\documentclass[notes=show,10pt]{beamer}
\usepackage{mathpazo}			%% Beamer stuff
\usepackage{amsmath}			%% AMS math packages
\usepackage{amsfonts}			%%	-	special fonts
\usepackage{amssymb}			%%	-	special symbols
\usepackage{graphicx}			%% graphics
\usepackage{tcolorbox}			%% color boxes

\usepackage{hyperref}			%% for web links
\hypersetup{
%    bookmarks=true,         		% show bookmarks bar?
    unicode=false,          		% non-Latin characters in Acrobat’s bookmarks
    pdftoolbar=true,        		% show Acrobat’s toolbar?
    pdfmenubar=true,        	% show Acrobat’s menu?
    pdffitwindow=false,     		% window fit to page when opened
    pdfstartview={FitH},    		% fits the width of the page to the window
    pdfnewwindow=true,      	% links in new PDF window
    colorlinks=true,       			% false: boxed links; true: colored links
    linkcolor=blue,          		% color of internal links (change box color with linkbordercolor)
    citecolor=PineGreen,    		% color of links to bibliography
    filecolor=magenta,      		% color of file links
    urlcolor=cyan           		% color of external links
}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		TiKZ Ploting Stuff
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage{tikz}				%% TikZ -graphics

\usetikzlibrary{arrows, positioning, calc, intersections}
\usetikzlibrary{decorations.pathreplacing, decorations.markings}

\usepackage{pgfplots}			%% fancy plotting

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	For 3d plotting, using the tikz-3dplot package so
%	that (x,y,z) coordinates are specified in normal order
%	and one can change the perspective for 3-d plots
%
%	To invoke, use
%
%	\begin{tikzpicture}[tdplot_main_coords,{other options}]
%		...
%	\end{tikzpicture}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage{tikz-3dplot}		%% Enhanced 3d plotting

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	adjustbox package to force minipage alignment
%
%	Two-column structure is
%
%	\adjustbox{valign=t}
%		{
%			\begin{minipage}{0.45\textwidth}
%			...
%			\end{minipage}
%		}
%	\adjustbox{valign=t}
%		{
%		
%			\begin{minipage}{0.45\textwidth}
%			...
%			\end{minipage}
%		}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage{adjustbox}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Set Beamer theme
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usetheme{Dresden}
\usecolortheme{dolphin}

%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Hack to remove subsection bar from our
%		friends at Tex.StackExchange.Com
%
%%%%%%%%%%%%%%%%%%%%%%%%%%
\makeatletter
\beamer@theme@subsectionfalse

\makeatother
%%%%%%%%%%%%%%%%%%%%%%%%%%
%		
%		Hack to make the nice little circles appear
%		from Tex.stackexchange.com
%
%		Reference: 
%
%		https://tex.stackexchange.com/questions/2072/beamer-navigation-circles-without-subsections
%
%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage{remreset}
\makeatletter
\@removefromreset{subsection}{section}
\makeatother
\setcounter{subsection}{1}

%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Get rid of triangles, put in circles for 
%		itemize items
%
%%%%%%%%%%%%%%%%%%%%%%%%%%
\setbeamertemplate{items}[circle]

%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Clean up fonts
%
%%%%%%%%%%%%%%%%%%%%%%%%%%

\usefonttheme{professionalfonts}
\usefonttheme{serif}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		tikz plotting options
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\pgfplotsset{my style/.append style={axis x line=middle, axis y line=
           middle, xlabel={}, ylabel={}, axis equal }}
           
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Math Operators
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\DeclareMathOperator{\comp}{comp}
\DeclareMathOperator{\proj}{proj}
\DeclareMathOperator{\Hess}{Hess}

\DeclareMathOperator{\curl}{curl}
\DeclareMathOperator{\diver}{div}
\DeclareMathOperator{\grad}{grad}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Document begins here...
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
           
\begin{document}

\input{213-Lecture-Macros}		%% 	macros, settings, et.c


\title{Math 213 - The Dot Product}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{August 30, 2019}
\maketitle


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%	
%		Due Date Reminders
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Reminders}

\begin{itemize}

\item	Access your WebWork account \emph{only through Canvas}!

