\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\ol}{\overline}
\parindent=0pt
\begin{document}

\begin{center}
{\bf Vector Algebra and Geometry}
\end{center}

{\bf Scalar and Vector Quantities}

A scalar quantity is some physical quantity with which is
associated a measure of its magnitude, but no idea of spatial
direction, examples being mass, volume, temperature, electrical
charge, atmospheric pressure.

A vector quantity is a physical quantity with which is associated
a magnitude and a direction in space, examples being velocity,
force, momentum.

Many such quantities combine with a similar law of combination, so
that the same algebra can be used to describe them.  Notice that
the physical quantities described above may have other properties
than magnitude and direction.  For example a force may have a
point of application in many situations.  However we shall be
concerned only with abstracting the properties of magnitude and
direction and modelling these.

We shall consider translations or displacements in 2 or 3
dimensions in order to formulate the idea of a vector, for such a
translation in space is completely characterised by its magnitude
and direction.  We can think of a translation as a function which
maps each point of space onto another point of space in a
particular way.  On a diagram or in space such a translation is
completely specified by giving the image of any one point.

So in the plane if I speak of the translation which maps $(1,2)$
onto $(3,4)$ then we automatically know what happens to every
other point in the plane, and similarly with a translation in
3-space.

If a translation takes $A$ to $B$ we can denote it by $\vec{AB}$.
Such a translation will map a point $P$ onto a point $Q$ say, so
that we might have $\vec{AB}=\vec{PQ}$.

This will be the case if the line segments $AB$ and $PQ$ have the
same length and direction.  We can further characterise the idea
of direction by imagining a directed line segment, so that
$\vec{AB}=\vec{PQ}$ if the length of $AB$ equals that of $PQ$,
$AB$ is parallel to $PQ$ and the sense of $AB$ is the same as that
of $PQ$.

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\put(0.75,1){\vector(2,1){1}}

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\put(3.5,0.4){\makebox(0,0){we sometimes say that}}

\put(3.5,0.25){\makebox(0,0){these are anti-parallel.}}
\end{picture}

Notice that $\vec{AB}$ does not stand for the directed line
segment.  If it did then clearly $\vec{AB}=\vec{PQ}$ is false as
the two line segments are different.  Rather $\vec{AB}$ denotes
the translation $T$ which takes $A$ to $B$.  Since $T$ also takes
$P$ to $Q$, $T$ can also be symbolised by $\vec{PQ}$.

We shall want to use symbols for vectors indendent of any
particular line segment used to represent such vectors and so we
shall use symbols such as ${\bf a}$, ${\bf b}$ etc.  The magnitude
of a vector is the distance of the translation, i.e. the length of
any representative line segment.  It is called the modulus of the
vector, and is denoted by $|{\bf a}|$ or just $a$.  In books you
will find vectors denoted by heavy type, and their magnitude using
the same letter in ordinary type.

Given a vector ${\bf a}$ and a vector ${\bf b}$ we form their sum
as follows.  Let $A$ be an arbitrary point in space.  Vector ${\bf
a}$ translates $A$ to some point $B$, so that ${\bf a}=\vec{AB}$.
${\bf b}$ then translates $B$ to some point $C$ so that ${\bf
b}=\vec{BC}$.  We then define ${\bf a}+{\bf b}=\vec{AC}$.  This is
easily remembered in the form $\vec{AB}+\vec{BC}=\vec{AC}$.

We use the symbol + only because the algebraic properties
correspond to those of number algebra, as we shall see.  It is
really translation ${\bf a}$ followed by translation ${\bf b}$.
For if we translate $A$ to $B$ then $B$ to $C$ this has the same
effect as translating $A$ straight to $C$.

\newpage

In a diagram we have

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\put(0.3,0.5){\makebox(0,0){$A$}}

\put(2,2.2){\makebox(0,0){$B$}}

\put(3.7,1.5){\makebox(0,0){$C$}}
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this is therefore sometimes called the triangle law of addition.

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\put(2,2){\vector(3,-1){1.5}}

\put(0.3,0.5){\makebox(0,0){$A$}}

\put(1.9,2.2){\makebox(0,0){$B$}}

\put(3.7,1.5){\makebox(0,0){$C$}}

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\multiput(2,2)(0.1,0.1){16}{\circle*{0.05}}

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\put(3.6,3.7){\makebox(0,0){$P$}}

\put(5.3,3){\makebox(0,0){$Q$}}

\put(6,3.5){\makebox(0,0)[l]{If we draw}}

\put(6,3.1){\makebox(0,0)[l]{$BP=AB$}}

\put(6,2.7){\makebox(0,0)[l]{$PQ=BC \,\,\, PQ//BC$}}

\put(6,2.3){\makebox(0,0)[l]{$CQ=AB \,\,\, CQ//AB$}}

\put(6,1.9){\makebox(0,0)[l]{then $BQ=AC$ and $BQ//AC$}}
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so that $\vec{BP}+\vec{BC}=\vec{BQ}$.

It is therefore also called the parallelogram law of addition.

The reason why vector algebra is useful in applied mathematics is
that many physical quantities are found experimentally to combine
according to this law.

