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%TCIDATA{<META NAME="Title" CONTENT="Vector Calculus">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 17:11:16 1997}
%TCIDATA{Language=American English}
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\input{tcilatex}
\begin{document}


\section{Vector Calculus\label{Vectors}}

In this chapter, we introduce vectors and the use of vectors in calculus.

\subsection{\ Notation for Vectors}

\label{Vector} 
\index{Vector@Vector}You can enter vectors in \textsl{Scientific Notebook}
in any one of the following ways.

\begin{itemize}
\item  $1$ $\times $ $n$ matrices: $\left[ 
\begin{array}{ccc}
1 & 2 & 3
\end{array}
\right] $, $\left[ 
\begin{array}{cccccc}
5 & -1 & 3 & 17 & -8 & 2
\end{array}
\right] $, $\left[ 
\begin{array}{cccc}
x_{1} & x_{2} & x_{3} & x_{4}
\end{array}
\right] $

\item  $n\times 1$ matrices: $\left[ 
\begin{array}{r}
1 \\ 
0 \\ 
-1
\end{array}
\right] $, $\left[ 
\begin{array}{c}
35 \\ 
-4
\end{array}
\right] $, $\left( 
\begin{array}{c}
x_{1} \\ 
x_{2}
\end{array}
\right) $

\item  $n$-tuples within parentheses or brackets: $(2,-1,0)$, $\left(
x_{1},x_{2},x_{3}\right) $, $\lbrack 3,2,1\rbrack $, $\left[
x_{1},x_{2},x_{3}\right] $
\end{itemize}

\strut

This flexibility allows you to use the output of previous work as input,
without undue worry about the shape of the output. For purposes of clear
exposition, you will find it preferable to use consistent notation.

\begin{quote}
$\blacktriangleright $ To create a vector in matrix form
\end{quote}

\begin{enumerate}
\item[1.]  Click%
\FRAME{itbpF}{0.3009in}{0.3009in}{0.0701in}{}{}{matrix.wmf}{\special%
{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file
"F";width 0.3009in;height 0.3009in;depth 0.0701in;original-width
17.8125pt;original-height 16.9375pt;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'Matrix.wmf';file-properties "XNPEU";}} or
choose \textsf{Insert + Matrix.}

\item[2.]  Set the number of rows (or columns) to $1$ and the other number
to the dimension you wish.

\item[3.]  Enter the values for the components in the input boxes.\medskip
\end{enumerate}

\subsection{\ Vector Sums\label{Vector sums}}

\begin{quote}
$\blacktriangleright $ To evaluate a vector expression
\end{quote}

\begin{enumerate}
\item[1.]  Type the expression in mathematics mode

\item[2.]  Choose \textsf{Evaluate}.
\end{enumerate}

$\blacktriangleright $ \textsf{Evaluate}

\begin{quotation}
$\left( x_{1},x_{2},x_{3}\right) +\left( y_{1},y_{2},y_{3}\right)
=\allowbreak \left( x_{1}+y_{1},x_{2}+y_{2},x_{3}+y_{3}\right) \medskip $

$(2,-1,0)+(2,3,-5)=\allowbreak \left( 4,2,-5\right) \medskip $

$\left[ 1,2,3,4.3\right] -\left[ 3,-5.1,6,0\right] =\allowbreak \left[
-2,7.\,1,-3,4.\,3\right] \medskip $

$\left( 
\begin{array}{c}
a \\ 
b
\end{array}
\right) +\left( 
\begin{array}{c}
c \\ 
d
\end{array}
\right) =\allowbreak \left( 
\begin{array}{c}
a+c \\ 
b+d
\end{array}
\right) \medskip $

$\left[ 
\begin{array}{c}
1 \\ 
2
\end{array}
\right] -\left[ 
\begin{array}{c}
-3 \\ 
1
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{c}
4 \\ 
1
\end{array}
\right] \medskip $

$\left[ 
\begin{array}{cc}
x & y
\end{array}
\right] +\left[ 
\begin{array}{cc}
w & z
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{cc}
x+w & y+z
\end{array}
\right] \medskip $
\end{quotation}

\strut

\subsection{\ Dot Product 
\index{Dot product@Dot product}\label{Dot product}}

The \emph{dot product} (or \emph{inner product}) 
\index{Inner product@Inner product}\label{Inner product}of two vectors $%
\left( a_{1},a_{2},...,a_{n}\right) $ and $\left(
b_{1},b_{2},...,b_{n}\right) $ is defined by 
\[
\left( a_{1},a_{2},...,a_{n}\right) \cdot \left(
b_{1},b_{2},...,b_{n}\right) =a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n} 
\]
The dot is available on the \textsf{Common Symbols} toolbar and in the 
\textsf{Binary Operations} drop-down panel below 
\FRAME{itbpF}{0.3009in}{0.3009in}{0.0701in}{}{}{binop.wmf}{\special{language
"Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width
0.3009in;height 0.3009in;depth 0.0701in;original-width
24.8125pt;original-height 23.3125pt;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'Binop.wmf';file-properties "XNPEU";}}.