\medskip

\item	Homework A1 on Section 12.1 is due tonight!

\medskip

\item	Applications for an alternate Exam 1 are due \textbf{no later than September 4}
		
		\medskip
		
		Review  your schedule and apply for all alternate exams at once by using the 
		\href{https://docs.google.com/forms/d/e/1FAIpQLSdAMcpIdWP54-vD2UU1F_M6p3DJMphjuXuEjgvGd-y5GSPcPQ/viewform}{Google Form} linked from 
		Canvas or the course home page.
		
	
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Unit Outline
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Unit I: Geometry and Motion in Space}
\begin{itemize}

\item[{12.1}]		Lecture 1: Three-Dimensional Coordinate Systems
\item[{12.2}] 		Lecture 2: Vectors in the Plane and in Space
\begingroup
\color{red}
\item[{12.3}]		Lecture 3: The Dot Product
\endgroup
\item[{12.4}]		Lecture 4: The Cross Product
\item[{12.5}]		Lecture 5: Equations of Lines and Planes, I
\item[{12.5}]		Lecture 6: Equations of Lines and Planes, II
\item[{12.6}]		Lecture 7: Surfaces in Space
\item[{13.1}]		Lecture 8: Vector Functions and Space Curves
\item[{13.2}]		Lecture 9 Derivatives and Integrals of Vector Functions
\item[{}]				Lecture 10:  Exam I Review
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Learning Goals
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Learning Goals}
\begin{itemize}

\item		Know how to compute the dot product $\bfa \cdot \bfb$ of two vectors and understand its geometric interpretation

\medskip

\item		Understand \emph{direction angles} and \emph{direction cosines} of a vector and how to compute
them using dot products

\medskip

\item		Understand what the \emph{projection} of one vector onto another vector is

\medskip

\item		Understand the connection between dot products and the work done by a given force $\bfF$ through
a displacement $\bfD$
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		New material begins here...
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Make font size on all slides small
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begingroup
\small

\section{The Dot Product}

\begin{frame}
\frametitle{The Dot Product}
\footnotesize
\begin{center}
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan} \textbf{Definition}}  If $\bfa = \langle a_1, a_2, a_3 \rangle$ and $\bfb = \langle b_1, b_2, b_3 \rangle$, the \textbf{dot product} of $\bfa$ and $\bfb$ is the \underline{number} $\bfa \cdot \bfb$ given by
$$ \bfa \cdot \bfb = a_1 b_1 + a_2 b_2 + a_3 b_3 $$
\end{minipage}
}
\end{center}

\medskip
There's a similar definition for the dot product of vectors in two dimensions. The dot product is also called the \emph{scalar product} of two vectors.

\medskip

\hrulefill

\medskip

Find $\bfa \cdot \bfb$ if $\ldots$
\begin{enumerate}
\item		$\bfa = \langle 1, 1 \rangle$ and $\bfb = \langle 1, -1 \rangle$
\item		$\bfa  = \bfb = \left\langle \frac{1}{\sqrt{2}},  \frac{1}{\sqrt{2}} \right\rangle$
\item		$\bfa = 3 \ii - 4 \jj + \kk$, $\bfb = 2\ii +5\jj$
\item		$\bfa = 2\ii +5\jj$ and $\bfb=3 \ii - 4 \jj$
\item		$\bfa = 2\ii- 3\jj +4\kk$ and $\bfb = \kk$
\end{enumerate}

\end{frame}

\begin{frame}
\frametitle{Properties of the Dot Product}

Fill in the blanks:

\bigskip

\begin{align*}
\bfa \cdot \bfa 			&=	\underline{\hphantom{|\bfa|^2}}								
								&	\bfa \cdot \bfb 	&=	\bfb \cdot \bfa	\\
\bfa \cdot(\bfb + \bfc)	&=	\underline{\hphantom{\bfa \cdot \bfb + \bfa \cdot \bfc}}
								&	(c\bfa) \cdot \bfb	&=	\underline{\hphantom{c}}\left(\bfa \cdot \bfb\right) 
																= \bfa \cdot(\underline{\hphantom{c \bf b}})\\
\mathbf{0} \cdot \bfa 	&=	\underline{\hphantom{0}}
\end{align*}

\bigskip

How can you check these identities?


