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\begin{document}

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


\title{Math 213 - Moving Around in  Space}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{August 28, 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 Sections 12.1-12.2  is due Friday August 30

\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
\begingroup
\color{red}
\item[{12.2}] 		Lecture 2: Vectors in the Plane and in Space
\endgroup
\item[{12.3}]		Lecture 3: The Dot Product
\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	Understand vectors as \emph{displacements}
\item	Understand how to combine vectors by addition, subtraction, and scalar multiplication
\item	Understand \emph{components} of vectors
\item	Understand \emph{unit vectors}, and know the standard basis vectors 
$\ii$, $\jj$, and $\kk$
\item	Use vectors to solve problems involving forces and velocities
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%		New material begins here...
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%		Make font size on all slides small
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\begingroup
\small


\section{Vectors}

\begin{frame}
A \textbf{vector} is a set of instructions for how to move from one location in space to another.  We've already seen this in our discussion of coordinate systems. 

\bigskip

\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(4,0,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,5,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,0,6);
\node[below] 	at		(0,0,6)	{\footnotesize{$x$}};
\node[right]	 at		(4,0,0)	{\footnotesize{$y$}};
\node[above]	 at		(0,5,0)	{\footnotesize{$z$}};
\uncover<2-4>
{
\draw[green,thick,->,>=stealth]		(0,0,0)	--	(0,0,2);
}
\uncover<3-4>
{
\draw[red,thick,->,>=stealth]		(0,0,2)	--	(4,0,2);
}
\uncover<4-4>
{
\draw[blue,thick,->,>=stealth]		(4,0,2)	--	(4,3,2);
}
\uncover<5->
{
\draw[green,dashed,->,>-stealth]					(0,0,0)	--	(0,0,2);
\draw[red,dashed,->,>=stealth,opacity=0.5]	(0,0,2)	--	(4,0,2);
\draw[blue,dashed,->,>=stealth,opacity=0.5]	(4,0,2)	--	(4,3,2);
\draw[dashed,opacity=0.5]							(0,0,0)	--	(4,0,2);
\draw[black,thick,->,>=stealth]		(0,0,0)	--	(4,3,2);
\node[above] at (4,3,2)		{$\mathbf{v}=\langle 2,4,3 \rangle$};
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
The vector $\bfv = \langle 2, 4, 3 \rangle$ is an instruction to move
\begin{itemize}
\uncover<2->
{\item $2$ units in the $x$ direction}
\uncover<3->
{\item $4$ units in the $y$ direction}
\uncover<4->
{\item $3$ units in the $z$ direction}
\end{itemize}
\bigskip
\uncover<5->
{
In this picture:

\begin{itemize}
\item 		the \textbf{initial point} of the vector is $(0,0,0)$
\item		the \textbf{final point} is $(2,4,3)$.
\end{itemize}

\medskip

We could also choose a different initial point$\ldots$
}

\end{minipage}
\end{frame}

\begin{frame}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(4,0,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,5,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,0,6);
\node[below] 	at		(0,0,6)	{\footnotesize{$x$}};
\node[right]	 at		(4,0,0)	{\footnotesize{$y$}};
\node[above]	 at		(0,5,0)	{\footnotesize{$z$}};
%%
\node[left] at (0,1,0)									{\footnotesize{$A=(0,0,1)$}};
\node[above] at (4,4,2)								{\footnotesize{$B=(2,4,4)$}};
\draw[black,fill=black]		(0,1,0) circle (0.10cm);
\draw[black,fill=black]		(4,4,2) circle (0.10cm);
%%
\uncover<2-4>
{
\draw[green,thick,->,>=stealth]		(0,1,0)	--	(0,1,2);
}
\uncover<3-4>
{
\draw[red,thick,->,>=stealth]		(0,1,2)	--	(4,1,2);
}
\uncover<4-4>
{
\draw[blue,thick,->,>=stealth]		(4,1,2)	--	(4,4,2);
}
\uncover<5->
{
%% the "3.9" instead of 4 is to avoid a collision of arrowheads!
\draw[green,dashed,->,>-stealth]					(0,1,0)	--	(0,1,2);
\draw[red,dashed,->,>=stealth,opacity=0.5]	(0,1,2)	--	(4,1,2);
\draw[blue,dashed,->,>=stealth,opacity=0.5]	(4,1,2)	--	(4,3.9,2);
\draw[dashed,opacity=0.5]							(0,1,0)	--	(4,1,2);
\draw[black,thick,->,>=stealth]		(0,1,0)	--	(4,3.9,2);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
Let's begin at $A=(0,0,1)$
The vector $\bfv = \langle 2, 4, 3 \rangle$ is an instruction to move
\begin{itemize}
\uncover<2->
{\item $2$ units in the $x$ direction}
\uncover<3->
{\item $4$ units in the $y$ direction}
\uncover<4->
{\item $3$ units in the $z$ direction}
\end{itemize}