If we let the following system come to equilibrium then we find
that the forces at P measured by the weights and direction of
string satisfy the triangle of forces.  The result of $w_1$ and
$w_2$ acting on P in the directions shown is equal and opposite to
$w_3$.

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\put(3.95,1.6){\line(0,1){0.4}}

\put(4.35,1.6){\line(0,1){0.4}}

\put(4.15,1.4){\makebox(0,0){$w_2$}}
\end{picture}

The diagrams below do not reflect the fact that the forces are
acting at the same point.  They only reflect the magnitude and
direction properties.

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\put(2,1){\line(-2,1){1}}

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\put(1.7,1.4){\makebox(0,0){$w_1$}}


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\put(5.4,0.1){\makebox(0,0){$w_1$}}

\put(5,0.5){\line(1,1){1}}
\end{picture}

We can see from the diagram that
$|\vec{AC}|<|\vec{AB}|+|\vec{BC}|$.

If however $A,\,B,\,C$ are in line then
$|\vec{AC}|=|\vec{AB}|+|\vec{BC}|$, so generally we have $|{\bf
a}+ {\bf b}| \leq |{\bf a}|+|{\bf b}|$.

We have a problem if $C$ coincides with $A$, for then $\vec{AC}$
does not make sense, there is no direction specified.  However to
make the algebra work we introduce the idea of a zero vector, {\bf
0} corresponding to the identity transformation, mapping each
point to itself.  We then have ${\bf a}+{\bf 0}={\bf 0}+{\bf
a}={\bf a}$ for all ${\bf a}$.

Also if ${\bf a}=\vec{AB}$ then $\vec{AB}+\vec{BA}={\bf 0}$, and
we write $\vec{BA}=-{\bf a}$, by analogy with numbers.

Since vector addition is a new operation with new objects we need
to see what the algebra is like

\begin{itemize}
\item[i)]
commutative law of addition


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\put(4.2,2.1){\makebox(0,0){{\bf a}}}

\put(0.5,1.5){\line(1,0){4}}
\end{picture}

Consider the parallelogram $PQRS$ where $\vec{PQ}={\bf a} \,\,\,
\vec{QR}={\bf b}$.  Then $\vec{PS}={\bf b}$ and $\vec{SR}={\bf
a}$.  Therefore applying the triangle law to the two separate
triangles $PQR$ and$PSR$ we have ${\bf a} +{\bf b}={\bf b}+{\bf
a}$.

\item[ii)]
the associative law of addition

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\put(0.8,1.6){\makebox(0,0){{\bf a}}}

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\put(1.5,1){\vector(4,-1){1}}

\put(2.5,0.5){\makebox(0,0){{\bf a+b}}}

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\put(2.4,3.2){\makebox(0,0){{\bf b+c}}}

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\put(1.6,1.6){\makebox(0,0){{\bf b}}}
\end{picture}

We can now add any number of vectors in any order we like to get
the same result.  $\vec{PR}={\bf a}+({\bf b+c})=({\bf a+b})+{\bf
c}$

\item[iii)]
subtraction

having already defined $-{\bf a}$ (negative {\bf a}) it seems
natural to define ${\bf b}-{\bf a}$ to be ${\bf b}+(-{\bf a})$.
Denoting this by ${\bf x}$ we see that ${\bf b}+(-{\bf a})={\bf
x}$ is equivalent to ${\bf x}+{\bf a}={\bf b}$.

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\put(4,3){\vector(4,-3){1}}

\put(5.1,2.5){\makebox(0,0){{\bf -a}}}
\end{picture}

So ${\bf b}-{\bf a}$ is the solution {\bf x} of the equation ${\bf
x}+{\bf a}={\bf b}$.
\end{itemize}


{\bf Multiplication by a number}

In ordinary algebra when we meet an expression like $x+x+x+y+y$ we
abbreviate it to $3x+2y$.  So with vectors we abbreviate ${\bf
a}+{\bf a}+{\bf a}$ to $3{\bf a}$.

In fact if ${\bf a}=\vec{PQ}=\vec{QR}=\vec{RS}$ then
$\vec{PS}=3{\bf a}$.

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\put(0.3,0.1){\makebox(0,0){$P$}}

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\put(4.3,2.1){\makebox(0,0){$R$}}

\put(6.3,3.1){\makebox(0,0){$S$}}
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In fact we see that $\vec{PS}$ is a vector with 3 times the
magnitude and the same sense and direction as {\bf a}.  Similarly
if we interpret $-3{\bf a}$ as $(-{\bf a})+(-{\bf a})+(-{\bf a})$
then from the same diagram we see that $-3{\bf a}=\vec{SP}$.  We
therefore adopt the following definition.

Given {\bf a} and any real number $k$, we define $k{\bf a}$ as
follows

\begin{itemize}
\item[1)]
$|k{\bf a}|=k|{\bf a}|$

\item[2)]
The direction of $k{\bf a}$ is that of {\bf a} if $k>0$.

\item[3)]
The direction of $k{\bf a}$ is opposite that of {\bf a} if $k<0$.

