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\begin{document}

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


\title{Math 213 - Equations of Lines and Planes, I}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{September 6, 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 tonight!

\medskip

\item	Applications for an alternate Exam 1 \textbf{were due Wednesday!}
		
		\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
\item[{12.4}]		Lecture 4:The Cross Product
\begingroup
\color{red}
\item[{12.5}]		Lecture 5:	Equations of Lines and Planes, I
\endgroup
\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		Learn how to write the  parametric equation of a line
\item		Learn how to write the symmetric equation of a line
\item		Learn how to write the vector equation of a plane
\item		Learn how to write the scalar equation of a plane
\end{itemize}
\end{frame}

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

\begingroup
\small



\section{Line - Parametric}

\begin{frame}
\frametitle{Line - Vector Equation}
\begin{minipage}{0.45\textwidth}
%%
%%	For this line, P_0 = (2,2,2)
%%	v = (0,0,-3) in TikZ coordinates
%%
\begin{tikzpicture}[scale=0.65]
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,0,5);
\draw[thick,->,>=stealth]	(0,0,0)	--	(5,0,0);
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,5,0);
%%
\node[below]	at	(0,0,5)	{\footnotesize{$x$}};
\node[right]	at	(5,0,0)	{\footnotesize{$y$}};
\node[above]	at	(0,5,0)	{\footnotesize{$z$}};
%%%%%%%%%%%%%%%%%%%%
%
%		Slide 1	-  	the line (dashed)
%
%%%%%%%%%%%%%%%%%%%%
\draw[cyan,dashed]  (2,3,6) -- (2,3,-6);
%%%%%%%%%%%%%%%%%%%%
%
%		Slide 2 	-	a point on the line
%
%%%%%%%%%%%%%%%%%%%%
\uncover<2->
{
\draw[black,fill=black]		 (2,3,0)	circle(0.1cm);
\node[right] at					(2,3,0)	{\scriptsize{$\bfr_0=(x_0,y_0,z_0)$}};
}
%%%%%%%%%%%%%%%%%%%%
%
%		Slide 3	-	a vector along the line
%
%%%%%%%%%%%%%%%%%%%%
%
%		the vector is emphasized in slide 3
%
\only<3>
{
\draw[very thick,red,fill=red,->,>=stealth]
%					(2,2,2)	--	(2,2.5,-2);
					(2,3,0)	--	++	(0,0,-3);
\node[right] at (2,3,-3)	{\scriptsize{$\bfv=\langle a,b,c \rangle$}};
}
%%%%%%%%%%%%%%%%%%%%
%
%		Slide 4 - a generic point r_0 + tv
%
%%%%%%%%%%%%%%%%%%%%
\uncover<4->
{
%%%%%
%
%		Mute the vector v
%%%%
\draw[very thick,red,fill=red,->,>=stealth,opacity=0.5]
					(2,3,0)	--	++	(0,0,-3);
\node[right,color=gray] at (2,3,-3)	{\scriptsize{$\bfv=\langle a,b,c \rangle$}};
%%%%
%
%		Show the vector addition
%
%%%%
\draw[black,fill=black]	(2,3,-6)	circle(0.1cm);
\node[right]	at	(2,3,-6)	{\scriptsize{$P$}};
%%
\draw	[purple,thick,->,>=stealth]
	(0,0,0)	--	(2,3,0);
\node[below]	at	(1,1,1)	{\scriptsize{$\bfr_0$}};
\draw[purple,thick,->,>=stealth]
	(2,3,0)	--	++(0,0,-6);
\node[left] at (2,3,-6)	{\scriptsize{$t\bfv$}};
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
A line $L$ in three-dimensional space is determined by
\begin{itemize}
\uncover<2->
{
\item		A point $\bfr_0 = (x_0,y_0,z_0)$ on the line
}
\uncover<3->
{
\item		A vector $\bfv=\langle a,b,c \rangle $ that gives the direction of the line
}
\end{itemize}
\vfill
\uncover<4->
{Any point $P$ on the line can be expressed as 
$$ \bfr_0 + t \bfv $$
for some real number $t$ called the \emph{parameter}
}
\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Line - Vector Equation}
\begin{minipage}{0.45\textwidth}
%%%%
%
%		This code was copied from the previous slide... maybe sloppy coding
%
%%%%
\begin{tikzpicture}[scale=0.65]
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,0,5);
\draw[thick,->,>=stealth]	(0,0,0)	--	(5,0,0);
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,5,0);
%%
\node[below]	at	(0,0,5)	{\footnotesize{$x$}};
\node[right]	at	(5,0,0)	{\footnotesize{$y$}};
\node[above]	at	(0,5,0)	{\footnotesize{$z$}};
%%
\draw[cyan,dashed]  (2,3,6) -- (2,3,-6);
%%
\draw[black,fill=black]		 (2,3,0)	circle(0.1cm);
\node[right] at					(2,3,0)	{\scriptsize{$\bfr_0=(x_0,y_0,z_0)$}};
%%
\draw[black,fill=black]	(2,3,-6)	circle(0.1cm);
\node[right]	at	(2,3,-6)	{\scriptsize{$P$}};
%%
\draw	[purple,thick,->,>=stealth]
	(0,0,0)	--	(2,3,0);
\draw[purple,thick,->,>=stealth]
	(2,3,0)	--	++(0,0,-6);
\node[left] at (2,3,-6)	{\scriptsize{$t\bfv$}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
If
$$ \bfr_0 = \langle x_0, y_0, z_0 \rangle,
\quad \bfv = \langle a,b,c \rangle, $$
the \emph{function}
$$ \bfr(t) = \bfr_0 + t \bfv $$
traces out a line through $$P =(x_0, y_0, z_0)$$ in the direction of $$\bfv = \langle a, b, c \rangle$$