For the following examples of dot products with $n=3$, define

\[
a=\left[ 
\begin{array}{ccc}
1 & 2 & 3
\end{array}
\right] \text{, }b=\left( 
\begin{array}{r}
1 \\ 
0 \\ 
-1
\end{array}
\right) \text{, }c=[3,2,1]\text{, and }d=(2,-1,0) 
\]
with \textsf{New Definition} from the \textsf{Define} submenu.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\left( u,v,w\right) \cdot \left( x,y,z\right) =ux+vy+wz$

$(1,2,3)\cdot (3,2,1)=10$

$a\cdot c=10$

$a\cdot b=-2$

$c\cdot d=\,4\medskip $
\end{quotation}

\subsection{\ Cross Product 
\index{Cross product@Cross product}\label{Cross product}}

The \emph{cross product} of three-dimensional vectors $a=\left(
a_{1},a_{2},a_{3}\right) $ and $b=\left( b_{1},b_{2},b_{3}\right) $ is
defined by 
\[
a\times b=\left(
a_{2}b_{3}-a_{3}b_{2},a_{3}b_{1}-a_{1}b_{3},a_{1}b_{2}-a_{2}b_{1}\right) 
\]
To enter the cross symbol, click the cross on the \textsf{Common Symbols}
toolbar or in the drop-down panel below 
\FRAME{itbpF}{0.3009in}{0.3009in}{0.0701in}{}{}{binop.wmf}{\special{language
"Scientific Word";type "GRAPHIC";display "PICT";valid-file "F";width
0.3009in;height 0.3009in;depth 0.0701in;original-width
24.8125pt;original-height 23.3125pt;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'Binop.wmf';file-properties "XNPEU";}}.

For the following examples, use the vectors $a=\left[ 
\begin{array}{ccc}
1 & 2 & 3
\end{array}
\right] $, $b=\left( 
\begin{array}{r}
1 \\ 
0 \\ 
-1
\end{array}
\right) $, $c=[3,2,1]$, and $d=(2,-1,0)$, as defined previously.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$a\times b=\left( -2,4,-2\right) $\qquad $a\times c=\left( -4,8,-4\right) $%
\qquad $c\times d=\left( 1,2,-7\right) \medskip $

$\left( 
\begin{array}{r}
.35 \\ 
-.73 \\ 
1.2
\end{array}
\right) \times \left( 
\begin{array}{r}
.85 \\ 
.32 \\ 
-.77
\end{array}
\right) =\allowbreak \left( 
\begin{array}{c}
.\,\allowbreak 178\,1 \\ 
1.\,\allowbreak 289\,5 \\ 
.\,\allowbreak 732\,5
\end{array}
\right) $\qquad $\left[ 
\begin{array}{ccc}
1 & -2 & 5
\end{array}
\right] \times \left[ 
\begin{array}{ccc}
5 & 3 & -5
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{ccc}
-5 & 30 & 13
\end{array}
\right] $
\end{quotation}

\subsubsection{Triple Products}

Since the cross product of two vectors produces another vector, it is
possible to string cross products together. Use the same vectors $a$, $b$, $%
c $, and $d$ as before for these 
\index{Triple vector product@Triple vector product}\emph{triple vector
products}. Note that different choices of parentheses generally produce
different results. The default order of operations for products is from left
to right.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$a\times (b\times c)=\left( 16,4,-8\right) \medskip $

$\left( a\times b\right) \times c=\left( 8,-4,-16\right) \medskip $

$a\times \left( \left( b\times c\right) \times d\right) =\left( 0,0,0\right)
\medskip $

$\left( a\times \left( b\times c\right) \right) \times d=\left(
-8,-16,-24\right) \medskip $

$\left( \left[ 
\begin{array}{ccc}
1 & -2 & 5
\end{array}
\right] \times \left[ 
\begin{array}{ccc}
5 & 3 & -5
\end{array}
\right] \right) \times \left[ 
\begin{array}{ccc}
-7 & 2 & 8
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{ccc}
214 & -51 & 200
\end{array}
\right] \medskip $

