\section{Dot Products}

\definition{}
We can also define the \textit{dot product} of two vectors.\footnotemark{} \\
The dot product maps two elements of $\mathbb{R}^n$ to one element of $\mathbb{R}$:

\footnotetext{
	\textbf{Bonus content. Feel free to skip.}

	Formally, we would say that the dot product is a map from $\mathbb{R}^n \times \mathbb{R}^n$ to $\mathbb{R}$. Why is this reasonable?

	\vspace{2mm}

	It's also worth noting that a function $f$ from $X$ to $Y$ can be defined as a subset of $X \times Y$, where for all $x \in X$ there exists a unique $y \in Y$ so that $(x, y) \in f$. Try to make sense of this definition.
}

$$
	a \cdot b = \sum_{i = 1}^n a_ib_i = a_1b_1 + a_2b_2 + ... + a_nb_n
$$

\problem{}
Compute $[2, 3, 4, 1] \cdot [2, 4, 10, 12]$

\vfill

\problem{}
Show that the dot product is
\begin{itemize}
	\item Commutative
	\item Distributive $a \cdot (b + c) = a \cdot b + a \cdot c$
	\item Homogenous: $x(a \cdot b) = xa \cdot b = a \cdot xb$ \\
	\note{$x \in \mathbb{R}$, and $a, b$ are vectors.}
	\item Positive definite: $a \cdot a \geq 0$, with equality iff $a = 0$ \\
	\note{$a \in \mathbb{R}^n$, and $0$ is the zero vector.}
\end{itemize}


\vfill
\pagebreak




\problem{}
Say you have two vectors, $a$ and $b$. Show that $a \cdot b$ = $||a||~||b||\cos(\alpha)$, \\
where $\alpha$ is the angle between $a$ and $b$. \\
\hint{What is $c$ in terms of $a$ and $b$?}
\hint{The law of cosines is $a^2 + b^2 - 2ab\cos(\alpha) = c^2$}
\hint{The length of $a$ is $||a||$}


\begin{center}
\begin{tikzpicture}[scale=1]

	\draw[->]
		(0,0) coordinate (o) -- node[above left] {$a$}
		(1,2) coordinate (a)
	;

	\draw[->]
		(o) -- node[below] {$b$}
		(3,0.5) coordinate (b)
	;

	\draw[
		draw = gray,
		text = gray,
		-
	] (a) -- node[above] {$c$} (b);

	\draw
		pic[
			"$\alpha$",
			draw = orange,
			text = orange,
			<->,
			angle eccentricity = 1.2,
			angle radius = 1cm
		]
		{ angle = b--o--a }
	;

\end{tikzpicture}
\end{center}

\vfill

\problem{}
If $a$ and $b$ are perpendicular, what must $a \cdot b$ be? Is the converse true?


\vfill
\pagebreak