\item[4)]
If $k=0$ we define $k{\bf a}={\bf 0}$.
\end{itemize}


{\bf Distributive laws}

\begin{itemize}
\item[i)]
$(k+l){\bf a}=(k{\bf a})+(l{\bf a})$

\item[ii)]
$k({\bf a}+{\bf b})=(k{\bf a})+(k{\bf b})$

\item[iii)]
$k(l{\bf a})=(kl){\bf a} \hspace{1in}$ for all $k,\,l,\,{\bf
a},\,{\bf b}.$

\item[i)]
suppose $k,l$ are both $>0$ and ${\bf a}\not={\bf 0}$.

then $|(k+l){\bf a}|=|k+l||{\bf a}|=(k+l)|{\bf a}|=k|{\bf
a}|+l|{\bf a}|=|k{\bf a}|+|l{\bf a}|$

Furthermore $k{\bf a}, \,\, l{\bf a}$ and $(k+l){\bf a}$ all have
the same direction and sense.  Thus $(k+l){\bf a}=k{\bf a}+l{\bf
a}$.

There are numerous other cases to consider, with $k,\,l$ either or
both negative or zero.

\item[ii)]
${}$

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\put(3.2,3.6){\makebox(0,0){$R$}}

\put(2.7,2.4){\makebox(0,0){{\bf b}}}

\put(5,3){\line(1,-1){2}}

\put(5,3){\vector(1,-1){1}}

\put(4.8,2.9){\makebox(0,0){$P'$}}

\put(7.1,0.7){\makebox(0,0){$Q'$}}

\put(6,1.8){\makebox(0,0){k{\bf a}}}

\put(5,3){\line(2,1){4}}

\put(7,1){\line(1,2){2}}

\put(7,1){\vector(1,2){1}}

\put(9.2,5.1){\makebox(0,0){$R'$}}

\put(8.2,2.8){\makebox(0,0){k{\bf b}}}
\end{picture}

Let $PQR$ be a triangle with $\vec{PQ}={\bf a} \,\,\,
\vec{QR}={\bf b} \,\,\, {\rm so \ } \vec{PR}={\bf a+b}$.

Let $P'Q'R'$ be a triangle with sides parallel to $PQR$ and


$\vec{P'Q'}=k{\bf a} \,\,\, \vec{Q'R'}=k{\bf b} \,\,\, {\rm then \
} \vec{P'R'}=k{\bf a}+k{\bf b}$.

But the triangles $PQR$ and $P'Q'R'$ are similar.  So
$|\vec{PR}|=|k||\vec{PR}|$ and so $\vec{P'R'}=k\vec{PR}$ thus
$k{\bf a}+k{\bf b}=k({\bf a+b})$.

This does not cover the cases where $k$, {\bf a} or {\bf b} is
zero, or where {\bf a} and {\bf b} are parallel.

\item[iii)]
left and right-hand sides both represent vectors with the same
magnitude, direction and sense and so are the same.
\end{itemize}

{\bf Summary}

The system $V$ of vectors with the system $R$ of real numbers have
operations of addition and multiplication by a real number defined
with the following properties

\begin{itemize}
\item[a)]
for all ${\bf a},{\bf b}\epsilon V \,\,\,\, {\bf a}+{\bf
b}\epsilon V\,\,\,\,$ (Closure)

\item[b)]
for all ${\bf a},{\bf b},{\bf c}\epsilon V \,\,\,\, ({\bf a}+{\bf
b})+{\bf c}={\bf a}+({\bf b}+{\bf c}) \,\,\,\,$ (Associativity)

\item[c)]
there exists ${\bf 0}\epsilon V,$ for all ${\bf a}\epsilon V
\,\,\,\, {\bf a}+{\bf 0}={\bf a} \,\,\,\,$ ({\bf 0} is an
identity)

\item[d)]
for all ${\bf a}\epsilon V$ there exists ${\bf b}\epsilon V
\,\,\,\, {\bf a}+{\bf b}={\bf 0} \,\,\,\,$ ({\bf b} is an inverse
for {\bf a} ({\bf b}=-{\bf a})

\item[e)]
for all ${\bf a},{\bf b}\epsilon V \,\,\,\, {\bf a}+{\bf b}={\bf
b}+{\bf a}\,\,\,\,$ (Commutativity)

\item[f)]
for all $k\epsilon{\bf R}$, for all$ {\bf a}\epsilon V \,\,\,\,
k{\bf a}\epsilon V$

\item[g)]
for all $k,l\epsilon{\bf R}$, for all ${\bf a}\epsilon V \,\,\,\,
k(l{\bf a})=(kl){\bf a}$

\item[h)]
for all ${\bf a}\epsilon V \,\,\,\, 1*{\bf a}={\bf a}$

\item[i)]
for all $k\epsilon{\bf R}$, for all ${\bf a},{\bf b}\epsilon V
\,\,\,\, k({\bf a+b})=(k{\bf a})+(k{\bf b})$ (distributive law)

\item[j)]
for all $k,l\epsilon{\bf R}$, for all ${\bf a}\epsilon V \,\,\,\,
(k+l){\bf a}=(k{\bf a})+(l{\bf a})$ (distributive law)
\end{itemize}

Notice that we did not include h) in our definition above.  We
haven't included $0*{\bf a}={\bf a}$.  In fact we can deduce this
algebraically from the rules above as follows.

