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\begin{document}

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


\title{Math 213 - The Cross Product}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{September 4, 2019}
\maketitle


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

\begin{frame}
\frametitle{Reminders}

\begin{itemize}

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

\medskip

\item	Homework A2 on Sections 12.3-12.4 is due Friday

\medskip

\item	Applications for an alternate Exam 1 are due \textbf{today!}
		
		\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
\item[{12.3}]		Lecture 3: The Dot Product
\begingroup
\color{red}
\item[{12.4}]		Lecture 4: The Cross Product
\endgroup
\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 determinants of orders $2$ and $3$

\item		Know how to compute the cross product $\bfa \times \bfb$ of two vectors using determinants
			and understand its geometric meaning

\medskip

\item		Learn the properties of the cross product

\medskip

\item		Understand the scalar triple product and its geometric meaning

\item		Understand how to use cross products to compute torque
\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{Determinants}

\begin{frame}
\frametitle{Determinants}
\footnotesize
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
$$
\begin{vmatrix}
a_1	&	a_2	&	a_3	\\
b_1	&	b_2	&	b_3	\\
c_1	&	c_2	&	c_3
\end{vmatrix}
=
a_1 
\begin{vmatrix} 
	b_2 & b_3 \\ c_2 & c_3 
\end{vmatrix}
%
{\color{red} -} a_2 
\begin{vmatrix}
	b_1	&	b_3	\\	c_1	&	c_3	
\end{vmatrix}
%
+ a_3
\begin{vmatrix}
	b_1	&	b_2	\\	c_1	&	c_2
\end{vmatrix}
$$
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}

\medskip

Find, using your favorite method:

$$
\begin{vmatrix}
1	&	-4	&	-2	\\
2	&	0	&	4	\\
-1	&	2	&	3	
\end{vmatrix}
$$
\end{frame}

\section{Cross Product}

\begin{frame}
\footnotesize
\begin{center}
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
If $\bfa = \langle a_1, a_2, a_3 \rangle$ and $\bfb = \langle b_1, b_2, b_3 \rangle$, then the \textbf{cross product} of $\bfa$ and $\bfb$ is the \underline{vector}
$$ \bfa \times \bfb = \begin{vmatrix}
\ii		&	\jj		&	\kk	\\
a_1	&	a_2	&	a_3	\\
b_1	&	b_2	&	b_3
\end{vmatrix} $$
\end{minipage}
}
\end{center}

\bigskip
\begin{minipage}{0.30\textwidth}
\begin{tikzpicture}[scale=0.7]
\draw[thick,black,->,>=stealth]	(0,0,0)	--	(3,0,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	(0,3,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	(0,0,3);
%%
\draw[cyan,thick,->,>=stealth]	(0,0,0)	--	(2,1,0);
\node[right]	at	(2,1,0)			{\footnotesize{$\bfa$}};
\draw[cyan,thick,->,>=stealth]	(0,0,0)	--	(1,1,1);
\node[above]	at	(1,1,1)			{\footnotesize{$\bfb$}};
%%
\draw[red,thick,->,>=stealth]	(0,0,0)	--	(1,-2,1);
\node[below]	at 	(1,-2,1)			{\footnotesize{$\bfa \times \bfb$}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.60\textwidth}
\footnotesize
Find $\bfa \times \bfb$ if $\bfa = \langle 1,0,2 \rangle$ and $\bfb = \langle 1, 1, 1 \rangle$. 
\begin{align*}
\uncover<2->
{
\bfa \times \bfb	&=	\begin{vmatrix}
									\ii		&	\jj		&	\kk	\\
									1		&	0		&	2		\\
									1		&	1		&	1
								\end{vmatrix}\\
}
\uncover<3->
{
						&= 	\ii \begin{vmatrix} 0 & 2 \\ 1 & 1 \end{vmatrix} {\color{red} -}
								\jj	\begin{vmatrix} 	1	&	2	\\	1	&	1	\end{vmatrix} 
						+
								\kk \begin{vmatrix}	1	&	0	\\	1	&	1	\end{vmatrix}\\
}
\uncover<4->
{
						&=	-2\ii +\jj + \kk
}
\end{align*}
\end{minipage}

