Vector
A vector is alternate notation for an
n tuple, so that \[ \begin{bmatrix} a_1 \\ a_2
\\ \vdots \\ a_n \end{bmatrix} \] is the n-tuple \( (a_1, a_2, \ldots, a_n ) \)
The length of a Vector
Given \( x \in \mathbb{R}^n \), then we define the it's length as
\[
\lVert x \rVert := \sqrt{ \sum _ { i = 1 } ^ n x _ i ^ 2 }
\]
Dot Product
Given two sequences of numbers of the same length: \( a = \left ( a_{1} , \ldots , a_{k} \right ) \)
and \( b = \left ( b_{1} , \ldots , b_{k} \right ) \), their dot product denoted and defined by
\[
a \cdot b = \sum_{i = 1}^{k} a_{i} \cdot b_{i}
\]
Note that \( \cdot : \mathbb{R}^{k} \times \mathbb{R}^{k} \to \mathbb{R} \)
Dot Product Geometric
\[
a \cdot b = \lVert a \rVert \lVert b \rVert \cos \left( \theta \right)
\]
where \( \theta \) is the angle between \( a, b \)
Norm Squared is the Dot Product
For any \( x \in \mathbb{ R } ^ n \) we have that
\[
\lVert x \rVert ^ 2 = x \cdot x
\]
Cross Product
Given two vectors \( a, b \in \mathbb{ R } ^ 3 \) we define their cross-product to be
\[
a \times b := \lVert a \rVert \lVert b \rVert \sin \left( \theta \right) n
\]
where \( \theta \) is the angle between \( a \) and \( b \) and \( n \) is a unit vector such that \( a , n , b \) is positively oriented.
Collinear
Given a vector space \( \left( V, F \right) \) then we say that two vectors \( x, y \in V \) are collinear if there exists an \( \alpha \in F \) such that
\[
x = \alpha y
\]
Parallelogram Law
Suppose \( x, y \in \mathbb{ R } ^ n \) then we have
\[
\lVert x + y \rVert ^ 2 + \lVert x - y \rVert ^ 2 = 2 \lVert x \rVert ^ 2 + 2 \lVert y \rVert ^ 2
\]
\[
\begin{align}
\lVert x + y \rVert ^ 2 + \lVert x - y \rVert ^ 2 &= \left( x + y \right) \cdot \left( x + y \right) + \left( x - y \right) \cdot \left( x - y \right) \\
&= x \cdot x + 2 x \cdot y + y \cdot y + x \cdot x - 2 x \cdot y + y \cdot y \\
&= 2 x \cdot x + 2 y \cdot y \\
&= 2 \lVert x \rVert ^ 2 + 2 \lVert y \rVert ^ 2
\end{align}
\]
The Norm of The Half of The Sum of Two Unit Vectors is One iff They are Equal
Suppose \( x, y \in \mathbb{ R } ^ n \) such that \( \lVert x \rVert = \lVert y \rVert = 1 \) then
\[
\left\lVert \frac{x + y}{2} \right\rVert = 1 \iff x = y
\]
\( \implies \) Suppose that \( \left\lVert \frac{x + y}{2} \right\rVert = 1 \) we'd like to prove that \( x = y \), thus it's equivalent to show that \( \lVert x - y \rVert = 0 \), recall that from the
parallelogram law that
\[
\lVert x - y \rVert = \sqrt{ 2 \lVert x \rVert ^ 2 + 2 \lVert y \rVert ^ 2 - \lVert x + y \rVert ^ 2 }
\]
thus it's enough to show that \( 2 \lVert x \rVert ^ 2 + 2 \lVert y \rVert ^ 2 - \lVert x + y \rVert ^ 2 = 0 \), which means we just have to show that \( \lVert x + y \rVert ^ 2 = 2 \lVert x \rVert ^ 2 + 2 \lVert y \rVert ^ 2 \), moreover \( x, y \) were unit vectors so we have \( \lVert x \rVert = \lVert y \rVert = 1 \) and we now just need to prove that \( \lVert x + y \rVert ^ 2 = 4 \). We've assumed that \( \left\lVert \frac{x + y}{2} \right\rVert = 1 \), this means that \( \lVert x + y \rVert = 2 \) thus we conclude that \( \lVert x + y \rVert ^ 2 = 4 \) as needed.
\( \impliedby \) Suppose that \( x = y \), then
\[
\left\lVert \frac{x + y}{2} \right\rVert = \left\lVert \frac{2x}{2} \right\rVert = \lVert x \rVert = 1
\]
as needed.
Cosine Law
Suppose that \( x, y \in \mathbb{ R } ^ n \) and \( \theta \) is the angle between them, then
\[
\lVert x + y \rVert ^ 2 = \lVert x \rVert ^ 2 + 2 \lVert x \rVert \lVert y \rVert \cos \left( \theta \right) + \lVert y \rVert ^ 2
\]
This follows from distributivity of the
geometric interpretation of the dot product
\[
\begin{align}
\lVert x + y \rVert ^ 2 &= \left( x + y \right) \cdot \left( x + y \right) \\
&= x \cdot x + 2 x \cdot y + y ^ 2 \\
&= \lVert x \rVert ^ 2 + 2 \lVert x \rVert \lVert y \rVert \cos \left( \theta \right) + \lVert y \rVert ^ 2
\end{align}
\]
Dot Product Is a Sum of Norms
\[
x \cdot y = \frac{\lVert x + y \rVert ^ 2 - \lVert x \rVert ^ 2 - \lVert y \rVert ^ 2}{2}
\]