Proposition: $ || \vec{u} + \vec{v} ||^2 = || \vec{u} ||^2 + || \vec{v} ||^2$
Proof: $ || \vec{u} + \vec{v} ||^2 = (\vec{u} + \vec{v}) \circ (\vec{u} + \vec{v}) $
$ = \vec{u} \circ \vec{u} + \vec{v} \circ \vec{u} +\vec{u} \circ \vec{v} + \vec{v} \circ \vec{v}$
$ = || \vec{u} ||^2 + || \vec{v} ||^2 $ (because $\vec{u} \circ \vec{v}$ is assumed to be 0)
Angle Theorem:
let $ \vec{u}, \vec{v} \in \mathbb{R},$ not parallel then:
$ \vec{u} \circ \vec{v} = || \vec{u} |||| \vec{v} || \cos \theta$
Proof: Apply cosine law.
$ c^2 = a^2 + b^2 - 2ab \cos \theta$
$ || \vec{u} - \vec{v} ||^2 = || \vec{v} ||^2 + || \vec{u} ||^2 - 2 || \vec{u} |||| \vec{v} ||
\cos \theta$
$ || \vec{u} - \vec{v} || = ( \vec{u} - \vec{v}) \circ ( \vec{u} - \vec{u}) $
$ = \vec{u} \circ \vec{u} + \vec{u} \circ \vec{v} + \vec{v} \circ \vec{u} + \vec{v} \circ \vec{v} $
$ = || \vec{u} ||^2 + || \vec{v} ||^2 - 2 \vec{u} \cdot \vec{v} $
$ \therefore \vec{u} \circ \vec{v} = || \vec{u} |||| \vec{v} || \cos \theta $
Inequalities:
Triangle inequality :=$ || \vec{u} + \vec{v} || \leq || \vec{u} || + || \vec{v} ||$
Cauchy–Schwarz inequality := $ | \vec{u} \cdot \vec{v} | \leq || \vec{u} |||| \vec{v} || $
Definition:
let $ u \in \mathbb{R} $ the a vector. the standard unit vector in the direction of $ \vec{u} $ in the vector.
$$ \frac{1}{|| \vec{u} ||} \vec{u} $$
let $ \vec{u}, \vec{v} \in \mathbb{R}^n $ the vectors. The distance between $\vec{u}$ and $\vec{v}$ denoted by $ d (\vec{u}, \vec{v})$ is $ d ( \vec{u}, \vec{v}) = || \vec{u} - \vec{v} ||$
let $\vec{u}, \vec{v} \in \mathbb{R}$. The projection of $\vec{u}$ onto $\vec{v}$ to the vector is denoted as:
$$ proj_\vec{u} \vec{v} = \frac{ \vec{u} \circ \vec{v}}{|| \vec{u} ||} \vec{u} $$
Proposition:
$ \vec{v} - proj_\vec{u} \vec{v} \perp proj_\vec{u} \vec{v} $
$ || \vec{v} - proj_\vec{u} \vec{v} || = d( \vec{u}, \vec{p}) $
A line, $L$, in $ \mathbb{R}^2 $ is determined by a direction vector $ \vec{d} $ and a special point, $p$, on $L$
let $ \vec{x} = (x, y)$ any point on $L$
let $ \vec{p} = (p_1, p_2)$ is a special point on $L$ then $ \vec{x} - \vec{p} = t\vec{d}, t \in \mathbb{R}$
A vector form of $L$ is $ \vec{x} = \vec{p} + \vec{d}$ with $ t \in \mathbb{R}$
let $\vec{n} = (a, b)$ any vector orthogonal to the $L$ \
$ \therefore \vec{n} \circ ( \vec{x} - \vec{p} ) = 0 $
Normal Form is $ \vec{n} \circ ( \vec{x} - \vec{p} ) = C$
General Form is $ a\vec{x} + b\vec{n} = C $
Parametric From is:
$ \vec{x} = (x, y)$
$ \vec{p} = (p_1, p_2)$
$ \vec{d} = (d_1, d_2)$
$ x = p_1 + d_1 $
$ y = p_2 + d_2 $
let $ \vec{x} = (x, y, y)$ the any point on $\delta$
let $ \vec{p} = (p_1, p_2, p_3 ) $ the special point
then $ \vec{x} - \vec{p} $ in a vector lying on $\delta$
let $ \vec{n} = (a, b, c) $ the normal vector for $\delta$ (aka, a is orthogonal to \delta)
$$ \vec{A} \circ ( \vec{x} - \vec{p}) = 0 $$
Any vector on $\delta$ will also be orthogonal to the normal vector ($ \vec{n} \circ \vec{x} = \vec{n} \circ \vec{p}$)