\end{frame}

\begin{frame}
\begin{minipage}{0.30\textwidth}
\begin{tikzpicture}[scale=0.7]
\draw[thick,black,->,>=stealth]	(0,0,0)	--	(3,0,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	(0,3,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	(0,0,3);
%%
\draw[cyan,thick,->,>=stealth]	(0,0,0)	--	(2,1,0);
\node[right]	at	(2,1,0)			{\footnotesize{$\bfa$}};
\draw[cyan,thick,->,>=stealth]	(0,0,0)	--	(1,1,1);
\node[above]	at	(1,1,1)			{\footnotesize{$\bfb$}};
%%
\only<1>
{
\draw[red,thick,->,>=stealth]	(0,0,0)	--	(1,-2,1);
}
\node[below]	at 	(1,-2,1)			{\footnotesize{$\bfa \times \bfb$}};
%%
\uncover<2->
{
\draw[purple,thick,->,>=stealth]	(0,0,0)	--	(-1, 2, -1);
\node[above] at (-1.2,2,-1)			{\footnotesize{$\bfb \times \bfa$}};
\draw[thick,dashed,red,->,>=stealth]	(0,0,0)	--	(1,-2,1);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.60\textwidth}
\footnotesize
\begin{align*}
\bfa	&=	\ii + 0\jj + 2\kk	\\
\bfb	&=	\ii + \jj + \kk 	\\
\bfa \times \bfb	&=	-2\ii + \jj + \kk 
\end{align*}
What are $\bfa \cdot (\bfa \times \bfb)$ and $\bfb \cdot (\bfa \times \bfb)$?

\bigskip

What is $\bfb \times \bfa$?
\end{minipage}
\end{frame}

\section{Properties}

\begin{frame}
\frametitle{Cross Product Properties}
\footnotesize
\begin{enumerate}
\item		The cross product $\bfa \times \bfb$ is perpendicular to $\bfa$ and $\bfb$ with direction given by the right-hand rule
\item		$|\bfa \times \bfb|$ is the area of the parallelogram spanned by $\bfa$ and $\bfb$, i.e.,
			$$ | \bfa \times \bfb | = |\bfa|\, |\bfb| \, \sin \theta $$
\item		Two vectors $\bfa$ and $\bfb$ are parallel if and only if $\bfa \times \bfb = 0$
\end{enumerate}

\medskip

\hrulefill

\medskip

\begin{enumerate}
\item	Find $\ii \times \jj$, $\ii \times \kk$, and $\jj \times \kk$
\item Use this result to find $(2\jj - 4 \kk) \times (-\ii + 3 \jj + \kk)$
\item	Which of the following expressions is meaningful?
		\begin{align*}
		\bfa \cdot (\bfb \times \bfc)	&&	\bfa \times (\bfb \times \bfc)	\\
		\bfa \times (\bfb \times \bfc)	&&	\bfa \cdot (\bfb \cdot \bfc)	\\
		(\bfa \cdot \bfb) \times (\bfc \cdot \bfd)	&&	(\bfa \times \bfb) \cdot (\bfc \times \bfd)
		\end{align*}
\end{enumerate}


\end{frame}

\begin{frame}
\frametitle{Cross Product Properties}
\footnotesize
Can you fill in the blanks?

\medskip

If $\bfa$, $\bfb$, and $\bfc$ are vectors and if $c$ is a scalar:
\begin{enumerate}
\item	$\bfa \times \bfb = -\bfb \times \bfa$
\item	$(c\bfa) \times \bfb = \underline{\hphantom{c}} (\bfa \times \bfb) =\bfa \times (\underline{\hphantom{c\bfb}})$
\item	$\bfa \times (\bfb + \bfc) = \underline{\hphantom{\bfa \times \bfb + \bfa \times \bfc}}$
\item	$(\bfa + \bfb) \times \bfc = \underline{\hphantom{\bfa \times \bfc + \bfb \times \bfc}}$
\item	$\bfa \cdot (\bfb \times \bfc) = (\bfa \times \bfb) \cdot \bfc$
\item	$\bfa \times (\bfb \times \bfc) = (\bfa \cdot \bfc) \bfb - (\bfa \cdot \bfb) \bfc$
\end{enumerate}
\end{frame}

\begin{frame}
\frametitle{Cross Product Properties}
\footnotesize
Can you fill in the blanks?