\end{frame}

\begin{frame}
\frametitle{The Law of Cosines}

\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}
\draw[thick]	(0,0)	--	(4,0)	--	(1,2)	-- (0,0);
%%
\node[left] 		at (0,0) 		{\footnotesize{A}};
\node[right]	at	(4,0)		{\footnotesize{B}};
\node[above] 	at (1,2)		{\footnotesize{C}};
%%
\node[below]	at	(2,0)		{\footnotesize{c}};
\node[left]		at	(0.5,1)	{\footnotesize{b}};
\node[right]	at	(2.4,1.2)	{\footnotesize{a}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
Recall from trigonometry:
$$ c^2 = a^2 + b^2 - 2ab \cos \theta $$
where
$$ \theta = m\angle ACB $$
\end{minipage}
\end{frame}

\begin{frame}
\frametitle{The Most Important Slide in this Lecture}
\footnotesize
\begin{center}
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan}\textbf{Theorem}} If $\theta$ is the angle between the vectors $\bfa$ and $\bfb$, then
$$ \bfa \cdot \bfb = | \bfa | \, |\bfb | \, \cos (\theta) $$
\end{minipage}
}
\end{center}

You can prove that this is true using the law of cosines to the triangle $OAB$:

\medskip

\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}[scale=0.8]
\draw[thick,->,>=stealth]	(0,0,0)	--	(4,0,0);			%% y
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,0,3);			%% x
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,3,0);			%% z
%%
\draw[cyan,thick,->,>=stealth]	(0,0,0)	--	(3,1,2);
\draw[cyan,thick,->,>=stealth]	(0,0,0)	--	(0,2,1);
\draw[cyan,thick,->,>=stealth]	(0,2,1)	--	(3,1,2);
%%
\node[left] at (0,1,0.5)				{\footnotesize{$\bfb$}};
\node[above] at (2.0,1.5,1.5)	{\footnotesize{$\bfa-\bfb$}};
\node[below]	at	(1.5,0,1)		{\footnotesize{$\bfa$}};
%%
\node[below]	at	(0,0,0)			{\footnotesize{O}};
\node[above]	at	(0,2,1)			{\footnotesize{B}};
\node[right]	at	(3,1,2)			{\footnotesize{A}};
%%
\node			at	(45:0.5)			{\footnotesize{$\theta$}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.5\textwidth}
\begin{align*}
|AB|^2 	&=	|OA|^2 + |OB|^2
			- 2 |OA| |OB| \cos \theta	\\
\intertext{so}
|\bfa-\bfb|^2	&=	|\bfa|^2 + |\bfb|^2 - 2|\bfa| |\bfb| \cos \theta \\
\end{align*}
Now express $|\bfa-\bfb|^2$ using the dot product.

\end{minipage}
\end{frame}

\begin{frame}
\frametitle{Why The Last Slide Was Important}
\footnotesize
$$ \underbrace{\,\, \bfa \cdot \bfb \,\, }_{\text{the dot product}} = \underbrace{|\bfa| \, |\bfb| \, \cos \theta}_{\text{its geometric meaning}} $$

\bigskip
\begin{itemize}
\item	To find the angle between two nonzero vectors $\bfa$ and $\bfb$, compute
		$$ \cos \theta = \frac{\bfa \cdot \bfb}{|\bfa| |\bfb|} $$
\item	Two nonzero vectors are orthogonal if and only if $\bfa \cdot \bfb = 0$
\end{itemize}