$\left[ 
\begin{array}{ccc}
1 & -2 & 5
\end{array}
\right] \times \left[ \left[ 
\begin{array}{ccc}
5 & 3 & -5
\end{array}
\right] \times \left[ 
\begin{array}{ccc}
-7 & 2 & 8
\end{array}
\right] \right] =\left[ 
\begin{array}{ccc}
-37 & 139 & 63
\end{array}
\right] $
\end{quotation}

$\allowbreak $

To obtain intermediate results, select a portion of the expression and press
the \textsc{ctrl} key down while evaluating. This technique does an in-place
computation.

\begin{quote}
$\blacktriangleright $ \textsc{ctrl }\textsf{+ Evaluate, Evaluate}
\end{quote}

\begin{quotation}
$a\times (b\times c)=a\times \left( 2,-4,2\right) =\left( 16,4,-8\right)
\medskip $

$\left( a\times b\right) \times c=\left( -2,4,-2\right) \times c=\left(
8,-4,-16\right) \medskip $

$\left( \left[ 
\begin{array}{ccc}
1 & -2 & 5
\end{array}
\right] \times \left[ 
\begin{array}{ccc}
5 & 3 & -5
\end{array}
\right] \right) \times \left[ 
\begin{array}{ccc}
-7 & 2 & 8
\end{array}
\right] =\left[ 
\begin{array}{ccc}
-5 & 30 & 13
\end{array}
\right] \times \left[ 
\begin{array}{ccc}
-7 & 2 & 8
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{ccc}
214 & -51 & 200
\end{array}
\right] \medskip $

$\left[ 
\begin{array}{ccc}
1 & -2 & 5
\end{array}
\right] \times \left[ \left[ 
\begin{array}{ccc}
5 & 3 & -5
\end{array}
\right] \times \left[ 
\begin{array}{ccc}
-7 & 2 & 8
\end{array}
\right] \right] =\left[ 
\begin{array}{ccc}
1 & -2 & 5
\end{array}
\right] \times \left[ 
\begin{array}{ccc}
34 & -5 & 31
\end{array}
\right] =\left[ 
\begin{array}{ccc}
-37 & 139 & 63
\end{array}
\right] $
\end{quotation}

\begin{quote}
$\blacktriangleright $ \textsc{ctrl }\textsf{+ Evaluate, }\textsc{ctrl}%
\textsf{\ + Evaluate, Evaluate}
\end{quote}

\begin{quotation}
$a\times \left( \left( b\times c\right) \times d\right) =a\times \left(
\left( 2,-4,2\right) \times d\right) =a\times \left( 2,4,6\right) =\left(
0,0,0\right) \medskip $

$\left( a\times \left( b\times c\right) \right) \times d=\left( a\times
\left( 2,-4,2\right) \right) \times d=\left( 16,4,-8\right) \times d=\left(
-8,-16,-24\right) \medskip $

$\left( a\times b\right) \times \left( c\times d\right) =\left(
-2,4,-2\right) \times \left( 1,2,-7\right) =\left( -24,-16,-8\right)
\medskip $

$\left( \left( a\times b\right) \times c\right) \times d=\left( \left(
-2,4,-2\right) \times c\right) \times d=\left( 8,-4,-16\right) \times
d=\left( -16,-32,0\right) \bigskip $

$\left( 
\begin{array}{r}
.35 \\ 
-.73 \\ 
1.2
\end{array}
\right) \times \left( \left( \left( 
\begin{array}{r}
.85 \\ 
.32 \\ 
-.77
\end{array}
\right) \times \left( 
\begin{array}{r}
1.35 \\ 
-.23 \\ 
1.26
\end{array}
\right) \right) \times \left( 
\begin{array}{r}
-1.85 \\ 
.57 \\ 
.375
\end{array}
\right) \right) \medskip $

$\qquad \qquad \qquad =\left( 
\begin{array}{r}
.35 \\ 
-.73 \\ 
1.2
\end{array}
\right) \times \left( \left( 
\begin{array}{c}
.\,\allowbreak 226\,1 \\ 
-2.\,\allowbreak 110\,5 \\ 
-.\,\allowbreak 627\,5
\end{array}
\right) \times \left( 
\begin{array}{r}
-1.85 \\ 
.57 \\ 
.375
\end{array}
\right) \right) \medskip $