$k{\bf a}=(0+k){\bf a}=0{\bf a}+k{\bf a} \hspace{0.95in}$ by j)

add $-(k{\bf a})$ to both sides

$\begin{array}{rclc}k{\bf a}+-(k{\bf a})&=&(0{\bf a}+k{\bf
a})+-(k{\bf a})\\ k{\bf a}+-(k{\bf a})&=&0{\bf a}+(k{\bf
a})+-(k{\bf a})&{\rm by\ b)}\\ {\bf 0}&=&0{\bf a}+{\bf 0}& {\rm
by\ d)}\\ {\bf 0}&=&0{\bf a}&{\rm by\ c)}\end{array}$

There are many algebraic systems having the above properties. Such
a system is a vector space.

Example

Let $V$ be the set of arithmetic progressions
$(a,\,a+d,\,a+2d,\,a+3d,\, ...)$.  We define addition term by
term, and multiplication by a real number term by term, then $V$
has all the properties (a)-(j).  i.e. $V$ is an example of a
vector space.

{\bf Independence}

Consider the three vectors {\bf a}, {\bf b}, {\bf c}, where ${\bf
a}=\vec{PQ}, \,\, {\bf b}=\vec{QR}, \,\, {\bf c}=\vec{RP}$


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then ${\bf a}+{\bf b}+{\bf c}={\bf 0}$, there is a linear
relationship between them.  However it is possible to choose
vectors where there is no such relationship.  Consider two vectors
{\bf a} and {\bf b}.  Choose points $PQR$ so that ${\bf
a}=\vec{PQ}, \,\, {\bf b}=\vec{PR}$.

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\multiput(2.5,2.5)(0.15,0){27}{\circle*{0.05}}

\multiput(2,0.5)(0.15,0){17}{\circle*{0.05}}

\multiput(4.55,0.5)(0.1,0.1){21}{\circle*{0.05}}

\multiput(2.5,2.5)(0.1025,-0.1){20}{\circle*{0.05}}

\put(0.3,0.5){\makebox(0,0){$P$}}

\put(1.3,1.6){\makebox(0,0){$Q$}}

\put(2,0.3){\makebox(0,0){$R$}}

\put(3,1.7){\makebox(0,0){$S$}}

\put(2.3,2.6){\makebox(0,0){$Q'$}}

\put(4.55,0.3){\makebox(0,0){$R'$}}

\put(6.8,2.6){\makebox(0,0){$S'$}}

\put(0.8,1.1){\makebox(0,0){{\bf a}}}

\put(1.3,0.3){\makebox(0,0){{\bf b}}}
\end{picture}

Let $\vec{PQ'}=k{\bf a}$ and $\vec{PR'}=l{\bf b}$, then
$\vec{PS'}={\bf a+b}$ and $\vec{PS'}=k{\bf a}+l{\bf b}$.  By the
parallelogram rule we can see that $S$, $S'$ lie in the plane
determined by $PQR$.  So if {\bf c} is a vector whose direction is
not parallel to this plane then {\bf c} will not be of the form
$k{\bf a}+l{\bf b}$.  {\bf c} or any multiple of it cannot be
expressed as a linear combination of {\bf a} and {\bf b}.  In fact
the equation $k{\bf a}+l{\bf b}+m{\bf c}={\bf 0}$ will only be
possible if $k=l=m=0$.  This leads us to frame the following
definitions.

The vectors ${\bf a}_1, \, ..., \, {\bf a}_n$ are linearly
independent if $k_1{\bf a}_1+k_2{\bf a}_2+...+k_n{\bf a}_n={\bf
0}$ only holds with $k_1=k_2=...=k_n=0$.

If there is a solution with some of the $k$'s non zero then the
vectors are linearly dependent.  If for example $k_1\not=0$ then
we can write

${\bf a}_1=-\frac{k_2}{k_1}{\bf a}_2-...-\frac{k_n}{k_1}{\bf
a}_n=m_2{\bf a}_2+...+m_n{\bf a}_n$.

${\bf a}_1$ is a linear combination of the vectors ${\bf a}_2, \,
..., \, {\bf a}_n$.  What we are interested in is finding a set of
vectors having the property that all other vectors can be obtained
as linear combinations of these vectors.  We would naturally look
for a set of independent vectors with this property.  Many vector
spaces have a finite set with this property, that they generate
the whole space through linear combinations.  Such a set is called
a basic set, or a basis.  It will not be unique, but it can be
proved that all bases contain the same number of vectors.  This
number is called the dimension of the space.

For example, the set of AP's has dimension 2, with basis

$(1,1,1,1,1,1,...)\,(0,1,2,3,4,5,...)$

{\bf Components}

A vector {\bf u} is called a unit vector if $|{\bf u}|=1$.  If
{\bf a} is any non-zero vector then there are two unit vectors
associated with {\bf a}, $\ds \hat{{\bf a}}=\frac{{\bf a}}{|{\bf
a}|}$ and $\ds -\hat{{\bf a}}=-\frac{{\bf a}}{|{\bf a}|}$.

