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 $\lVert x \rVert ^ 2 = x \cdot x$
vector form of a line
Suppose that $$d , p$$ are vectors, then we say that the set $$l := \left \lbrace x : x = t d + p , \text{ for some } t \in \mathbb{R} \right \rbrace$$ 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 $$a, b \in \mathbb{ R } ^ 3$$ 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.
Cross Product Vector Component Formula
$\left( a _ 1 , a _ 2, a _ 3 \right) \times \left( b _ 1, b _ 2, b _ 3 \right) = \left( a _ 2 b _ 2 - a _ 3 b _ 2, a _ 3 b _ 1 - a _ 1 b _ 3, a _ 1 b _ 2 - a _ 2 b _ 1 \right)$
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
$x \cdot y = \frac{\lVert x + y \rVert ^ 2 - \lVert x \rVert ^ 2 - \lVert y \rVert ^ 2}{2}$
Rearrange this.