\medskip

\hrulefill

\medskip

\begin{enumerate}
\item	Are the vectors $\bfa = \langle 9,3 \rangle$ and $\bfb = \langle -2,6 \rangle$ parallel, orthogonal, or neither?
\item	Find the three angles of the triangle with vertices $P(2,0)$, $Q(0,3)$, $R(3,4)$
\item	Is the triangle with vertices $P(1,-3,-2)$, $Q(2,0,-4)$, $R(6,-2,-5)$ a right triangle?
\end{enumerate}
\end{frame}


\begin{frame}
\frametitle{Why Two Slides Ago Was Important}
\footnotesize
\begin{itemize}
\item	Two nonzero vectors are orthogonal if and only if $\bfa \cdot \bfb = 0$
\end{itemize}

\medskip

\hrulefill

\medskip

\begin{enumerate}
\item[3.]	Is the triangle with vertices $P(1,-3,-2)$, $Q(2,0,-4)$, $R(6,-2,-5)$ a right triangle?
\end{enumerate}

\begin{tikzpicture}[scale=0.4]
\draw[thick,->,>=stealth]	(0,0,0)	--	(5,0,0);			%% y
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,0,5);			%% x
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,5,0);			%% z
%%
\draw[black,fill=black]	(1,-3,-2)	circle(0.1cm);
\node[above] at	(1,-3,-2)			{\footnotesize{$P$}};
\draw[black,fill=black]	(2,0,-4)	circle(0.1cm);
\node[above] at	(2,0,-4)			{\footnotesize{$Q$}};
\draw[black,fill=black]	(6,-2,-5)	circle(0.1cm);
\node[above]	at	(6,-2,-5)			{\footnotesize{$R$}};
%%
\draw[cyan,dashed]			(1,-3,-2)	--	(2,0,-4)	--	(6,-2,-5)	--	(1,-3,-2);
\end{tikzpicture}

\end{frame}

\begin{frame}
\frametitle{Puzzlers}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.7]
\draw[cyan,thick,->,>=stealth]	(0,0)	--	(3,0);
\draw[cyan,thick,->,>=stealth]	(3,0)	--	++(120:3);
\draw[cyan,thick,->,>=stealth]	(1.5,2.598)	--++(240:3);
%%
\node[below] at	(1.5,0)	{\footnotesize{$\bfw$}};
\node[left]		at	(0.75,1.3)	{\footnotesize{$\bfu$}};
\node[right]	at	(2.25,1.3)	{\footnotesize{$\bfv$}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
At left is an equilateral triangle made of of vectors $\bfu$, $\bfv$, and $\bfw$.
If $\bfu$ is a unit vector, find $\bfu \cdot \bfv$ and $\bfu \cdot \bfw$
\end{minipage}

\vfill

\hrulefill

\vfill

\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.7]
\draw[thick,->,>=stealth]	(0,0)	--	(0,4.5);
\draw[thick,->,>=stealth]	(0,0)	--	(4.5,0);
\draw[help lines]	(0,0)	grid	(4,4);
\begin{scope}
\clip (0,0)	rectangle (4,4);
\draw[thick,cyan]	plot[domain=0:4]	({\x},{2*\x-3});
\draw[thick,cyan]	plot[domain=0:4]	({\x},{7-3*\x});
\end{scope}
\uncover<2->
{
\draw[thick,purple,->,>=stealth]	(2,1)	--	(1,4);
\draw[thick,purple,->,>=stealth]	(2,1)	--	(3,3);
\draw[black,fill=black] 			(2,1)	circle(0.05cm);
\draw[black,fill=black]			(1,4)	circle(0.05cm);
\draw[black,fill=black]			(3,3)	circle(0.05cm);
\node[left] at (2,1)						{\footnotesize{$(2,1)$}};
\node[above] at (1,4)					{\footnotesize{$(1,4)$}};
\node[right] at (3,3)						{\footnotesize{$(3,3)$}};
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
Find the acute angle between the lines
\begin{align*}
2x-y	&=	3\\
3x+y	&=	7
\end{align*}
\end{minipage}
\end{frame}