\bigskip

\uncover<5->
{
In this picture:

\begin{itemize}
\item		 the \textbf{initial point} of the vector is $A=(0,0,1)$ 
\item		the \textbf{final point} is $B=(2,4,4)$.
\end{itemize}

\medskip

Another name for the vector $\bfv$ is $$\overrightarrow{AB}.$$
}

\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Puzzler}
\footnotesize
Can you name all of the equal vectors in the parallelogram shown below?

\bigskip
\begin{center}
\begin{tikzpicture}
\draw[cyan,thick,->,>=stealth]	(0,0)	--	++(1,3);					%%	DA
\draw[cyan,thick,->,>=stealth]	(1,3)	--	++(4,0);					%%	AB
\draw[cyan,thick,->,>=stealth]	(4,0)	--	++(1,3);					%%	BC
\draw[cyan,thick,->,>=stealth]	(0,0)	--	++(4,0); 				%%	DC
%%
\draw[cyan,thick,->,>=stealth]	(0,0)	--	++(2.5,1.5);			%%  DE
\draw[cyan,thick,->,>=stealth]	(2.5,1.5)	--	++(2.45,1.45);	%%  EB
\draw[cyan,thick,->,>=stealth]	(4,0)	--	(2.5,1.5);				%%	CE
\draw[cyan,thick,->,>=stealth]	(2.5,1.5)	--	(1,3)	;			%%	EA
%%
\node[left] at (0,0)			{\footnotesize{$D$}};
\node[left]	at	(1,3)			{\footnotesize{$A$}};
\node[right] at (5,3)			{\footnotesize{$B$}};
\node[right] at (4,0)			{\footnotesize{$C$}};
\node[above] at (2,1.5)	{\footnotesize{$E$}};
\end{tikzpicture}
\end{center}
\end{frame}

\section{Combining Vectors}

\begin{frame}
\frametitle{Vector Addition - Triangle Law}
\footnotesize
\begin{center}
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan} \textbf{Vector Addition} }
If $\bfu$ and $\bfv$ are vectors positioned so that the initial point of $\bfv$ is at the terminal point of $\bfu$, then the \text{sum} $\bfu + \bfv$ is the vector from the initial point of $\bfu$ to the terminal point of $\bfv$
\end{minipage}
}
\end{center}

\medskip

\begin{center}
\begin{tikzpicture}[scale=0.8]
\draw[thick,cyan,->,>=stealth]	(0,0)	--	(3,1);
\draw[thick,cyan,->,>=stealth]	(3,1) 	-- 	(5,4);
\draw[thick,red,->,>=stealth]	(0,0)	--	(5,4);
\node[below] 	at	(1.6,0.5)		{\footnotesize{$\mathbf{u}$}};
\node[right]	at	(4,2.5)			{\footnotesize{$\mathbf{v}$}};
\node[left]		at	(2.4,2)			{\footnotesize{$\mathbf{u}+\mathbf{v}$}};
\node[below]  at	(2.5,-0.5)		{The Triangle Law};
\end{tikzpicture}
\end{center}

