
\magnification=\magstep1
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\def\bE{\Bbb E}
\def\bR{\Bbb R}
\def\vv{\vec v}
\def\vw{\vec w}
\def\vu{\vec u}
\def\v0{\vec 0}
\def\vr{\vec r}
\def\vA{\vec A}
\def\vB{\vec B}
\def\vC{\vec C}
\def\vD{\vec D}
\def\vi{\vec i}
\def\vj{\vec j}
\def\vk{\vec k}
\def\vb{\vec b}
\def\vB{\vec B}
\def\vX{\vec X}
\def\vY{\vec Y}
\def\va{\vec a}
\def\vx{\vec x}
\def\vy{\vec y}
\baselineskip=12pt
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\rightline{Math 401}
\rightline{Spring 1997}
\rightline{D.~Gottlieb}
\bigskip
\centerline{\bf Vectors}
\bigskip\noindent
{\bf I.\ Definitions and Properties}




Let $\bE$ be the space studied by Euclid and the Greeks in 200 BC.
Let $x$ be a point in $\bE$.
Let $\bE_x$ be the set of all possible velocities of a particle at $x$.



\definition{Definition 1} The {\it exponential map} associates to 
every velocity $\vec v_x\in \bE_x$, the point in $\bE$ which a 
particle at $x$ reaches after one ``second'' after traveling in a 
straight line with velocity $\vec v_x$.
\enddefinition

\noindent
{\bf Proposition 2}. The exponential map is a one to one 
correspondence between $\bE_x$ and $\bE$.

\definition{Definition 3} Let $a\in \bR$ be a real number and $\vec 
v\in \bE_x$.
We define {\it scalar multiplication} $a\vec v$ to be that velocity 
so that a particle at $x$ travelling at velocity $a\vec v$ reaches a 
distance $|a|$ times the distance the particle travelling at velocity 
$\vec v$ .
The two particles travel on the same straight line, in the same 
direction if $a > 0$, in the opposite direction if $a < 0$, and if 
$a=0$, the particle moving with velocity $0\vec v=\vec 0$ does not 
move.
\enddefinition

\definition{Definition 4} Let $\vec v$ and $\vec w$ be in $\bE_x$. We define
{\it vector addition}
$\vec v+\vec w$ as that velocity which takes a particle to the point 
$z\in \bE$ in one second, where $z$ the point reached in two seconds 
by a particle which travels with velocity $\vv$ for one second to 
reach the point $y$, and then travels for one second along a velocity 
$\vw\in \bE_y$ for one second to reach $z$.
Here $\vw\in \bE_x$ is {\it equivalent} to $\vw\in \bE_y$ by the 
following definition.
\enddefinition

\definition{Definition 5} $\vv\in \bE_x$ is {\it equivalent} to 
$\vw\in \bE_y$ if the line segments swept out by the two particles 
are parallel, of the same length, and are directed in the same way.
That means it is possible to construct a parallelogram with the two 
line segments as opposite sides and the line segment from $x$ to $y$.
If the two line segments lie on a straight line, the motion of the 
two particles should be in the same direction.
\enddefinition

\noindent
{\bf Proposition 6}. Let $a,b\in \bR$ and $\vu,\vv,\vw\in \bE_x$.

\item{a)}\line{$\vu+\vv=\vv+\vu$\hfill commutivity of addition}

\item{b)}\line{$(\vu+\vv)+\vw=\vu+(\vv+\vw)$\hfill associativity of addition}

\item{c)}$\vu+\v0=\vu$

\item{d)}$\vu+(-\vu)=\v0$

\item{e)}$0\vu=\v0,\ 1\vu=\vu,\ (-1) \vu=-\vu$

\item{f)}\line{$a(b\vu)=(ab)\vu$\hfill associativity of scalar multiplication}

\item{g)}\line{$a(\vu+\vv)=a\vu+a\vv$\hfill distributivity}

\item{h)}\line{$(a+b)\vu=a\vu+b\vu$\hfill distributivity}

\proclaim{Theorem 7} (The Fundamental Theorem)