\section{Direction Angles, Direction Cosines}

\begin{frame}
\frametitle{Direction Angles and Direction Cosines}
\begin{minipage}{0.40\textwidth}
\begin{tikzpicture}[scale=0.6]
\draw[thick,->,>=stealth]	(0,0,0)	--	(5,0,0);			%% y
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,0,5);			%% x
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,5,0);			%% z
%%
\draw[thick,red,->,>=stealth]	(0,0,0)	--	(4,3,4.1);
\draw[dashed]	(0,0,0)	--	(4,0,0)	--	(4,3,0)	--	(0,3,0)	--	(0,0,0);
\draw[dashed]	(0,0,0)	--	(0,3,0)	--	(0,3,4.1)	--	(0,0,4.1)	--	(0,0,0);
\draw[dashed]	(0,0,0)	--	(0,0,4.1)	--	(4,0,4.1)	--	(4,0,0)	--	(0,0,0);
\draw[dashed]	(0,0,4.1)	--	(0,3,4.1)	--	(4,3,4.1)	--	(4,0,4.1)	--	(0,0,4.1);
\draw[dashed]	(4,3,4.1)	--	(4,3,0);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%
%%		the following is a completely reprehensible hack to draw
%%		circles representing the direction angles that go from the coordinate
%%		axis to the vector
%%
%%		the idea is to draw and clip a circle rather than actually trying to 
%%		work out where the circles should be
%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%alpha
\uncover<2->
{
\begin{scope}
\clip	(0,0,0)	--	(0,0,5)	--	(4,0,5)	--	(4,3,4)	--	(0,0,0);
\draw		(0,0)	circle(1cm and 0.8cm);
\node[below] at (0.25,0,0)		{\footnotesize{$\alpha$}};
\end{scope}
}
%%beta
\uncover<3->
{
\begin{scope}
\clip	(0,0,0)	--	(4,3,4)	--	(5,0,0)	--	(0,0,0);
\draw		(0,0)	circle(2cm);
\node[above]	at	(1.5,0,0)		{\footnotesize{$\beta$}};
\end{scope}
}
%%gamma
\uncover<4->
{
\begin{scope}
\clip	(0,0,0)	--	(4,3,4)	--	(0,5,0)	--	(0,0,0);
\draw		(0,0)	circle(1.2cm);
\node[right]	at	(0,0.5,0)		{\footnotesize{$\gamma$}};
\end{scope}
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.50\textwidth}
\footnotesize
The \emph{direction angles} associated to a vector $\bfv$ are shown in the picture at left. They can be computed by
\begin{gather*}
\uncover<2->
{
\cos \alpha		=	 \frac{\bfv \cdot \ii}{|\bfv|},
}
 \qquad
\uncover<3->
{
\cos \beta		=	 \frac{\bfv \cdot \jj}{|\bfv|}\\
}
\uncover<4->
{
\cos \gamma	=	 \frac{\bfv \cdot \kk}{|\bfv|}
}
\end{gather*}

\uncover<5->
{
\noindent
The numbers $\cos \alpha$, $\cos \beta$, and $\cos \gamma$ are called the \emph{direction cosines} of $\bfv$.
}

\medskip

\uncover<6->
{

\hrulefill

\medskip

\begin{enumerate}
\item
Find the direction cosines of the vector $\langle 2,1, 2 \rangle$

\item
Find the direction cosines of the vector $\langle c, c, c \rangle$ if $c>0$.
\end{enumerate}
}
\end{minipage}