\end{frame}

\begin{frame}
\frametitle{Vector Addition - Parallelogram Law}
\footnotesize

\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.8]
\node[below]  at	(2.5,-0.5)							{The Parallelogram Law};
\uncover<2->
{
\draw[thick,cyan,->,>=stealth]	(0,0)	--	(3,1);											%% 	u
\node[below] 	at	(1.6,0.5)							{\footnotesize{$\bfu$}};		%% label u
}
\uncover<3->
{
\draw[thick,black,->,>=stealth]	(3,1) 	-- 	(4.95,3.95);								%% 	v from u - 
																										%%	pull back to avoid arrow collison
\node[right]	at	(4,2.5)								{\footnotesize{$\bfv$}};		%%	label v from u
}
\uncover<4->
{
\draw[thick,red,->,>=stealth]	(0,0)	--	(5,4);											%% resultant
\node[below,rotate=40]		at	(2.4,2)			{\footnotesize{$\bfu+\bfv$}};
}
\uncover<6->
{
\draw[thick,cyan,->,>=stealth]	(0,0) 	--	(2,3);											%% 	v
\node[left]		at	(1,1.5)								{\footnotesize{$\bfv$}};		%% 	label v
}
\uncover<7->
{
\draw[thick,black,->,>=stealth]	(2,3)	--	(4.95,3.95);								%%	u from v - 
																										%%	pull back to avoid arrow collision
\node[above]	at	(3.5,3.5)							{\footnotesize{$\bfu$}};		%%	label u
}
\uncover<8->
{
\node[above,rotate=40]		at	(2.4,2)			{\footnotesize{$\bfv+\bfu$}};	%%	label resultant
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
To add $\bfu$ and $\bfv$, we can either:
\begin{itemize}
\uncover<2->
{
\item	Begin with $\bfu$
}
\uncover<3->
{
\item	Displace by $\bfv$
}
\uncover<4->
{
\item	Obtain $\bfu+\bfv$
}
\end{itemize}
\uncover<5->
{\medskip
OR
\medskip
}
\begin{itemize}
\uncover<6->
{\item	Begin with $\bfv$}
\uncover<7->
{\item	Displace by $\bfu$}
\uncover<8->
{\item	Obtain $\bfv+\bfu$}
\end{itemize}

\bigskip

\uncover<9->
{Notice that 
$$\bfu+\bfv=\bfv+\bfu$$
}
\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Vector Addition - Spoiler}
You can compute $\bfu + \bfv$ by \emph{adding components}:

\bigskip

\begin{center}
\begin{tikzpicture}[scale=0.8]
\uncover<2->
{
\draw[thick,cyan,->,>=stealth]	(0,0)	--	(3,1);
\node[below] 	at	(1.8,0.5)		{\footnotesize{$\mathbf{u}=\langle 3,1 \rangle$}};
}
\uncover<3->
{
\draw[thick,cyan,->,>=stealth]	(3,1) 	-- 	(5,4);
\node[right]	at	(4,2.5)			{\footnotesize{$\mathbf{v}=\langle 2, 3 \rangle$}};
}
\uncover<4->
{
\draw[thick,red,->,>=stealth]	(0,0)	--	(5,4);
\node[left]		at	(2.5,2.4)			{\footnotesize{$\mathbf{u}+\mathbf{v}=\langle 5,4 \rangle$}};
}
\end{tikzpicture}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Scalar Multiplication}
\footnotesize
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan} \textbf{Scalar Multiplication}} If $c$ is a scalar and $\bfv$ is a vector, then the \textbf{scalar multiple} $c\bfv$ is a vector $|c|$ times the length of $\bfv$ and whose direction is:
\begin{itemize}
\uncover<2->
{
\item	The \emph{same} as $\bfv$, if $c>0$
}
\uncover<3->
{
\item	\emph{Opposite} to $\bfv$, if $c<0$, 
}
\uncover<4->
{
\item	The \emph{zero vector} $\mathbf{0}$ if $c=0$
}
\end{itemize}
\end{minipage}
}


\vfill

\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw[cyan,thick,->,>=stealth]	(0,0)	--	++(1,1);
\node[left] at (0.5,0.6)				{\footnotesize{$\bfv$}};
%%
\uncover<2->
{
\draw[red,thick,->,>=stealth]	(2,0)	--	++(2,2);
\node[left] at	(3,1.5)				{\footnotesize{$2 \bfv$}};
%%
\draw[red,thick,->,>=stealth]	(4.5,0)	--	++(0.5,0.5);
\node[left]	at	(4.75,0.25)			{\footnotesize{$\frac{1}{2}\bfv$}};
}
\uncover<3->
{
%%
\draw[green,thick,->,>=stealth]	(7,1)	--	(6,0);
\node[left]	at	(6.5,0.5)			{\footnotesize{$-\bfv$}};
%%
\draw[green,thick,->,>=stealth]	(10,2)	--	(8,0);
\node[left] at (9,1)					{\footnotesize{$-2\bfv$}};
}
\uncover<4->
{
\draw[black,fill=black]	(11,0)	circle(0.10cm);
\node[left] at (11,0)					{\footnotesize{$\mathbf{0}$}};
}
\end{tikzpicture}
\end{center}
\end{frame}


\begin{frame}
\frametitle{Scalar Multiplication - Spoiler}
You can compute $c \bfv$ by \emph{componentwise multiplication}:

