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_{2} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1})

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{aligned} ‖ x + y ‖^{2} + ‖ x - y ‖^{2} & = (x + y) \cdot (x + y) + (x - y) \cdot (x - y) \\ = x \cdot x + 2 x \cdot y + y \cdot y + x \cdot x - 2 x \cdot y + y \cdot y \\ = 2 x \cdot x + 2 y \cdot y \\ = 2 ‖ x ‖^{2} + 2 ‖ y ‖^{2} \end{aligned}

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{aligned} ‖ x + y ‖^{2} & = (x + y) \cdot (x + y) \\ = x \cdot x + 2 x \cdot y + y^{2} \\ = ‖ x ‖^{2} + 2 ‖ x ‖ ‖ y ‖ \cos (θ) + ‖ y ‖^{2} \end{aligned}

Dot Product Is a Sum of Norms

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

Rearrange this.

− >< ! − − 2 3 ] + [ 3 2 1 ] , where

[- - ><! - - \begin{matrix} 3 \\ 2 \\ 1 \end{matrix}]

(the red dot) is a part of the line when

t = 0

, and--> − >< ! − − ]

-->