\end{frame}

\section{Projections}

\begin{frame}
\frametitle{Projections}
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.7]
%% the vector b
\draw[thick,->,>=stealth]	(0,0)	--	(-2,3);
\node[left] at	(-1,1.5)		{\footnotesize{$\bfb$}};
%% the vector a
\draw[thick,->,>=stealth]	(0,0)	--	(0,5);
\node[right] at	(0,4)		{\footnotesize{$\bfa$}};
%% the projection line
\draw[dashed]	(-2,3)		--	(0,3);
%%
\uncover<2->
{
\draw[purple,thick,->,>=stealth]	(0,0)	--	(0,3);
\node[right]	at	(0,3)	{\color{purple} \footnotesize{$\proj_{\bfa} \bfb$}};
}
\uncover<3->
{
\node at		(-0.3 ,1)		{\footnotesize{$\theta$}};
\begin{scope}
\clip	(0,0)	--	(-2,3)		--	(0,3)	--	(0,0);
\draw		(0,0)	circle (1.5cm);
\end{scope}
}
\end{tikzpicture}
\end{minipage}  
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
Finally we can use the dot product to find  the \emph{vector projection} of a vector $\bfb$ onto another vector $\bfa$, 
denoted 
$$ \proj_{\bfa} \bfb $$

\medskip

\uncover<2->
{
To the left is a visual of what the projection means.
}

\medskip

\uncover<3->
{
The projection of $\bfb$ onto $\bfa$ is a vector in the direction of $\bfa$ having (signed) magnitude
$$  \comp_{\bfa} \bfb = \bfb \cdot \frac{\bfa}{|\bfa|} = |\bfb| \cos \theta $$
}

\medskip

\uncover<4->
{
So,
$$ \proj_{\bfa} \bfb = \left( \bfb \cdot \frac{\bfa}{|\bfa|} \right) \frac{\bfa}{|\bfa|} =\left( \frac{\bfa \cdot \bfb}{|\bfa|^2} \right) \bfa  $$
}

\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Projection Puzzler}
\begin{minipage}{0.45\textwidth}
%% Stack Two Figures
\begin{minipage}{0.95\textwidth}
\begin{tikzpicture}[scale=0.2]
\draw[thick,->,>=stealth]	(-6,0)		--	(6,0);
\draw[thick,->,>=stealth]	(0,0)		--	(0,13);
\draw[help lines]	(-5,0)	grid	(5,12);
%%
\draw[red,thick,->,>=stealth]		(0,0)	--	(4,6);
\draw[red,thick,->,>=stealth]		(0,0)	--	(-5,12);
%%
\node[above] at (-5,12)	{\footnotesize{$\bfa=\langle -5,12 \rangle$}};
\node[right] at (5,6)		{\footnotesize{$\bfb=\langle 4,6 \rangle$}};
%%
\end{tikzpicture}
\end{minipage}

\vskip 1cm


\begin{minipage}{0.95\textwidth}
\begin{tikzpicture}[scale=0.3]
\draw[thick,->,>=stealth]	(-5,0)		--	(5,0);
\draw[thick,->,>=stealth]	(0,-4)		--	(0,4);
\draw[help lines]				(-4,-3)	grid	(4,3);
%%
\draw[red,thick,->,>=stealth]	(0,0)	--	(1,3);
\draw[red,thick,->,>=stealth]	(0,0)	--	(-3,-2);
%%
\node[left] at (-2.8,-1.5)		{\footnotesize{$\bfb$}};
\node[right] at (0.9,2.6)		{\footnotesize{$\bfa$}};

\end{tikzpicture}
\end{minipage}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
Recall the \emph{scalar projection}
$$  \comp_{\bfa} \bfb = \bfb \cdot \frac{\bfa}{|\bfa|} = |\bfb| \cos \theta $$
and the \emph{vector projection}
$$ \proj_{\bfa} \bfb = \left( \frac{\bfa \cdot \bfb}{|\bfa|^2} \right) \bfa $$

\medskip

\hrulefill

\medskip

\begin{enumerate}
\item	Find the scalar and vector projections of  $\bfb = \langle 4,6 \rangle$ onto $\bfa=\langle -5,12 \rangle$ 
\item	In the second figure shown, is the scalar projection of $\bfb$ onto $\bfa$ a positive number, or a negative number?
\end{enumerate}
\end{minipage}
\end{frame}

\section{Dot Products and Work}

\begin{frame}
\frametitle{Dot Products and Work}
\footnotesize
The work done by a force $\bfF$ acting through a displacement $\bfD$ is
$$ W= \bfF \cdot \bfD $$