\medskip

If $\bfa$, $\bfb$, and $\bfc$ are vectors and if $c$ is a scalar:
\begin{enumerate}
\item	$\bfa \times \bfb = -\bfb \times \bfa$
\item	$(c\bfa) \times \bfb = \underline{{c}} (\bfa \times \bfb) =\bfa \times ({c\bfb})$
\item	$a \times (\bfb + \bfc) = \underline{{\bfa \times \bfb + \bfa \times \bfc}}$
\item	$(\bfa + \bfb) \times \bfc = \underline{{\bfa \times \bfc + \bfb \times \bfc}}$
\item	$\bfa \cdot (\bfb \times \bfc) = (\bfa \times \bfb) \cdot \bfc$
\item	$\bfa \times (\bfb \times \bfc) = (\bfa \cdot \bfc) \bfb - (\bfa \cdot \bfb) \bfc$
\end{enumerate}
\end{frame}

\begin{frame}
\frametitle{Cross Product Puzzlers}
\footnotesize
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.6]
\draw[thick,->,>=stealth]		(0,0,0)	--	(0,0,6);
\draw[thick,->,>=stealth]		(0,0,0)	--	(0,7,0);
\draw[thick,->,>=stealth]		(0,0,0)	--	(6,0,0);
%%
\draw[black,fill=black]		(1,0,2)	circle(0.1cm);
\draw[black,fill=black]		(3,3,3)	circle(0.1cm);
\draw[black,fill=black]		(7,5,8)	circle(0.1cm);
\draw[black,fill=black]		(5,2,7)	circle(0.1cm);
%%
\draw[dashed]	(1,0,2)	--	(5,2,7)	--	(7,5,8)	--	(3,3,3)	--	(1,0,2);
%%
\node[below,fill=white] at (1,-0.25,2)		{\scriptsize{$P(1,0,2)$}};
\node[above,fill=white] at (3,3.25,3)			{\scriptsize{$Q(3,3,3)$}};
\node[above,fill=white] at (7.25,5.25,8)			{\scriptsize{$R(7,5,8)$}};
\node[below,fill=white] at (5.3,1.75,7)		{\scriptsize{$S(5,2,7)$}};
\end{tikzpicture}
\end{minipage}
\begin{minipage}{0.45\textwidth}
\footnotesize
Find the area of the parallelogram with vertices 
$$P(1,0,2), \quad Q(3,3,3), $$ 
$$R(7,5,8), \quad S(5,2,7).$$

\vfill

\uncover<2->
{
Note that:

$$\overrightarrow{PQ} = \langle 2, 3, 1 \rangle$$
$$\overrightarrow{PS} = \langle 4,2,5 \rangle$$
}


\end{minipage}
\end{frame}

\begin{frame}
\frametitle{Cross Product Puzzlers}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]		(0,0,0)	--	(0,0,5);
\draw[thick,->,>=stealth]		(0,0,0)	--	(0,5,0);
\draw[thick,->,>=stealth]		(0,0,0)	--	(5,0,0);
%%
\node[left] at (0,0,5)			{\scriptsize{$x$}};
\node[right] at (5,0,0)		{\scriptsize{$y$}};
\node[above] at (0,5,0)		{\scriptsize{$z$}};

%%
\draw[black,fill=black]	(0,1,1)	circle(0.1cm);
\node[left] at				(0,1,1)				{\scriptsize{$P(1,0,1)$}};
\draw[black,fill=black]	(1,3,-2)	circle(0.1cm);
\node[right] at 			(1,3,-2)				{\scriptsize{$Q(-2,1,3)$}};
\draw[black,fill=black]	(2,5,4)	circle(0.1cm);
\node[left]	 at 			(1.5,5,4)				{\scriptsize{$R(4,2,5)$}};
%%
\draw[dashed]			(0,1,1)	--	(1,3,-2)	--	(2,5,4)	--	(0,1,1);
%%
\uncover<2->
{
\draw[cyan,thick,->,>=stealth]	(0,1,1)	--	(2,5,4);
\draw[cyan,thick,->,>=stealth]	(0,1,1)	--	(1,3,-2);
}
\uncover<3->
{
\draw[thick,red,->,>=stealth]	(0,1,1,)	--	(4,-1,1);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
Find a nonzero vector orthogonal to to the plane through $P(1,0,1)$, $Q(-2,1,3)$, and $R(4,2,5)$ and find the area of triangle $PQR$