If $\vu,\vv,\vw\in \bE_x$ do not lie in a plane, then every vector 
$\vr\in \bE_x$ can be written {\it uniquely} as a linear combination, 
$a\vu+b\vv+c\vw=\vr$.
\endproclaim

\definition{Definition 8} Three velocity vectors in $\bE_x$ which do 
not lie in a plane are a {\it basis} for $\bE_x$.
\enddefinition

\bigskip
\noindent
{\bf II.\ Using Vectors in Euclidean Geometry}

To use vectors to solve Euclidean Geometry problems, the first thing 
is to {\it choose} a point $x$ for your $\bE_x$.
You can choose it off the plane if you want to preserve generality 
and if you are looking for symmetry.
You choose $x$ at a point on the figure if you want to eliminate an equation.

You think of your vectors as velocity vectors in $\bE_x$ or as points 
in $\bE$, whichever seems more suitable.
You pass from one point of view to the other by the exponential map.

The most important elements of Euclidean geometry are planes, lines, 
line segments and triangles.

{\bf Planes}:\ Let $\Pi$ be a plane passing through points $A, B, C$ 
(which are not on a line) and suppose we choose a point $x$ 
\underbar{not} in the plane.
let $\vA, \vB, \vC\in \bE_x$ be the vectors associated with $A, B, C$ 
under the exponential maps.
Note $\vA, \vB, \vC$ form a basis of $\bE_x$.
Then any point in the plane $\Pi$ is associated under the exponential 
map {\it uniquely} to a vector of the form
$$
a\vA+b\vB+c\vC\qquad\text{ where }a+b+c=1.\tag1
$$

Those points on or inside the triangle $ABC$ correspond {\it 
uniquely} (under the exponential map) to a vector of form (1) with 
the condition
$$
a\geq 0, b\geq 0, c\geq 0.\tag2
$$

If $x$ were chosen \underbar{on} the plane, then $\vA, \vB$, and 
$\vC$ would no longer form a basis.

\noindent
{\bf Lines}:\ Let $\ell$ be a line passing through two distinct 
points $A$ and $B$, and suppose we choose a point $x$ \underbar{not} 
on the line.
Then any point on the line $\ell$ corresponds uniquely to
$$
a\vA+b\vB \qquad \text{ where }a+b=1.\tag3
$$
If the point lies on the line segment between $A$ and $B$, then
$$
a\geq 0,\ b\geq 0.\tag4
$$
Note that if $a=0$, then the corresponding point is $B$, and if 
$b=0$, then the corresponding point is $A$.

Another way to describe the line $\ell$ is by the set of points which 
corresponds to
$$
\vA+t\vv\qquad \text{ for arbitrary }t\text{ and}\tag5
$$
$\vv\in \bE_x$ which is parallel to the line $\ell$.
A useful choice of $\vv$ is $\vB-\vA$.
Then we can regard the particle as being at $A$ when $t=0$ and at $B$ 
when $t=1$.
Thus $\ell$ is given by
$$
\vA+t(\vB-\vA).\tag6
$$

Let $x$ be chosen at a key point in the diagram, then that point 
corresponds to the zero vector $\v0$.
Then the above equations lose their uniqueness clauses and their symmetry.
On the other hand an extra symbol is not needed.
The choice of  $x$  is the first step in solving the problem.

Another approach is to label the sides of the diagram as vectors.
Then the diagrams correspond to relations.

For example figure 1 corresponds to $\vA+\vB=\vC$ and figure 2
corresponds to $\vA+\vB+\vC=\v0$.

\bigskip\noindent
Figure 1

\vskip.75truein
\noindent
Figure 2

\vskip.75truein

To interpret the results of the algebra we note that
$\vA+\vB+\vC=\v0$ means the sides and lengths can form a triangle.
Also $\vA+\vB+\vC+\vD=\v0$ means that the sides labeled by
$A$, $B$, $C$, $D$ can form a quadrilateral.
Also $\vA=\displaystyle{-1\over 3} \vC$ means side $C$ is parallel to 
side A and 3 times bigger.
The minus sign can sometimes
be ignored. It means that the vector is directed in the
opposite direction, which sometimes isn't important in Euclidean Geometry.