\bigskip

\begin{center}
\begin{tikzpicture}[scale=0.6]
\uncover<2->
{
\draw[cyan,thick,->,>=stealth]	(0,0)	--	++(1,1);
\node[left] at (0.5,0.6)				{\footnotesize{$\bfv=\langle 1,1 \rangle$}};
%%
\draw[red,thick,->,>=stealth]	(4,0)	--	++(2,2);
\node[left] at	(4,0)				{\footnotesize{$2 \bfv = \langle 2,2 \rangle $}};
%%
\draw[red,thick,->,>=stealth]	(10,0)	--	++(0.5,0.5);
\node[left]	at	(10,0)			{\footnotesize{$\frac{1}{2}\bfv = \langle \frac{1}{2}, \frac{1}{2} \rangle$}};
}
\end{tikzpicture}

\bigskip

\begin{tikzpicture}[scale=0.8]
\uncover<3->
{
\draw[cyan,thick,->,>=stealth]	(0,0)	--	++(1,1);
\node[left] at (0.5,0.6)				{\footnotesize{$\bfv=\langle 1,1 \rangle$}};
%%
%%
\draw[green,thick,->,>=stealth]	(5,1)	--	(4,0);
\node[left]	at	(4,0)			{\footnotesize{$-\bfv = \langle -1, -1 \rangle$}};
%%
\draw[green,thick,->,>=stealth]	(10,2)	--	(8,0);
\node[left] at (8,0)					{\footnotesize{$-2\bfv = \langle -2, -2 \rangle$}};
}
\end{tikzpicture}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Vector Subtraction}
$$ \bfu - \bfv = \bfu + (-1)\bfv $$

\bigskip

\begin{center}
\begin{tikzpicture}
\draw[cyan,thick,->,>=stealth]		(0,0)	--	(-1,2);
\node[left] at 	(-0.5,1)					{\footnotesize{$\bfv$}};
\draw[cyan,thick,->,>=stealth]		(0,0)	--	(3,1);
\node[above] at (1.5, 0.5)			{\footnotesize{$\bfu$}};
%%
\draw[green,thick,->,>=stealth]		(0,0)	--	(1,-2);
%%
\draw[black,thick]						(1,-2)		--	++(3,1);
\draw[black,thick]						(3,1)		--	++(1,-2);
%%
\draw[red,thick,->,>=stealth]		(0,0)		--	(4,-1);
%%
\node[above] at (2.2,-0.5)				{\footnotesize{$\bfu-\bfv$}};
\end{tikzpicture}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Vector Subtraction - Spoiler}
\footnotesize
You can compute $\bfu - \bfv$ by \emph{componentwise subtraction}:

\bigskip

\begin{center}
\begin{tikzpicture}
\draw[cyan,thick,->,>=stealth]		(0,0)	--	(-1,2);
\node[left] at 	(-0.5,1)					{\footnotesize{$\bfv=\langle -1, 2 \rangle$}};
\draw[cyan,thick,->,>=stealth]		(0,0)	--	(3,1);
\node[above] at (1.5, 0.7)			{\footnotesize{$\bfu=\langle 3, 1 \rangle$}};
%%
\draw[green,thick,->,>=stealth]		(0,0)	--	(1,-2);
\node[left] at (0.5,-1)					{\footnotesize{$-\bfv$}};
%%
\draw[black,thick]						(1,-2)		--	++(3,1);
\draw[black,thick]						(3,1)		--	++(1,-2);
%%
\draw[red,thick,->,>=stealth]		(0,0)		--	(4,-1);
%%
\node[above] at (2.2,-0.5)				{\footnotesize{$\bfu-\bfv$}};
%%
\node[right] at (4,-1)					{\footnotesize{$\bfu - \bfv = \langle 4, -1 \rangle$}};
%%
\node[below] at (2.5,-3)				{\footnotesize{$\langle 3,1 \rangle - \langle -1,2 \rangle = \langle 4,-1 \rangle$}};
\end{tikzpicture}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Vector Algebra}
\footnotesize
We've seen three operations on vectors:  addition, scalar multiplication, and subtraction. Here are some basic rules for how these operations interact (see your text, p. 802, and know these properties!)