Let ${\bf u}=\vec{OP}$ be a unit vector, and let $\Pi$ be a plane
through 0 perpendicular to {\bf u}.  Let {\bf a} be an arbitrary
vector not parallel to $\Pi$ and let ${\bf a}=\vec{OA}$ (so $A$ is
not in $\Pi$).  Choose a point $Q$ in $\Pi$ such that $QA$ is
parallel to {\bf u}, so $\vec{QA}=a_1{\bf u}$ for some number
$a_1$.

Then $\vec{OA}=\vec{OQ}+\vec{QA}$

so   ${\bf a}={\bf q}+a_1{\bf u} \hspace{0.5in}({\bf q}=\vec{OQ})$

This gives a decomposition for {\bf a}.  With a fixed unit vector
{\bf u} this decomposition is unique, for if

${\bf a}=a_1{\bf u}+{\bf q}=a_1'{\bf u}+{\bf q}'$

then $(a_1-a_1'){\bf u}={\bf q}'-{\bf q}$

The LHS is in the direction of {\bf u} and the RHS is
perpendicular {\bf u}, unless both are zero, so $a_1=a_1'$ and
${\bf q}={\bf q}'$.

The number $a_1$ is called the component of {\bf a} in the
direction of {\bf u}.

Now choose three unit vectors at right angles and call them ${\bf
u}_1, \,{\bf u}_2, \,{\bf u}_3$.

Write ${\bf u}_1=\vec{OL}, \, {\bf u}_2=\vec{OM}, \, {\bf
u}_3=\vec{ON}$.

Given an arbitrary vector ${\bf a}=OA$

we have $\vec{OA}=\vec{OQ}+\vec{QP}+\vec{PA}=a_1{\bf u}_1+a_2{\bf
u}_2+a_3{\bf u}_3$.

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\put(2,2){\line(1,0){2.5}}

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\put(4,2){\line(-1,-1){1.25}}

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\put(2,2){\line(1,2){0.75}}

\put(2,2){\vector(1,2){0.5}}

\put(2.3,3){\makebox(0,0){{\bf a}}}

\put(2.75,3.5){\circle*{0.1}}

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Again the expression is unique, for $a_2{\bf u}_2+a_3{\bf u}_3$ is
in the plane $OMN$.

This means that ${\bf u}_1, \, {\bf u}_2, \, {\bf u}_3$ is a
basis, and $a_1, \, a_2, \, a_3$ are called the components of {\bf
a} (relative to the basis ${\bf u}_1, \, {\bf u}_2, \, {\bf u}_3$)

We can write ${\bf a}=(a_1, \, a_2, \, a_3)$

Suppose now we have vectors

${\bf a}=a_1{\bf u}_1+a_2{\bf u}_2+a_3{\bf u}_3 \hspace{0.75in}
{\bf b}=b_1{\bf u}_1+b_2{\bf u}_2+b_3{\bf u}_3$

Then by the laws of vector algebra.

${\bf a}+{\bf b}=(a_1+b_1){\bf u}_1+(a_2+b_2){\bf
u}_2+(a_3+b_3){\bf u}_3$

$k{\bf a}=(ka_1){\bf u}_1+(ka_2){\bf u}_2+(ka_3){\bf u}_3$

i.e. $(a_1,a_2,a_3)+(b_1,b_2,b_3)=(a_1+b_1,a_2+b_2,a_3+b_3)$

$k(a_1,a_2,a_3)=(ka_1,ka_2,ka_3)$

Now it can be verified that the set of all triples with operations
defined above is a vector space.  It is therefore isomorphic to
the space of displacement vectors.

Since the three axes are orthogonal, Pythagoras' theorem gives

$OA^2=OQ^2+QP^2+PA^2$

$|{\bf a}|^2=a_1^2+a_2^2+a_3^2$

In place of ${\bf u}_1, \, {\bf u}_2, \, {\bf u}_3$ a notation
widely used in three dimensions for three orthogonal unit vectors
is ${\bf i}, \, {\bf j}, \, {\bf k}$, and instead of $a_1{\bf
u}_1+a_2{\bf u}_2+a_3{\bf u}_3$, we commonly use $x{\bf i}+y{\bf
j}+z{\bf k}$.  These notations have the disadvantage that they do
not extend to more than 3 dimensions, but in this course we shall
be working in 3 dimensions.

Notice that since the components of a vector are uniquely
specified, the equation ${\bf a}_1={\bf a}_2$ is equivalent to
$x_1=x_2, \, y_1=y_2, \, z_1=z_2$.  i.e. in 3 dimensions a vector
equation is equivalent to three numerical equations.

{\bf Position Vector}

Let $O$ be a point fixed in space.  The displacement vector
$\vec{OP}$ is called the position vector of the point $P$
(relative to $O$).  If in addition to $O$ we have three orthogonal
unit vectors ${\bf i}, \, {\bf j}, \, {\bf k},$ the standard
notation for a position vector is {\bf r} so if $P$ has
co-ordinates $(x,y,z)$ we have ${\bf r}=x{\bf i}+y{\bf j}+z{\bf
k}$.