\end{minipage}
\end{frame}

\begin{frame}
\frametitle{Line - Parametric Equation}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.65]
%%%%%%%%
%
%		Another copy of the same code...
%
%%%%%%%
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,0,5);
\draw[thick,->,>=stealth]	(0,0,0)	--	(5,0,0);
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,5,0);
%%
\node[below]	at	(0,0,5)	{\footnotesize{$x$}};
\node[right]	at	(5,0,0)	{\footnotesize{$y$}};
\node[above]	at	(0,5,0)	{\footnotesize{$z$}};
%%
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,0,5);
\draw[thick,->,>=stealth]	(0,0,0)	--	(5,0,0);
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,5,0);
%%
\node[below]	at	(0,0,5)	{\footnotesize{$x$}};
\node[right]	at	(5,0,0)	{\footnotesize{$y$}};
\node[above]	at	(0,5,0)	{\footnotesize{$z$}};
%%
\draw[cyan,dashed]  (2,3,6) -- (2,3,-6);
%%
\draw[black,fill=black]		 (2,3,0)	circle(0.1cm);
\node[right] at					(2,3,0)	{\scriptsize{$\bfr_0=(x_0,y_0,z_0)$}};
%%
\draw[black,fill=black]	(2,3,-6)	circle(0.1cm);
\node[right]	at	(2,3,-6)	{\scriptsize{$P$}};
%%
\draw	[purple,thick,->,>=stealth]
	(0,0,0)	--	(2,3,0);
\draw[purple,thick,->,>=stealth]
	(2,3,0)	--	++(0,0,-6);
\node[left] at (2,3,-6)	{\scriptsize{$t\bfv$}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\medskip
If $\bfr(t) = \langle x(t), y(t), z(t) \rangle$ then
\begin{align*}
x(t)	&=	x_0	+ at\\
y(t)	&=	y_0	+bt\\
z(t) 	&=	z_0	+ct
\end{align*}
\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Line - Parametric Equation}
\footnotesize
If $\bfr(t) = \langle x(t), y(t), z(t) \rangle$ then
\begin{align*}
x(t)	&=	x_0	+ at\\
y(t)	&=	y_0	+bt\\
z(t) 	&=	z_0	+ct
\end{align*}
gives the parametric equations for a line through $P(x_0,y_0,z_0)$ in direction $\langle a,b,c \rangle$