\pagebreak
\bigskip

\noindent
{\bf III.\ Linear Transformations}

Mathematicians have learned in the last 50 years that functions are 
the fundamental concept of Mathematics.
Via functions, we can describe how two things can be the same and yet 
different at the same time.

A {\it function} (alias map, mapping, transformation, ...) is a rule 
which assigns to each element in a set, called the {\it source} (or 
domain),
an element in another set, which is called the {\it target} (or range) set.
The notation $f: A @>>> B$ is read:\ The function $f$ which maps the 
source $A$ into the target $B$.

The key class of functions to use with vectors are linear transformations.

\definition{Definition 9}
A {\it linear transformation} is a mapping $T:\bE @>>> \bE$ such that
$$
\align
& T(a\vv)=a T(\vv)\\
& T(\vv+\vw)=T(\vv)+T(\vw).\tag7
\endalign
$$
\enddefinition

\definition{Proposition 10} Let $T$ be a linear transformation.
\item{a)}$T(a\vA+b\vB+c\vC)=a\vA'+b\vB'+c\vC'$ where
$$
\vA'=T(\vA),\ \vB'=T(\vB)\text{ and }\vC'=T(\vC).
$$

\item{b)}If two linear transformations agree on a basis, then they 
agree for every vector.
That is, if $\vA, \vB, \vC$ is a basis and if $T(\vA)=S(\vA),\ 
T(\vB)=S(\vB)$ and $T(\vC)=S(\vC)$, then $T(\vv)=S(\vv)$ for any 
vector.

\item{c)}Let $\vA, \vB, \vC$ be a basis, and let $\vA', \vB', \vC'$ 
be any three vectors.
There is a linear transformation $T$ so that $\vA'=T(\vA),\ 
\vB'=T(\vB),\ \vC'=T(\vC)$.
If $\vA', \vB', \vC'$ also are a basis, then $T$ is an {\it isomorphism}.
\enddefinition

\item{d)}$T(\v0) = \v0 $ if $T$ is a linear transformation.

\definition{Definition 11} $T:\bE @>>> \bE'$ is an {\it isomorphism} if
\item{a)}For every $\vw\in \bE'$, there is a $\vv\in \bE$ so that 
$T(\vv)=\vw$ ,
and
\item{b)}If $T(\vv)=T(\vw)$, then $\vv=\vw$.
\enddefinition

If there is an isomorphism between two spaces, we say they are {\it 
isomorphic}.
The exponential map is an isomorphism, so $\bE_x$ and $\bE$ are isomorphic.
  Parallel translation gives an isomorphism between $\bE_x$ and $\bE_y$.

\bigskip\noindent
{\bf IV.\ \ Using Isomorphism to Solve Geometry Problem}

To prove a theoretical theorem involving a triangle, $\Delta ABC$, 
choose a particularly nice triangle $\Delta A' B' C'$ for the problem 
and define the unique linear transformation which takes $A @>>> A',\ 
B @>>> B'$, and $C @>>> C'$.
Suppose the theorem is easily true for triangle $\Delta A' B' C'$.
Then it may be true for $\Delta ABC$ under the {\it right} circumstances.

The {\it wrong} circumstances involve equal angles or equal lengths.
The {\it right} circumstances include midpoints, parallel lines, 
intersecting lines.

The reason this works follows from the following Proposition.

\definition{Proposition 12} Let $T:\bE @>>> \bE$ be an isomorphism.
\item{a)}$T$ sends a plane not passing through $\v0$ into a plane not 
passing through $\v0$.

\item{b)}$T$ sends a line into a line.

\item{c)}$T$ sends line segments into line segments, and midpoints 
into midpoints, and preserves any ratio of lengths along the line.

\item{d)}$T$ sends parallel lines into parallel lines and 
intersecting lines into intersecting lines.