\bigskip

\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan} \textbf{Properties of Vectors}} 
If $\bfa$, $\bfb$, and $\bfc$ are vectors, and $c$, $d$ are scalars:


\begin{align*}
\bfa + \bfb &= \bfb + \bfa		&		\bfa + (\bfb + \bfc)	&=	(\bfa + \bfb) + \bfc \\
\bfa + \mathbf{0}	&=	\bfa	&		\bfa + (-\bfa)			&=	\mathbf{0}	\\
c(\bfa + \bfb)		&=	c\bfa + c\bfb	&	(c+d)\bfa		&=	c \bfa + d \bfa	\\
(cd)\bfa				&=	c(d\bfa)			&			1\bfa		&=	\bfa
\end{align*}
\end{minipage}
}
\end{frame}

\section{Components}

\begin{frame}
\frametitle{Components}
\footnotesize
Two- and three-dimensional vectors can be specified by their \emph{components}:

\bigskip

\uncover<2->
{
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.7]
\draw[thick,->,>=stealth]	(-1,0)	--	(4,0);
\draw[thick,->,>=stealth]	(0,-1)	--	(0,4);
%%
\node[left] at (0,4)			{\footnotesize{$y$}};
\node[below]	at (4,0)		{\footnotesize{$x$}};
\node at (-0.5,-0.5)			{\footnotesize{$O$}};
%%
\draw[thick,red,->,>=stealth]	(0,0)	--	(3,1.45);
\draw[black,fill=black]	(3,1.5)	circle(0.05cm);
\node[left] at (1.7,0.95)	{\footnotesize{$\mathbf{a}$}};
\node[above] at (3,1.5)	{\footnotesize{$\langle a_1, a_2 \rangle$}};
%%
\draw[dashed]	(3,0)		--	(3,1.5);
\draw[dashed]	(0,1.5)	--	(3,1.5);
\node[left]	at	(0,1.5)		{\footnotesize{$a_2$}};
\node[below] at (3,0)		{\footnotesize{$a_1$}};
%%
\node[below]	at	(1.5,-3)	{\footnotesize{$\bfa = \langle a_1, a_2 \rangle$}};
\end{tikzpicture}
\end{minipage}
}
\qquad
\uncover<3->
{
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.7]
\draw[thick,->,>=stealth]	(0,0,0)	--	++(4,0,0);
\draw[thick,->,>=stealth]	(0,0,0)	--	++	(0,4,0);
\draw[thick,->,>=stealth]	(0,0,0)	--	++(0,0,6);
%%
\node[left] 		at (0,0,6)		{\footnotesize{$x$}};
\node[right]	at	(4,0,0)		{\footnotesize{$y$}};
\node[above] 	at (0,4,0)		{\footnotesize{$z$}};
%%
\node[below]	at	(0,0,0)		{\footnotesize{$O$}};
%%
\draw[thick,red,>=stealth]	(0,0,0)	--	(2.9,3.9,2.9);
\draw[black,fill=black]			(3,4,3)	circle(0.05cm);
\node[above] at (3,4,3)		{\footnotesize{$\langle a_1, a_2, a_3 \rangle$}};
%%
\draw[dashed]					(0,0,0)	--	(3,0,3);
\draw[dashed]					(3,0,3)	--	(3,4,3);	
\node[below]	at	(3,0,3)		{\footnotesize{$\bfa = \langle a_1, a_2, 0 \rangle$}};
%%
\node[below]	at	(2,-3)			{\footnotesize{$\bfa = \langle a_1, a_2, a_3 \rangle$}};
\end{tikzpicture}
\end{minipage}
}
\end{frame}