\bigskip

\uncover<2->
{
Two sides of the triangle are spanned by:
\begin{align*}
\overrightarrow{PR}	&=	\langle 3, 2, 4 \rangle\\
\overrightarrow{PQ}	&=	\langle -3,1,2 \rangle
\end{align*}
}

\uncover<3->
{
The cross product is:
$$
\overrightarrow{PQ}\times \overrightarrow{PR} = \langle 0, 18, -9 \rangle
$$
(not drawn to scale!)
}
\end{minipage}
\end{frame}



\section{Scalar Triple Product}

\begin{frame}
\frametitle{Scalar Triple Product}
\footnotesize
The \emph{scalar triple product} of three vectors $\bfa$, $\bfb$, and $\bfc$ is the determinant
$$ \bfa \cdot (\bfb \times \bfc) = 
\begin{vmatrix}
a_1	&	a_2	&	a_3	\\
b_1	&	b_2	&	b_3	\\
c_1	&	c_2	& 	c_3
\end{vmatrix}
$$

\bigskip

\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.6]
\draw[red,thick,->,>=stealth]	(0,0,0)	--	(4,0,0);		%%	b
\draw[thick,red,->,>=stealth]	(0,0,0)	--	(1,3,0);		%%	a
\draw[thick,red,->,>=stealth]	(0,0,0)	--	(0,0,-2);		%%	c
%%
\node[right] at		(0.60,2,0)		{\footnotesize{$\bfa$}};
\node[right] at		(0.25,0,-1)			{\footnotesize{$\bfc$}};
\node[below] at 	(2,0,0)			{\footnotesize{$\bfb$}};
%%
\draw[dashed]	(0,0,-2)	--	(4,0,-2);
\draw[dashed]	(0,0,-2)	--	(1,3,-2);
%%
\draw					(1,3,0)	--	(5,3,0);
\draw					(4,0,0)	--	(5,3,0);
\draw					(4,0,0)	--	(4,0,-2);
\draw					(1,3,0)	--	(1,3,-2);
\draw					(1,3,-2)	--	(5,3,-2);
\draw					(5,3,-2)	--	(5,3,0);
\draw					(5,3,-2)	--	(4,0,-2);
%%
\draw[cyan,thick,->,>=stealth]	(0,0,0)	--	(0,8,0);
\node[right] at (0,5,0)			{\footnotesize{$\bfb \times \bfc$}};
%%
\begin{scope}
\clip	(0,0,0)	--	(0,8,0)	--	(2,6,0)	--	(0,0,0);
\draw		(0,0,0)	circle(1.95cm);
\end{scope}
\node at (0.225,1.5)			{\footnotesize{$\theta$}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
The volume of the parallelepiped formed by the vectors $\bfa$, $\bfb$, and $\bfc$ is given by
$$ |\bfa \cdot (\bfb \times \bfc) |. $$

\bigskip

What happens if the vectors $\bfa$ , $\bfb$, and $\bfc$ are coplanar?
\end{minipage}

\end{frame}



\begin{frame}
\frametitle{Triple Product Puzzler}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]		(0,0,0)	--	(0,0,5);
\draw[thick,->,>=stealth]		(0,0,0)	--	(0,5,0);
\draw[thick,->,>=stealth]		(0,0,0)	--	(6,0,0);
%%
\node[left] at (0,0,5)			{\scriptsize{$x$}};
\node[right] at (6,0,0)			{\scriptsize{$y$}};
\node[above] at (0,5,0)		{\scriptsize{$z$}};
%%
\draw[black,fill=black]	(1,0,-2)	circle(0.1cm);
\draw[black,fill=black]	(3,2,2)	circle(0.1cm);
\draw[black,fill=black]	(4,-1,1)	circle(0.1cm);
\draw[black,fill=black]	(6,1,3)	circle(0.1cm);
%%
\draw[cyan,->,>=stealth]		(1,0,-2)	--	(3,2,2);
\draw[cyan,->,>=stealth]		(1,0,-2)	--	(4,-1,1);
\draw[cyan,->,>=stealth]		(1,0,-2)	--	(6,1,3);
%%
\node[left] at (1,0,-2)			{\scriptsize{$P$}};
\node[above] at (3,2,2)		{\scriptsize{$Q$}};
\node[below] at	(4,-1,1)		{\scriptsize{$R$}};
\node[below]	at	(6,1,3)		{\scriptsize{$S$}};
%%
\draw[densely dashed,gray]	(3,2,2)	--	++(3,-1,3)	--	++(5,1,5)	--	++(-3,1,-3)	--	++(-5,-1,-5);
\draw[densely dashed,gray]	(4,-1,1)	--	++(2,2,4);
\draw[densely dashed,gray]	(4,-1,1)	--	++(5,1,5)	--	++(-3,1,-3);
\draw[densely dashed,gray]	(4,-1,1)	--	++(5,1,5)	--	++(2,2,4);
\draw[densely dashed,gray]	(6,1,3)	--	++(2,2,4);
%%
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
Find the volume of the parallelepiped with adjacent edges $PQ$, $PR$, and $PS$ if
$$P=P(-2,1,0), \quad Q=Q(2,3,2),$$ 
$$R=R(1,4,-1) \quad S=S(3,6,1)$$