$\qquad \qquad \qquad =\left( 
\begin{array}{r}
.35 \\ 
-.73 \\ 
1.2
\end{array}
\right) \times \left( 
\begin{array}{c}
-.\,\allowbreak 433\,76 \\ 
1.\,\allowbreak 076\,1 \\ 
-3.\,\allowbreak 775\,5
\end{array}
\right) \allowbreak =\allowbreak \left( 
\begin{array}{c}
1.\,\allowbreak 464\,8 \\ 
.\,\allowbreak 800\,91 \\ 
.0\,\allowbreak 599\,9
\end{array}
\right) $
\end{quotation}

\strut

\begin{quotation}
\emph{Tip\quad }Parentheses are important. As always, careful and consistent
use of mathematical notation is in order. When in doubt, add an extra set of
parentheses to clarify an expression. \bigskip
\end{quotation}

When mixing cross products with scalar products, use parentheses for clarity.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$(1,0,1)\cdot \left( (1,2,3)\times (3,2,1)\right) =\,-8$

$\left( (1,0,1)\times (1,2,3)\right) \cdot (3,2,1)=\,-8\medskip $
\end{quotation}

Without these parentheses, you obtain different results for the first
product.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$(1,0,1)\cdot (1,2,3)\times (3,2,1)=\,4\left[ 
\begin{array}{c}
3 \\ 
2 \\ 
1
\end{array}
\right] $

$(1,0,1)\times (1,2,3)\cdot (3,2,1)=\,-8$

$(1,2,3)\times (3,2,1)\cdot (1,0,1)=\,-8\medskip $
\end{quotation}

\strut

\begin{quotation}
\emph{Note\quad } 
\index{Parallelepiped@Parallelepiped}\label{Parallelpiped}The triple scalar
product has an interesting geometric interpretation. The volume of the
parallelepiped spanned by three vectors $A$, $B$, and $C$ is equal to $%
\left| A\cdot \left( B\times C\right) \right| $. Thus, the volume of the
parallelepiped spanned by $\left( 1,1,0\right) $, $\left( 1,0,1\right) $,
and $\left( 0,1,1\right) $ is given by 
\[
\left| \left( 1,1,0\right) \cdot \left[ \left( 1,0,1\right) \times \left(
0,1,1\right) \right] \right| =\,2 
\]
In particular, this value does not depend on the order of the vectors in the
triple scalar product. 
\begin{eqnarray*}
\left| \left( 1,0,1\right) \cdot \left[ \left( 1,1,0\right) \times \left(
0,1,1\right) \right] \right| &=&2 \\
\left| \left( 0,1,1\right) \cdot \left[ \left( 1,1,0\right) \times \left(
1,0,1\right) \right] \right| &=&2
\end{eqnarray*}
\end{quotation}

\strut

\subsection{\ Vector Norms\label{Vector norms}}

You can compute 
\index{Vector@Vector!norm@norm}vector norms in \textsl{Scientific Notebook}
for every positive integer $n$, where 
\[
\Vert v\Vert _{n}=\left( \sum \left| v_{i}\right| ^{n}\right) ^{%
\frac{1}{n}} 
\]
and for $\infty $, with 
\[
\left\| v\right\| _{\infty }=\max \left( \left| v_{i}\right| \right) 
\]
as illustrated by the following examples.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\Vert \left( a,b,c\right) \Vert _{1}=\allowbreak \left| a\right| +\left|
b\right| +\left| c\right| \vspace{6pt}$

$\Vert \left( a,b,c\right) \Vert _{3}=\allowbreak \sqrt[3]{\left( \left|
a\right| ^{3}+\left| b\right| ^{3}+\left| c\right| ^{3}\right) }\vspace{6pt}$%
\ 

$\Vert \left( -1,2,1\right) \Vert _{5}=\allowbreak \sqrt[5]{34}\vspace{6pt}$

$\Vert \left[ a,b,c\right] \Vert _{\infty }=\,\max \left( \left| a\right|
,\left| b\right| ,\left| c\right| \right) \vspace{6pt}$\ 

$\Vert \left[ 8,-10,2\right] \Vert _{\infty }=\allowbreak 10\vspace{6pt}$

$\left\| \left( 
\begin{array}{c}
a \\ 
b
\end{array}
\right) \right\| _{2}\vspace{6pt}=\sqrt{\left( \left| a\right| ^{2}+\left|
b\right| ^{2}\right) }$

$\left\| \left( 
\begin{array}{ccc}
a & b & c
\end{array}
\right) \right\| _{4}=\sqrt[4]{\left( \left| a\right| ^{4}+\left| b\right|
^{4}+\left| c\right| ^{4}\right) }\medskip $
\end{quotation}

\begin{quote}
$\blacktriangleright $ To enter the two pairs of vertical lines used in the
norm symbol
\end{quote}

\begin{itemize}
\item  Click \FRAME{itbpF}{0.3009in}{0.3009in}{0.0701in}{}{}{brackets1.wmf}{%
\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file
"F";width 0.3009in;height 0.3009in;depth 0.0701in;original-width
20.3125pt;original-height 18.8125pt;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'brackets1.wmf';file-properties "XNPEU";}} or
choose \textsf{Insert + Brackets}, and select the norm symbols.
\end{itemize}

