pmb a = a_1\pmb i + a_2\pmb j + a_3\pmb k$ then the scalars $a_1$, $a_2$, and $a_3$ are the
*components* of the vector (relative to the base vectors).

### Vector Projections

The *vector projection* of $\pmb a$ onto $\pmb n$ is given by

$$(\pmb a \cdot \pmb n)\pmb n$$

![](./images/vimscrot-2022-02-18T21:40:15,724449945+00:00.png)

They look like the same as [vector components](#components-of-a-vector) to me...
no idea what the difference is but uh StackExchange says
([permalink](https://physics.stackexchange.com/a/537690)):

> As pointed out, the projection and component actually refers to the same thing.
> To solve a problem like this it useful to introduce a coordinate system, as you mentioned yourself
> you project onto the x-axis.
> As soon as you introduce a coordinate system you can talk about the components of some vector.

## Position Vectors

If an origin $O$ is fixed, then any point $P$ in space may be represented by the vector $\pmb r$
which has a magnitude and direction given by the line $\overrightarrow{OP}$.

A point $(x, y, z)$ in Cartesian space has the position vector $r = x\pmb i + y\pmb j + z\pmb k$.

## Angle Between Vectors

By rearranging the [dot product equation](#the-scalar-product-inner-product-dot-product) you can get
an equation to find the angle, $\theta$, between two vectors:

$$\cos\theta = \frac{\pmb a \cdot \pmb b}{|\pmb a||\pmb b|} = \frac{a_1b_1 + a_2b_2 + a_3b_3}{|\pmb a ||\pmb b|}$$

# Applications of Vectors

## Application of Vectors to Geometry

### Equation of a Straight Line

A straight line can be specified by

- two points it passes
- one point it passes and a direction

If $\pmb a$ and $\pmb b$ are the position vectors of two distinct points, then the position vectors
of an arbitrary point on the line joining these points is:

$$\pmb r = \pmb a + \lambda(\pmb b - \pmb a)$$

where $\lambda \in \Re$ is a parameter.

![](./images/vimscrot-2022-02-18T21:55:30,367159917+00:00.png)

Suppose $O$ is an origin and $\pmb a$, $\pmb b$, and $\pmb r$ are position vectors on the line such
that

\begin{align*}
\pmb a &= (x_0, y_0, z_0) \\
\pmb b &= (x_1, y_1, z_1) \\
\pmb r &= (x, y, z)\\
\\
(x, y, z) &= (x_0, y_0, z_0) + \lambda((x_1, y_1, z_1) - (x_0, y_0, z_0)) \\
\\
x &= x_0 + \lambda(x_1-x_0) \\
y &= y_0 + \lambda(y_1-y_0) \\
z &= z_0 + \lambda(z_1-z_0) \\
\\
\lambda &= \frac{x-x_0}{x_1-x_0} = \frac{y-y_0}{y_1-y_0} = \frac{z-z_0}{z_1-z_0}
\end{align*}

In the above, the vector $\pmb b - \pmb a$ is in the direction of the line.
Thus the equation of a line can be specified by giving a point it passes through ($\pmb a$, say) and
the direction of the line ($\pmb d = (d_1, d_2, d_3)$, say).

The vector equation is then

$$\pmb r = \pmb a + \lambda\pmb d$$

#### The Cartesian Equation

$$\frac{x-x_0}{d_1} = \frac{y-y_0}{d_2} = \frac{z-z_0}{d_3}$$

### Equation of a Plane

A *plane* can be defined by specifying either:

- three points (as long as they're not in a straight line)
- a point on the plane and two directions (useful for a parametric form)
- specifying a point on the plane and the normal vector to the plane

#### Specifying a Point and a Normal Vector

Let $\pmb a$ be the position vector of a point on the plane, and $\pmb n$ a normal vector to the
plane.

If $\pmb r$ is the position vector of an arbitrary point on the plane, then $\pmb r - \pmb a$ is a
vector lying *in* the plane, so

$$(\pmb r - \pmb a) \cdot \pmb n = 0$$

So the *vector equation* of the plane is

$$\pmb r \cdot \pmb n = \pmb a \cdot \pmb n = \pmb d$$

where $\pmb r = (x, y, z)$ and the vectors $\pmb a$ and $\pmb n$ are known.

Suppose $\pmb a$, $\pmb n$, and $\pmb r$ are given by

\begin{align*}
\pmb a &= (x_0, y_0, z_0) \\
\pmb n &= (l, m, p) \\
\pmb n &= (x, y, z)\\
\text{then } 0 &= ((x, y, z) - (x_0, y_0, z_0))\cdot(l, m, p)
\end{align*}

#### Specifying Three Points on a Plane

If we specify three points on a plane with position vectors $\pmb a$, $\pmb b$, and $\pmb c$ the
vectors $\pmb c - \pmb a$ and $\pmb c - \pmb b$ lie *in* the plane.
(The vectors $\pmb a$, $\pmb b$, and $\pmb c$ do not necessarily lie *in* the plane;
rather they take you from $O$ **to** the plane.)

The normal to the plane, $\pmb n$, is then parallel to

$$(\pmb c - \pmb a)\times(\pmb c - \pmb b)$$

and so the equation of the plane is

$$(\pmb r - \pmb a)\cdot((\pmb c - \pmb a)\times(\pmb c - \pmb b)) = 0$$


#### The Angle Between Two Planes

... is the same as the angle between their normal vectors





[^d_associative]: The grouping of elements in in an operation do not matter (e.g. scalar addition: $a+(b+c) = (a+b)+c$)
[^d_commutative]: The order of elements in an operation do not matter (e.g. scalar addition: $a+b = b+a$)
[^d_distributive]: Easiest to explain with examples. Scalar multiplication is said to be distributive because $(a+b)c = ac + bc$