title: MMME1026 // Vectors

uuid: ef1669d9-b8c3-4813-8b33-5764a2f57a55 (permalink)

tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, vectors ]

written by: Akbar Rahman

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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 .

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 $\pmb{b}$ :

Vector addition is associative¹ and commutative².

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 commutative³
The dot product is distributive⁴
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
1. The vectors are orthogonal
2. 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:

$\pmb a \times \pmb b = |\pmb a||\pmb b|\sin\theta \hat{\pmb n}$

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:

$\pmb a \times \pmb b = -\pmb b \times \pmb a$ (the vector product is anti-commutative⁵)
If $\pmb a \times \pmb b = 0$ then either
1. The vectors are parallel
2. 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 (and co-planar vectors)

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

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:

The Unit Vector

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

Components of a Vector

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

$\pmb a \cdot \hat{\pmb n}$

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.

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 $\hat{\pmb n}$ is given by

$(\pmb a \cdot \hat{\pmb n})\hat{\pmb n}$

They look like the same as vector components to me… no idea what the difference is but uh StackExchange says (permalink):

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 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

Area of a Parallelogram

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

Volume of a Parallelepiped

$volume = | [ \pmb a, \pmb, b, \pmb c ] |$

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.

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

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

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

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

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

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

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