\smallskip

Unit Reminders:

\medskip

\begin{center}
{\renewcommand{\arraystretch}{1.4}
\begin{tabular}{|l|l|l|l|l|}
\hline
\textbf{Quantity} 	&	\textbf{Type}		&	\textbf{MKS Unit}	&	\textbf{FPS Unit}	
\\
\hline
Force					&	Vector				&	Newton					&	Pound					\\
\hline
Displacement			&	Vector				&	Meter					&	Foot					\\
\hline
Work						&	Scalar				&	Joule (Nt-m)			&	Foot-pound	\\
\hline
\end{tabular}
}
\end{center}
\medskip

\hrulefill

\bigskip

A boat sails south with the help of a wind blowing in the direction  S $36^\circ$E with magnitude $400$ lb. Find the work done by the wind as the boat moves $120$ ft.

\bigskip

\end{frame}

\begin{frame}
\frametitle{For Next Time: Determinants}
\footnotesize
Next time we'll define the \emph{cross product} of two vectors, and we'll need to know how to compute the \emph{determinant} of a $2 \times 2$ or $3 \times 3$ matrix.

\medskip

A \textbf{determinant of order $2$} is defined by
$$
\begin{vmatrix}
a_1	&	a_2	\\
b_1	&	b_2
\end{vmatrix} 
=
a_1 b_2 - a_2 b_1
$$

\medskip

\hrulefill	

\medskip	

Find the following determinants:
$$
\begin{vmatrix}
2	&	1	\\	4	&	-6
\end{vmatrix},
\quad
\begin{vmatrix}
4	&	-6	\\
2	&	1
\end{vmatrix},
\quad
\begin{vmatrix}
1	&	2\\
2	&	4
\end{vmatrix}
$$
\end{frame}

\begin{frame}
\frametitle{Determinants, Continued}
\footnotesize
A \textbf{determinant of order $3$} is defined by
$$
\begin{vmatrix}
a_1	&	a_2	&	a_3	\\
b_1	&	b_2	&	b_3	\\
c_1	&	c_2	&	c_3
\end{vmatrix}
=
a_1 
\begin{vmatrix} 
	b_2 & b_3 \\ c_2 & c_3 
\end{vmatrix}
%
{\color{red} -} a_2 
\begin{vmatrix}
	b_1	&	c_3	\\	c_1	&	c_3	
\end{vmatrix}
%
+ a_3
\begin{vmatrix}
	b_1	&	b_2	\\	c_1	&	c_2
\end{vmatrix}
$$
For an illustration of this formula, see this \href{https://www.khanacademy.org/math/algebra-home/alg-matrices/alg-determinants-and-inverses-of-large-matrices/v/finding-the-determinant-of-a-3x3-matrix-method-2}{Khan Academy Video}

\medskip

For a shortcut method that many students like, see this \href{https://www.khanacademy.org/math/algebra-home/alg-matrices/alg-determinants-and-inverses-of-large-matrices/v/finding-the-determinant-of-a-3x3-matrix-method-1}{Khan Academy Video}
\end{frame}

\begin{frame}
\frametitle{Lecture Review}
\begin{itemize}
\item	We defined the \emph{dot product} of two vectors and found its geometric meaning
\item	We defined \emph{direction angles} and \emph{direction cosines} and computed them using dot products
\item	We used the dot product to compute the projection of one vector onto another
\item	We computed the work done by a force $\bfF$ through a displacement $\bfD$ using dot products
\end{itemize}
\end{frame}


\begin{frame}
\frametitle{Homework}
\begin{itemize}
\item	Review section 12.3 
\item Complete  homework A1 due Friday; begin work on homework A2
\item	Review or learn how to compute the determinant of a $3 \times 3$ matrix
\item	Read and study section 12.4 for next Wednesday's lecture
\item	Enjoy Labor Day weekend!
\end{itemize}
\end{frame}
		

\endgroup

\end{document}