\item{e)}$T$ sends triangles into triangles, quadrilaterals into 
quadrilaterals, pentagons into pentagons.
\enddefinition

To show how this method works, consider the proposition that the 
medians of a triangle meet in a common point.
Let $\Delta ABC$ be an arbitrary triangle and $\Delta A' B' C'$ be an 
equilateral triangle.
The medians obviously meet in a point in a equilateral triangle.
Now the isomorphism which takes $\Delta ABC$ into $\Delta A' B' C'$ 
takes medians to medians and intersections to intersections.
So the medians of $\Delta ABC$ also meet in a point.

We summarize the method: Any proposition whose hypotheses and 
conclusions are invariant under linear transformations is true for 
all cases if it is true for one case.
\bigskip\noindent
{\bf V.\ Axiomatics:\ Level 0 of Notation}

Euclid wrote down his 10 axioms, after much thought and trial and error
by Euclid himself and probably by other mathematicians as well.
He knew, and his contemporaries knew, that the choice of axioms was 
arbitrary, subject only to the conditions that they were obviously 
true, and that they implied most things known about geometry.

Over the centuries, though, the axioms became set in stone.
No one dreamed of tampering with them.
Only the parallel axiom was played with, to try to show that it was 
implied by the other nine.

Nowadays mathematicians can propose all kinds of axioms.
We do it as follows:\ We define objects, such as {\it vector spaces}, 
by listing their properties.
For example,
Proposition 6 is transformed into a list of axioms for a vector space.
Two vector spaces are the ``same'' ({\it isomorphic}) via Definition 11.

Now the description of vectors as velocities and the choice of an $x$ 
makes $\bE$ isomorphic to a vector space in which Theorem 7 holds as 
an additional axiom.
So in effect, we have chosen a different set of axioms for Euclidean 
Geometry (Proposition 6, Theorem 7).
Does this choice imply all the theorems of Euclidean Geometry?

No!
Because we haven't defined the idea of a circle or sphere.
That is, we haven't defined the notion of the ``length'' of a vector.
Similarly, we haven't defined Area or Volume.
We will add these concepts to our vector spaces by means of two 
choices:\ A standard length and a standard orientation.

The choice of a standard length, which is called a
{\it metric}, leads to the dot product in Level $-1$ of notation.
The additional choice of orientation leads to the cross product in 
Level $-2$ of notation.

The Greeks, by the way, never chose a unit of length or area or 
volume in their theoretical work.
On the one hand this is very {\it good}, since different choices 
couldn't cause confusion.
A bee knows what it means for one flower to be twice as far or twice 
as big as another, without the aid of inches or centimeters.
So length does not depend on a metric.

On the other hand, picking a metric, and thus expressing length, 
area, or volume as numbers leads to very convenient ways of 
expressing information.

For example, the famous formula $A=\pi r^2$ expresses the relation of 
the area of a circle to its radius much better (even in words) than 
Euclid's ``The areas of two circles are proportional to the squares 
of their radii.''
On the other hand, replacing
$a:b=c:d$ by ${a\over b}={c\over d}$ loses something too.
For example, the statement ``A man is to a woman as a bull is to a 
cow'' makes more sense than ``A man divided by a woman is equal to a 
bull divided by a cow''.
\bigskip\noindent
{\bf VI.\ The Dot Product:\ Level $-1$}

We choose a unit of length.
Then every vector $\vv\in \bE_x$ has a length $\|\vv\|=v\geq 0$.
We define the {\it dot product} of two vectors $\vv$ and $\vw\in 
\bE_x$ as follows:

\definition{Definition 13} $\vv\cdot \vw=vw\cos\theta$ where $\theta$ 
is the angle between $\vv$ and $\vw$.
\enddefinition

\noindent
{\bf Proposition 14.}

\item{a)}\line{$\vv\cdot \vw=\vw\cdot \vv$\hfill commutativity}

\item{b)}$a(\vv\cdot \vw)=(a\vv)\cdot \vw=\vv\cdot (a\vw)$

\item{c)}\line{$(\vv+ \vw)\cdot \vu=\vv\cdot \vu+\vw\cdot\vu$\hfill 
distributivity}

\item{d)}\line{$\vv\cdot \vv=v^2\geq 0$, and if $\vv\cdot \vv=0$, 
then $\vv=\v0$\hfill positive definiteness}