\medskip

\hrulefill

\medskip

\begin{enumerate}
\item		Find the parametric equations of a line $L$ through the points $P(1,2,-1)$ and $Q(2,3,4)$.
\item		Find the parametric equations of the line $L$ through the point $(1,2,3)$ and parallel to the vector 
$\langle 2, -3, 4 \rangle$
\end{enumerate}
\end{frame}

\section{Line - Symmetric}

\begin{frame}
\frametitle{Line - Symmetric Equation}
\footnotesize
If we begin with the parametric equations of a line:
\begin{align*}
x(t)	&=	x_0	+ at	\\
y(t)	&=	y_0 	+  bt 	\\
z(t)	&=	z_0	+	ct
\end{align*}
\uncover<2->
{
we can eliminate the parameter to get the 
\emph{symmetric equation of a line};
$$
\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}
$$
The numbers $(a,b,c)$ are the \emph{direction numbers}
of the line.
}

\medskip

\hrulefill

\medskip

\uncover<3->
{

\begin{enumerate}
\item	Find the parametric and symmetric equations of the line through the origin and the point $(4,3,-1)$

\item	Find the parametric and symmetric equations of the line through $(2,1,0)$ and perpendicular to both
$ \ii+\jj$ and $\jj+\kk$.
\end{enumerate}
}
\end{frame}

\section{Plane - Vector}

\begin{frame}
\frametitle{Plane - Vector Equation}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,0,7);
\draw[thick,->,>=stealth]	(0,0,0)	--	(7,0,0);
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,7,0);
%%
\node[below]	at	(0,0,7)	{\footnotesize{$x$}};
\node[right]	at	(7,0,0)	{\footnotesize{$y$}};
\node[above]	at	(0,7,0)	{\footnotesize{$z$}};
%%
\draw[dashed,fill=cyan!20,opacity=0.5]	(0,6,0)	--	(0,0,6)	--	(6,0,0)	--	(0,6,0);
%%
\draw[black,fill=black]	(4,1,1)	circle(0.1cm);
\node[above] at			(4,1,1)	{\scriptsize{$Q$}};
%%
\uncover<2->
{
\draw[black,fill=black]	(2,2,2)	circle(0.1cm);
\node[above]	at	(1.8,2,2)	{\scriptsize{$P_0$}};
}
%%
\uncover<3->
{
\draw[cyan,thick,->,>=stealth]	(2,2,2)	--	(6,6,6);
\node[right] at (6,6,6)		{\scriptsize{$\bfn$}};
\draw[thick,red,->,>=stealth]	(2,2,2) -- (4,1,1);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
A \emph{plane} is the collection of all points $Q$:
\begin{itemize}
\uncover<2->
{
\item		Passing through  given point $P_0(x_0,y_0,z_0)$
}
\uncover<3->
{
\item		Having the property that $\overrightarrow{P_0Q}$ is perpendicular to a vector $\bfn$, the \emph{normal vector}

\smallskip

}
\uncover<4->
{That is
$$ \bfn \cdot \overrightarrow{P_0Q}=0 $$
}

\uncover<5->
{
If $\bfr_0 = \overrightarrow{OP_0}$, $\bfr = \overrightarrow{OQ}$, then$\ldots$
}
\end{itemize}
\end{minipage}
\footnotesize

\medskip
\uncover<6->
{
\begin{center}
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.5\textwidth}
$ \bfn \cdot (\bfr - \bfr_0) = 0 \quad $ OR 
$ \quad\bfn \cdot \bfr = \bfn \cdot \bfr_0$
\end{minipage}
}
\end{center}
}

\end{frame}

\section{Plane - Scalar}

\begin{frame}
\frametitle{Plane - Scalar Equation}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,0,7);
\draw[thick,->,>=stealth]	(0,0,0)	--	(7,0,0);
\draw[thick,->,>=stealth]	(0,0,0)	--	(0,7,0);
%%
\node[below]	at	(0,0,7)	{\footnotesize{$x$}};
\node[right]	at	(7,0,0)	{\footnotesize{$y$}};
\node[above]	at	(0,7,0)	{\footnotesize{$z$}};
%%
\draw[dashed,fill=cyan!20,opacity=0.5]	(0,6,0)	--	(0,0,6)	--	(6,0,0)	--	(0,6,0);
%%
\draw[black,fill=black]	(4,1,1)	circle(0.1cm);
\node[above] at			(4,1,1)	{\scriptsize{$Q$}};
%%