We can now combine this idea with that of independence to give us
a parametric equation for the plane.  We have seen that if $S'$ is
an arbitrary point in the plane $PQR$ its position vector {\bf r}
can be expressed as ${\bf r}=\vec{OP}+k\vec{PQ}+l\vec{PR}$ for
some values of the parameters $k,\,l$.

In these calculations we use the triangle law in the form

\setlength{\unitlength}{0.5in}
\begin{picture}(8,3)
\put(0.5,0.5){\vector(2,1){2}}

\put(0.5,0.5){\vector(1,2){0.95}}

\put(1.5,2.5){\vector(1,-1){0.95}}

\put(0.3,0.3){\makebox(0,0){$O$}}

\put(1.5,2.7){\makebox(0,0){$P$}}

\put(2.7,1.5){\makebox(0,0){$Q$}}
\end{picture}

given $\vec{OP}$ and $\vec{OQ}$, since
$\vec{OP}+\vec{PQ}=\vec{OQ}$

we have $\vec{PQ}=\vec{OQ}-\vec{OP}$

Example

Find the parametric equation for the plane through the points

$P(1,2,3), \,\, Q(3,4,5), \,\, R(-1,-2,4)$

$\vec{PQ}=(2,2,2), \,\, \vec{PR}=(-2,-4,1)$ so the parametric
equations are

${\bf r}=(1,2,3)+k(2,2,2)+l(-2,-4,1)$

e.g. $k=2,\, l=-1$ gives

${\bf r}=(1,2,3)+(4,4,4)+(2,4,-1)=(7,10,6)$ a point in the plane.

{\bf The Ratio Theorem}

Suppose we have fixed points $P$, $Q$, and a point $X$ which lies
on the line $PQ$ and which satisfies $PX:XQ=k:l$

Notice that to specify $X$ we use a sign convention so that $X$ is
specified by the ratio

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\put(0.5,0.5){\circle*{0.1}}

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\put(6.5,0.5){\circle*{0.1}}

\put(0.5,0.8){\makebox(0,0){$P$}}

\put(2.5,0.8){\makebox(0,0){$X$}}

\put(6.5,0.8){\makebox(0,0){$Q$}}

\put(1.5,0.3){\makebox(0,0){$1"$}}

\put(4.5,0.3){\makebox(0,0){$2"$}}

\put(9,0.5){\makebox(0,0){$PX:XQ=1:2$}}
\end{picture}

\setlength{\unitlength}{0.5in}
\begin{picture}(8,1)
\put(0.5,0.5){\line(1,0){6}}

\put(0.5,0.5){\circle*{0.1}}

\put(3.5,0.5){\circle*{0.1}}

\put(6.5,0.5){\circle*{0.1}}

\put(0.5,0.8){\makebox(0,0){$X$}}

\put(3.5,0.8){\makebox(0,0){$P$}}

\put(6.5,0.8){\makebox(0,0){$Q$}}

\put(2,0.3){\makebox(0,0){$1.5"$}}

\put(4.5,0.3){\makebox(0,0){$1.5"$}}

\put(9,0.5){\makebox(0,0){$PX:XQ=-1:2$}}
\end{picture}

Now $\vec{PX}=\frac{k}{l}\vec{XQ}$ and so if $P$, $X$, $Q$ have
position vectors {\bf p}, {\bf x}, {\bf q} relative to some $O$,
then

\begin{eqnarray*} l({\bf x}-{\bf p})&=&k({\bf q}-{\bf x})\\
(k+l){\bf x}&=&k{\bf q}+l{\bf p}\\ {\bf x}&=&\frac{k{\bf q}+l{\bf
p}}{k+l}\end{eqnarray*}

In particular if $k=l=1$ then the position vector of the mid point
of $PQ$ is

$\ds {\bf x}=\frac{1}{2}({\bf p}+{\bf q})$

Examples \begin{itemize}
\item [i)]

The centoid of a triangle.

Let $A,\,B,\,C,\,L,\,M,\,N$ have position vectors {\bf a}, {\bf
b}, {\bf c}, {\bf l}, {\bf m}, {\bf n}.

Then ${\bf l}=\frac{1}{2}({\bf b}+{\bf c}), \,{\bf
m}=\frac{1}{2}({\bf c}+{\bf a}), \,{\bf n}=\frac{1}{2}({\bf
a}+{\bf b})$

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\begin{picture}(8,3)
\put(0.5,0.5){\line(1,1){2}}

\put(0.5,0.5){\line(6,1){3}}

\put(2.5,2.5){\line(2,-3){1}}

\multiput(1.5,1.5)(0.2,-0.05){11}{\circle*{0.05}}

\multiput(3,1.75)(-0.2,-0.1){13}{\circle*{0.05}}

\multiput(2,0.75)(0.05,0.175){11}{\circle*{0.05}}

\put(0.3,0.5){\makebox(0,0){$A$}}

\put(2.5,2.7){\makebox(0,0){$B$}}

\put(3.7,1){\makebox(0,0){$C$}}

\put(3.2,1.75){\makebox(0,0){$L$}}

\put(2,0.5){\makebox(0,0){$M$}}

\put(1.2,1.5){\makebox(0,0){$N$}}
\end{picture}

Consider the point $G$ on $AL$ such that $AG:GL=2:1$.  This has
position vector

$\frac{1}{3}({\bf a}+2{\bf l})=\frac{1}{3}({\bf a}+{\bf b}+{\bf
c})$

also $\frac{1}{3}({\bf b}+2{\bf m})=\frac{1}{3}({\bf a}+{\bf
b}+{\bf c})=\frac{1}{3}({\bf c}+2{\bf n})$