Before doing the next set of examples, make the following definition.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Define + New Definition}
\end{quote}

\begin{quotation}
$v=[3,2,1]\medskip $
\end{quotation}

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate, Evaluate Numerically}
\end{quote}

\begin{quotation}
$\left\| v\right\| _{1}=\,6\vspace{6pt}$

$\left\| v\right\| _{2}=\,\sqrt{14}=\,3.7417\vspace{6pt}$

$\left\| v\right\| _{6}=\,\sqrt[6]{794}=\,3.043\vspace{6pt}$

$\left\| v\right\| _{20}=\,\sqrt[20]{34878\,32978}=\,3.000045103\vspace{6pt}$

$\left\| v\right\| _{\infty }=\,3\medskip $
\end{quotation}

This series of examples suggests that for a vector $v$, 
\[
\lim _{n\rightarrow \infty }\left\| v\right\| _n=\left\| v\right\| _\infty 
\]

\label{Euclidean norm}The default $\left\| v\right\| $ is the 2-norm, which
is also known as the 
\index{Euclidean norm@Euclidean norm}\textsl{Euclidean norm}.\medskip
\bigskip

\begin{example}
\label{Parallelogram}The area of the parallelogram in the plane with
vertices $\left( 0,0\right) $, $\left( a_{1},a_{2}\right) $, $\left(
b_{1},b_{2}\right) $, and $\left( a_{1}+b_{1},a_{2}+b_{2}\right) $ is given
by 
\[
\Vert \left( a_{1},a_{2},0\right) \times \left( b_{1},b_{2},0\right) \Vert 
\]
In particular, the area of the parallelogram spanned by the two vectors $%
\left( 1,2\right) $ and $\left( 2,1\right) $ is given by 
\[
\left\| \left( 1,2,0\right) \times \left( 2,1,0\right) \right\| =\,3 
\]
\end{example}

Since 
\[
A\cdot B=\left\| A\right\| \left\| B\right\| \cos \theta 
\]
where $\theta $ is the angle between the vectors $A$ and $B$, you can use
the dot product to find the angle between two vectors.\medskip

\begin{example}
Define $A=\left( 1,2,-3\right) $ and $B=\left( -2,1,2\right) $ and solve the
equation $A\cdot B=\left\| A\right\| \left\| B\right\| \cos \theta $, to get 
\[
\text{Solution is: }\left\{ \theta =\pi -\arccos \left( \frac{1}{7}\sqrt{14}%
\right) \right\} \vspace{6pt} 
\]
Apply \textsf{Evaluate Numerically} to get $\theta =\,2.1347.\medskip $
\end{example}

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Gradient, Divergence, and Curl}{}{}{DM9-2.tex#Gradient}

\item  \hyperref{Plots of Vector Fields and Gradients}{}{}{DM9-3.tex#Vector
field}

\item  \hyperref{Scalar and Vector Potentials}{}{}{DM9-4.tex#Scalar potential%
}

\item  \hyperref{Matrix-Valued Operators}{}{}{DM9-5.tex#Hessian}

\item  \hyperref{Plots of Complex Functions}{}{}{DM9-6.tex#Conformal plot}

\item  \hyperref{Exercises and Solutions}{}{}{DM9-7.tex#Exercises}
\end{itemize}

\rule{0.67in}{0.01in}

\hyperref{\FRAME{itbpF}{0.5275in}{0.3399in}{0.0519in}{}{}{index.wmf}{\special%
{language "Scientific Word";type "GRAPHIC";display "PICT";valid-file
"F";width 0.5275in;height 0.3399in;depth 0.0519in;original-width
36.875pt;original-height 22.5625pt;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'index.wmf';file-properties "XNPEU";}}}{}{}{%
DMIndex.tex#Math Index}\quad {\small Index entries: \hyperref{Cross product}{%
}{}{DMIndex.tex#Cross product}, \hyperref{Inner product}{}{}{%
DMIndex.tex#Inner product}, \hyperref{Matrices, \hyperref{Norm}{}{}{%
DMIndex.tex#Norm}}{}{}{DMIndex.tex#Matrices}, \hyperref{Vector}{}{}{%
DMIndex.tex#Vector}}

\end{document}