{
\draw[black,fill=black]	(2,2,2)	circle(0.1cm);
\node[above]	at	(1.8,2,2)	{\scriptsize{$P_0$}};
}
%%
{
\draw[cyan,thick,->,>=stealth]	(2,2,2)	--	(6,6,6);
\node[right] at (6,6,6)		{\scriptsize{$\bfn$}};
\draw[thick,red,->,>=stealth]	(2,2,2) -- (4,1,1);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
A \emph{plane} is the collection of all points $Q$:
\begin{itemize}
{
\item		Passing through  given point $P_0(x_0,y_0,z_0)$
}
{
\item		Having the property that $\overrightarrow{P_0Q}$ is perpendicular to a vector $\bfn$, the \emph{normal vector}
\medskip
}
{That is
$$ \bfn \cdot \overrightarrow{P_0Q}=0 $$
}

\uncover<2->
{
If $Q = (x,y,z)$, $\bfn = \langle a, b, c \rangle$, then $\ldots$
}
\end{itemize}
\end{minipage}
\footnotesize

\medskip

\uncover<3->
{
\begin{center}
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.5\textwidth}
$a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$
\end{minipage}
}
\end{center}
}

\end{frame}

\begin{frame}
\frametitle{Plane Puzzlers}
\footnotesize
Let
$$ P_0=P_0(x_0,y_0,z_0), \quad
\bfn = \langle a,b,c \rangle, \quad P=P(x,y,z) $$

\smallskip

The \textbf{vector equation} of the plane through $P_0$ with normal $\bfn$ is
$$ \bfn \cdot \bfr = \bfn \cdot \bfr_0 $$
where $\bfr$ and $\bfr_0$ are position vectors for $P$ and $P_0$ respectively.

\smallskip
The \textbf{scalar equation} of the plane through $P_0$ with normal $\bfn$ is
$$ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 $$



\hrulefill

\smallskip

\begin{enumerate}
\item	Find the vector equation of a plane through the origin and perpendicular to the vector $\langle -1, 2, 5 \rangle$

\item	Find the scalar equation of the plane through $(1,-1,-1)$ and parallel to the plane $5x-y-z=6$

\item	Find the equation of the plane that contains the line $x=1+t, y=2-t, z=4-3t$ and is parallel to the plane $5x+2y+z=1$
\end{enumerate}
\end{frame}

\section{Review}

\begin{frame}
\frametitle{Summary}
\footnotesize
\begin{itemize}
\item	We learned that a line is determined by a point $P_0=(x_0,y_0,z_0)$ on the line, and a vector $\bfv = \langle a,b,c \rangle $ that points along the line
\begin{itemize}
\item	The \emph{parametric} equations of a line are
		$$
		x(t) 	=	x_0 + at, \quad
		y(t) 	=	y_0 + bt, \quad
		z(t) 	=	z_0 + ct
		$$
\item	The \emph{symmetric} equations of a line are
		$$ \frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} $$
\end{itemize}
\item	We learned that a plane is determined by a point $P_0=(x_0,y_0,z_0)$ in the plane and a vector $\bfn = \langle a,b,c \rangle$ \emph{normal} to the plane
\begin{itemize}
\item	The \emph{vector} equation of a plane is
		$$ \bfn \cdot ( \bfr - \bfr_0 ) = 0 $$
\item	The \emph{scalar} equation of a plane is
		$$ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 $$
\end{itemize}
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Homework}

\begin{center}
\fbox{
\begin{minipage}{0.8\textwidth}
\begin{itemize}
\item Webwork A2 is due tonight!
\item Re-read section 12.5, pp.\  823--830
\end{itemize}

\end{minipage}
}
\end{center}

\end{frame}


\endgroup

\end{document}