Thus the point $G$ lies on $AL$, $BM$, $CN$.  It is called the
centroid of the triangle.  In general if we have points ${\bf
a}_1, \ldots, {\bf a}_n$ we define their centroid to be
$\frac{1}{n}({\bf a}_1+ \ldots +{\bf a}_n)$.

\newpage
\item[ii)]
${}$

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\put(0,0){\line(1,0){4}}

\put(-.3,0){\makebox(0,0){$D$}}

\put(2,0){\circle*{.15}}

\put(2,-.4){\makebox(0,0){$T$}}

\put(0,0){\line(4,1){8}}

\put(8.3,2){\makebox(0,0){$B$}}

\put(4,1){\circle*{.15}}

\put(4.2,1.4){\makebox(0,0){$S$}}

\put(0,0){\line(1,3){2}}

\put(2,6.4){\makebox(0,0){$A$}}

\put(1,3){\circle*{.15}}

\put(0.8,3.2){\makebox(0,0){$U$}}

\put(4,0){\line(-1,3){2}}

\put(3,3){\circle*{.15}}

\put(3.4,3){\makebox(0,0){$R$}}

\put(4,0){\line(2,1){4}}

\put(6,1){\circle*{.15}}

\put(6,0.5){\makebox(0,0){$Q$}}

\put(2,6){\line(3,-2){6}}

\put(5,4){\circle*{.15}}

\put(5,4.5){\makebox(0,0){$P$}}

\put(4,-0.4){\makebox(0,0){$C$}}

\end{picture}

Let {\bf a}, {\bf b}, {\bf c}, {\bf d} be position vectors of $A$,
$B$, $C$, $D$.

${\bf p}=\frac{1}{2}({\bf a}+{\bf b}) \hspace{0.5in} {\bf
t}=\frac{1}{2}({\bf c}+{\bf d})$

The mid-point of $PT$ has position vector $\frac{1}{4}({\bf
a}+{\bf b}+{\bf c}+{\bf d})$.  By symmetry (or verification) this
is also the mid-point of $QU$ and $RS$.  So the lines joining the
mid-points of opposite sides of a tetrahedron are concurrent at
the centroid.
\end{itemize}

{\bf Equations of a line}

\begin{itemize}
\item[i)]
Suppose we are given two points {\bf p} and {\bf q} and we want
the equation of a line through $P$ and $Q$.  If $R$ is an
arbitrary point on the line then for some $k, \, l$ we have

$\ds {\bf r}=\frac{k{\bf q}+l{\bf p}}{k+l}=\frac{k}{k+l}{\bf
q}+\frac{l}{k+l}{\bf p}$

(Note the convention of {\bf r} for the position vector of a
variable point.)

Let $\ds \frac{k}{k+l}=t$ then $\ds \frac{l}{k+l}=(1-t)$.

So ${\bf r}=t{\bf q}+(1-t){\bf p} \hspace{0.2in} t\epsilon{\bf R}$

Note that $0<t<1$ gives the region between $P$ and $Q$.

\item[ii)]
If $O$ is the origin and $\vec{OP}={\bf p},\,\, \vec{OQ}={\bf q}$
then from the triangle law $\vec{PQ}={\bf q}-{\bf p}$.

Given

\begin{eqnarray*} {\bf r}&=&t{\bf q}+(1-t){\bf p}\\ {\bf r}&=&{\bf
p}+t({\bf q}-{\bf p})\end{eqnarray*}

If we are given a point $A$ with position vector {\bf a} and a
vector {\bf b} then the line through $A$ in direction {\bf b} has
equation ${\bf r}={\bf a}+t{\bf b} \hspace{0.2in} t\epsilon{\bf
R}$.

\item[iii)]
Three points $A$, $B$, $C$ are collinear.  So $C$ lies on the line
through $A$ and $B$, so for some $t\epsilon{\bf R}$

${\bf c}=t{\bf a}+(1-t){\bf b}$

$t{\bf a}+(1-t){\bf b}-{\bf c}={\bf 0}$

Notice that the sum of the coefficients is zero.

Now suppose $\alpha{\bf a}+\beta{\bf b}+\gamma{\bf c}={\bf 0}$,
with $\alpha+\beta+\gamma=0$, and not all three zero.

Suppose without loss of generality that $\gamma\not=0$, then $\ds
\frac{\alpha}{\gamma}{\bf a}+\frac{\beta}{\gamma}{\bf b}+{\bf
c}=0$

so $\ds c=-\frac{\alpha}{\gamma}{\bf a}-\frac{\beta}{\gamma}{\bf
b}$.