\bigskip

\uncover<2->
{
\begin{align*}
\overrightarrow{PQ}	&=	\langle 4,2,2 \rangle	\\
\overrightarrow{PR}	&=	\langle	3,3,-1 \rangle \\
\overrightarrow{PS}	&=	\langle	5,5,1 \rangle
\end{align*}
}
\end{minipage}
\end{frame}

\section{Torque}

\begin{frame}
\frametitle{Torque}
If a force $\bfF$ acts on a rigid body at a position $\bfr$, the \textbf{torque} $\tau$ is given by
$$\tau= \bfr \times \bfF $$
The torque vector points along the axis of rotation

\bigskip

Two friends sit on a see-saw. Friend 1 weighs 100 N and sits 2 m to the left of the fulcrum, while friend 2 was 50N and sits 5 m to the right of the fulcrum. Find the torques and predict which way the see-saw will go. 

\end{frame}

\begin{frame}

\frametitle{Dot Product, Cross Product, Triple Product}

\medskip

\footnotesize

{\renewcommand{\arraystretch}{1.6}
\begin{tabular}{llll}
	&	\textbf{Dot Product} & \textbf{Cross Product} & \textbf{Scalar Triple Product}\\
	\\
\textbf{Type}	&	Scalar $\bfa \cdot \bfb$	
					&	Vector $\bfa \times \bfb$	
					& Scalar $\bfa \cdot (\bfb \times \bfc)$\\
\textbf{Magnitude}
					& $|\bfa | | \bfb| \cos\theta$	
					& $|\bfa| |\bfb| \sin \theta$\\
\textbf{Symmetry}	
					&$\bfa \cdot \bfb = \bfb \cdot \bfa$	
					& $\bfa \times \bfb = -\bfb \times \bfa$ 
					& Antisymmetric \\
\textbf{Direction}
					&	None!	
					&	Right-hand rule 
					& None!\\
\textbf{In Physics}		
					&	Work		
					&	Torque	
					&	\\
\textbf{In Geometry}	
					&	Projection	
					&	Parallelogram Area	
					&	Parallelepiped Area
\end{tabular}
}
\end{frame}

\section{Review}

\begin{frame}
\frametitle{Lecture Review}
\begin{itemize}
\item	We reviewed rules for computing $2 \times 2$ and $3 \times 3$ determinants
\item	We defined the \emph{cross product} of two vectors, found its properties, and  found its geometric meaning (right hand rule, area of parallelogram)
\item  We defined the \emph{scalar triple product} of three vectors and found its geometric meaning (signed volume of a parallelipiped)
\item	  We reviewed basics of the dot product (scalar) and the cross product (vector)
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Homework}

\begin{center}
\fbox{
\begin{minipage}{0.8\textwidth}
\begin{itemize}
\item Webwork A2 is due Friday
\item Re-read section 12.4, pp.\  814--821
\item	Prepare for your recitation and study for Quiz \#1 on sections 12.1--12.2
\item Remember to access WebWork \emph{only through Canvas!}
\end{itemize}

Also, read section 12.5, pp.\ 823--830 for Friday. We'll spend two lectures on this material
\end{minipage}
}
\end{center}

\end{frame}


\endgroup

\end{document}