\begin{frame}
\frametitle{Vector Operations in Components}
\footnotesize
\begin{itemize}
\item		The vector $\overrightarrow{AB}$ from $A(x_1,y_1,z_1)$ to $B(x_2,y_2,z_2)$ has components
			$$ \overrightarrow{AB}	= \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle $$
			\smallskip
\item		The \textbf{length} of a two-dimensional vector $\bfa = \langle a_1, a_2 \rangle$ is
			$$	\left| \bfa \right| = \sqrt{a_1^2 + a_2^2}$$
			\smallskip
\item		The \textbf{length} of a three-dimensional vector $\bfa = \langle a_1, a_2, a_3 \rangle$ is
			$$\left| \bfa \right|  = \sqrt{a_1^2 + a_2^2 + a_3^2 }$$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Vector Operations in Components}
\footnotesize
If $\bfa = \langle a_1, a_2 \rangle$ and $\bfb = \langle b_1, b_2 \rangle$, then:
\begin{align*}
\bfa + \bfb 	&=	\langle a_1 + b_1, a_2 + b_2 \rangle\\
\bfa -  \bfb	&=	\langle	a_1 - b_1, a_2 - b_2 \rangle\\
c \bfa 		&=	\langle c a_1, ca_2 \rangle
\end{align*}

\bigskip
What are the corresponding rules for three-dimensional vectors?

\bigskip
If $\bfa = \langle 2, 1, 2 \rangle$ and $\bfb = \langle 3, -1, 5 \rangle$, find:
\begin{itemize}
\item	$\bfa - \bfb$
\item	$2\bfa + 3 \bfb$
\item	$\left| \bfa - \bfb \right|$
\end{itemize}
\end{frame}

\section{Unit Vectors}

\begin{frame}
\frametitle{Standard Basis Vectors}
Every three-dimensional vector can be expressed in terms of the \textbf{standard basis vectors}
$$
\ii = \langle 1, 0, 0 \rangle, \quad
\jj = \langle 0, 1, 0 \rangle, \quad
\kk = \langle 0, 0, 1 \rangle
$$

\bigskip
If $\bfa = \langle a_1, a_2, a_3 \rangle$, then another way of writing $\bfa $ is
$$ a_1 \ii + a_2 \jj + a_3 \kk $$

\bigskip
The vectors $\ii$, $\jj$, $\kk$ have length $1$.  Any such vector is called a \emph{unit vector}.

\end{frame}

\begin{frame}
\frametitle{Unit Vectors}

\bigskip
You can make any nonzero vector a unit vector if you scalar multiply by the inverse of its length.

\bigskip
Find a unit vector in the direction of the vector $\ii + 2 \jj$

\bigskip
Find a unit vector in the direction of the vector $\ii  + \jj + \kk$

\end{frame}

\section{Applications}

\begin{frame}
\footnotesize

\begin{enumerate}
\item	A quarterback throws a football with an angle of elevation of $40^\circ$ and a speed of $60$ ft/sec. Find the horizontal and vertical components of the velocity.

\item	A crane suspends a 500 lb steel beam horizontally by support cables. Each support cable makes an angle of $60^\circ$ with the beam. The cables can withstand a tension of up to $275$ pounds. Would you feel safe standing below this rig?

\item	A boatman wants to cross a canal that is $3$ km wide and wants to land at a point $2$ km upstream from his starting point. The current in the canal flows at $3.5$ km/hr and the speed of his boat is $13$ km/hr.  
\begin{enumerate}
\item[(a)]	In what direction should he steer?
\item[(b)]	How long will the trip take?
\end{enumerate}

\end{enumerate}

\end{frame}

\begin{frame}
\frametitle{Lecture Review}
\begin{itemize}
\item		We saw that vectors are \emph{displacements} or instructions for moving from one point to another in the plane or in space
\item		We learned the operations of vector addition, vector subtraction, and scalar multiplication
\item		We learned how to express vectors in terms of \emph{components}
\item		We learned about the unit vectors $\ii$, $\jj$, and $\kk$ and how to form a \emph{unit vector} from any nonzero vector $\bfv$: multiply $\bfv$ by the reciprocal of its length
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Homework}
\begin{itemize}
\item	Review section 12.2 and prepare for your Thursday recitation.
\item	Continue working on homework A1 due Friday
\item	Read and study section 12.3 for Friday's lecture
\end{itemize}
\end{frame}
		

\endgroup

\end{document}