Let $\ds -\frac{\alpha}{\gamma}=t$ then $\ds
-\frac{\beta}{\gamma}=1-t$ since $\alpha+\beta+\gamma=0$.

i.e. $A$, $B$, and $C$ are collinear iff $\alpha{\bf a}+\beta{\bf
b}+\gamma{\bf c}={\bf 0}$, with $\alpha+\beta+\gamma=0$ and not
all three are zero.
\end{itemize}


Example

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\put(3.5,1){\line(-1,1){1.25}}

\put(0.5,0.5){\line(3,2){2.3}}

\put(2.5,2.5){\line(1,-6){0.27}}

\put(0.3,0.5){\makebox(0,0){$C$}}

\put(2.5,2.7){\makebox(0,0){$A$}}

\put(3.7,1){\makebox(0,0){$B$}}

\put(2.9,2.1){\makebox(0,0){$N$}}

\put(2.8,0.8){\makebox(0,0){$L$}}

\put(2.1,2.3){\makebox(0,0){$M$}}

\put(2.3,2.5){\makebox(0,0){1}}

\put(3.3,1.6){\makebox(0,0){1}}

\put(2,0.6){\makebox(0,0){1}}

\put(1.5,1.7){\makebox(0,0){$\mu$}}

\put(3.2,0.8){\makebox(0,0){$\lambda$}}

\put(2.8,2.3){\makebox(0,0){$\upsilon$}}
\end{picture}

Suppose $BL:LC=\lambda:1 \hspace{0.3in} CM:MA=\mu:1 \hspace{0.3in}
AN:NB=\upsilon:1$

In many problems the algebra is simplified by choosing a special
origin.  In this case choose $C$ to be the origin and let the
position vectors of $A$ and $B$ be {\bf a} and {\bf b}.  Then

$\ds M=m{\bf a}$ where $\ds m=\frac{\mu}{\mu+1} \hspace{0.3in}
L=l{\bf b}$ where $\ds l=\frac{1}{\lambda+1}$

$\ds N=\frac{\upsilon{\bf b}+{\bf a}}{\upsilon+1}$

Let {\bf r} be the position vector of the point $R$ where $AL$
meets $BM$.

Since $R$ lies on $AL, \hspace{0.2in} {\bf r}={\bf a}+k(l{\bf
b}-{\bf a}) \hspace{0.3in}$ for some $k\epsilon {\bf R}$

Since $R$ lies on $MB, \hspace{0.2in} {\bf r}=m{\bf a}+h({\bf
b}-m{\bf a}) \hspace{0.3in}$ for some $h\epsilon {\bf R}$

So ${\bf a}+k(l{\bf b}-{\bf a})=m{\bf a}+h({\bf b}-m{\bf a})$

${\bf a}(1-k-m+mh)={\bf b}(h-kl)$

Now {\bf a} and {\bf b} are non-zero and not parallel, so $h=kl$
and $1-k=m(1-h)$.

So $1-k=m-mkl \hspace{0.2in}$ i.e. $\ds k=\frac{1-m}{1-lm}
\hspace{0.2in} lm\not=1$

Thus $\ds {\bf r}={\bf a}+\frac{1-m}{1-lm}(l{\bf b}-{\bf
a})=\frac{m(1-l){\bf a}+l(1-m){\bf b}}{1-lm}$

Now ${\bf n}=\frac{\upsilon{\bf b}+{\bf a}}{\upsilon+1}$ and $CN$
passes through $R$ iff ${\bf n}=p{\bf r}$ for some $p\epsilon{\bf
R}$

So $\ds \frac{1}{\upsilon+1}{\bf a}+{\upsilon}{\upsilon+1}{\bf
b}=p\left(\frac{m(1-l)}{1-lm}{\bf a}+\frac{l(1-m)}{1-lm}{\bf
b}\right)$

Now compare coefficients and eliminate $p$

$\ds \iff \upsilon=\frac{l(1-m)}{m(1-l)} \hspace{0.2in}
1-m=\frac{1}{\mu+1} \hspace{0.2in} 1-l=\frac{\lambda}{\lambda+1}$

$\ds \iff \upsilon=\frac{1}{\lambda+1} \frac{1}{\mu+1}
\frac{\mu+1}{\mu} \frac{\lambda+1}{lambda}$

$\ds \iff \lambda\mu\upsilon=1$

This result is know as Ceva's Theorem, and is often written as

$\ds \frac{BL}{LC}\frac{CM}{MA}\frac{AN}{NB}=1$

Notice that the configuration can also appear in the form below

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\put(3.5,0.5){\line(-1,2){0.6}}

\put(2.9,1.7){\line(3,-2){1.8}}

\put(0.5,0.5){\line(6,1){3.35}}

\put(0.4,0.5){\makebox(0,0){$C$}}

\put(2.8,1.8){\makebox(0,0){$A$}}

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\put(4,1.2){\makebox(0,0){$R$}}

\put(3,1.1){\makebox(0,0){$N$}}

\put(3.5,0.3){\makebox(0,0){$B$}}

\put(4.7,0.3){\makebox(0,0){$L$}}

\put(3.8,1.08){\circle*{0.1}}
\end{picture}


\end{document}
