Vectors

Vector

A vector is alternate notation for an n tuple, so that

[\begin{matrix} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{matrix}]

is the n-tuple

(a_{1}, a_{2}, \dots, a_{n})

The length of a Vector

Given

x \in ℝ^{n}

, then we define the it's length as

‖ x ‖ : = \sqrt{\sum_{i = 1}^{n} x_{i}^{2}}

Dot Product

Given two sequences of numbers of the same length:

a = (a_{1}, \dots, a_{k})

and

b = (b_{1}, \dots, b_{k})

, their dot product denoted and defined by

a \cdot b = \sum_{i = 1}^{k} a_{i} \cdot b_{i}

Note that

\cdot : ℝ^{k} \times ℝ^{k} \to ℝ

Dot Product Geometric

a \cdot b = ‖ a ‖ ‖ b ‖ \cos (θ)

where

θ

is the angle between

a, b

Norm Squared is the Dot Product

For any

x \in ℝ^{n}

we have that

{‖ x ‖}^{2} = x \cdot x

Vector Form of a Line

Suppose that

d, p

are vectors, then we say that the set

l : = {x : x = t d + p, for some t \in ℝ}

is a line and say that the vector equation for the line is

x = t d + p

We also say that

d

is the direction vector for

l

Cross Product

Given two vectors

a, b \in ℝ^{3}

we define their cross-product to be

a \times b : = ‖ a ‖ ‖ b ‖ \sin (θ) n

where

θ

is the angle between

a

and

b

and

n

is a unit vector such that

a, n, b

is positively oriented.

Cross Product Vector Component Formula

(a_{1}, a_{2}, a_{3}) \times (b_{1}, b_{2}, b_{3}) = (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1})

An easy way to remember this is ab 231 where you then get a2b3 - ..., a3b1 - ..., a1b2

Equation for a Line

We say the equation of a line

ℓ

is given by

a x + b y + c = 0

where

a, b, c \in ℝ

where the line is given by the set

ℓ = {(x, y) \in ℝ^{2} : a x + b y + c = 0}

Equation of a Line as a 3d Dot Product

Suppose that

ℓ

is given by

a x + b y + c = 0

then we have

ℓ = {(x, y, 1) \in ℝ^{3} : (x, y, 1) \cdot (a, b, c) = 0}

TODO: Add the proof here.

Equation of a Line Given by the Cross Product

Suppose that

p, q \in ℝ^{2}

and consider

L = (a, b, c) = p \times q

then the equation of the line that passes through

p, q

is given by

a x + b y + c = 0

We know that

a x + b y + c = 0

defines a line, but we need to verify that

p, q

are both on that line, but since

p \cdot L = 0

and

q \cdot L = 0

we know that they are on the line, as needed.

Collinear

Given a vector space

(V, F)

then we say that two vectors

x, y \in V

are collinear if there exists an

α \in F

such that

x = α y

Parallelogram Law

Suppose

x, y \in ℝ^{n}

then we have

{‖ x + y ‖}^{2} + {‖ x - y ‖}^{2} = 2 {‖ x ‖}^{2} + 2 {‖ y ‖}^{2}

\begin{matrix} {‖ x + y ‖}^{2} + {‖ x - y ‖}^{2} & a m p; = (x + y) \cdot (x + y) + (x - y) \cdot (x - y) \\ a m p; = x \cdot x + 2 x \cdot y + y \cdot y + x \cdot x - 2 x \cdot y + y \cdot y \\ a m p; = 2 x \cdot x + 2 y \cdot y \\ a m p; = 2 {‖ x ‖}^{2} + 2 {‖ y ‖}^{2} \end{matrix}

The Norm of The Half of The Sum of Two Unit Vectors is One iff They are Equal

Suppose

x, y \in ℝ^{n}

such that

‖ x ‖ = ‖ y ‖ = 1

then

‖ \frac{x + y}{2} ‖ = 1 ⟺ x = y

⟹

Suppose that

‖ \frac{x + y}{2} ‖ = 1

we'd like to prove that

x = y

, thus it's equivalent to show that

‖ x - y ‖ = 0

, recall that from the parallelogram law that

‖ x - y ‖ = \sqrt{2 {‖ x ‖}^{2} + 2 {‖ y ‖}^{2} - {‖ x + y ‖}^{2}}

thus it's enough to show that

2 {‖ x ‖}^{2} + 2 {‖ y ‖}^{2} - {‖ x + y ‖}^{2} = 0

, which means we just have to show that

{‖ x + y ‖}^{2} = 2 {‖ x ‖}^{2} + 2 {‖ y ‖}^{2}

, moreover

x, y

were unit vectors so we have

‖ x ‖ = ‖ y ‖ = 1

and we now just need to prove that

{‖ x + y ‖}^{2} = 4

. We've assumed that

‖ \frac{x + y}{2} ‖ = 1

, this means that

‖ x + y ‖ = 2

thus we conclude that

{‖ x + y ‖}^{2} = 4

as needed.

$⟸$ Suppose that $x = y$ , then $‖ \frac{x + y}{2} ‖ = ‖ \frac{2 x}{2} ‖ = ‖ x ‖ = 1$ as needed.

Cosine Law

Suppose that

x, y \in ℝ^{n}

and

θ

is the angle between them, then

{‖ x + y ‖}^{2} = {‖ x ‖}^{2} + 2 ‖ x ‖ ‖ y ‖ \cos (θ) + {‖ y ‖}^{2}

This follows from distributivity of the geometric interpretation of the dot product

\begin{matrix} {‖ x + y ‖}^{2} & a m p; = (x + y) \cdot (x + y) \\ a m p; = x \cdot x + 2 x \cdot y + y^{2} \\ a m p; = {‖ x ‖}^{2} + 2 ‖ x ‖ ‖ y ‖ \cos (θ) + {‖ y ‖}^{2} \end{matrix}

Dot Product Is a Sum of Norms

x \cdot y = \frac{{‖ x + y ‖}^{2} - {‖ x ‖}^{2} - {‖ y ‖}^{2}}{2}

Rearrange this.