\definition{Definition 15} If $\vv\cdot \vw=0$, we say that $\vv$ and 
$\vw$ are {\it orthogonal}.
If $\|\vv\|=1$, we say that $\vv$ is a {\it unit vector}.
We say $\vi, \vj, \vk$ is an {\it orthonormal basis} if each vector 
is a unit vector and $\vi\cdot \vj=\vj\cdot \vk=\vk\cdot\vi=0$.
\enddefinition

\pagebreak
\noindent
{\bf Proposition 16.}

\item{a)}Let $\vv$ and $\vw$ be expressed in terms of an orthonormal 
basis by $\vv=x\vi+y\vj+z\vk$ and $\vw=x' \vi+y'\vj+z'\vk$.
Then $\vv\cdot \vw=xx'+yy'+zz'$.

\item{b)}If the basis in 16a were just an ordinary basis, then
$$
\vv\cdot \vw=(x,y,z)\pmatrix \vi\cdot \vi & \vi\cdot \vj & i\cdot \vk\\
\vj\cdot \vi & \vj\cdot \vj & \vk\cdot \vk\\
\vk\cdot \vi & \vk\cdot \vj & \vk\cdot \vk\endpmatrix \ \
\pmatrix x'\\ y'\\ z'\endpmatrix
$$

Proposition 16a gives us the Pythagorean Theorem.
It forms the foundation of matrix notation.

   From the point of view of axiomatics, we let Proposition 14 be 
additional axioms which are added to the axioms of Proposition 6.
The result is what we call an ``inner product space.''

Two inner product spaces are {\it isometric} (the same) if there is 
an {\it isometry} between them.
An {\it isometry} $T:\bE_x @>>> \bE_x$ is an isomorphism which 
preserves the metric.
That is,
$$
(T\vv)\cdot (T\vw)=\vv\cdot \vw.\tag7
$$

This implies that $T$ preserves angles and lengths.
If we choose an $x\in \bE$ to be our origin, and choose a unit length 
in $\bE$, then $\bE$ is isometric to an innerproduct space satisfying 
Theorem 7.
Thus an axiom system comprising Proposition 6, Theorem 7, and 
Proposition 14 would give us Euclidean Geometry.
\bigskip\noindent
{\bf VII.\ The Cross Product:\ Level $-2$}

We choose an {\it orientation} of $\bE$, either the right hand or the 
left hand.

\definition{Definition 17} Let $\vv$ and $\vw\in \bE_x$.
We define the cross product $\vv\times \vw$ to be a vector of length 
$vw|\sin\theta|$ orthogonal to the plane defined by $\vv$ and $\vw$ 
with the direction fixed by the orientation.
(Line the fore finger along $\vv$ and the middle finger along $\vw$, 
then $\vv\times \vw$ points in the direction of the thumb.)

Unlike the properties in propositions 6 and 14, the cross product is 
not susceptible to generalization to different dimensions directly.
It is a creature of $\bE$.

\noindent
{\bf Proposition 18}.
Elementary Algebra.

\item{a)}\line{$\vA\times \vB=-\vB\times \vA$\hfill anti--commutivity}

\item{b)}$a(\vA\times \vB)=(a\vA)\times \vB=\vA\times a\vB$

\item{c)}\line{$(\vA+\vB)\times \vC=\vA\times \vC+\vB\times 
\vC$\hfill distributivity}

\item{d)}$\vA\times \vB=\v0$ if and only if $\vA$ and $\vB$ are parallel.

\pagebreak
\noindent
{\bf Proposition 19}.
Geometrical Facts.

\item{a)}$\vA\times \vB$ is orthogonal to $\vA$ and to $\vB$.

\item{b)}$\vA\times \vB=$ (area of $\vA \vB$ parallelogram) (oriented 
normal unit vector)

\item{c)}$(\vA\times \vB)\cdot \vC=\pm$ (volume of the parallelopiped 
with sides $\vA, \vB, \vC$)

The $+$ sign is taken if $\vA, \vB, \vC$ agree with orientation convention.

\noindent
{\bf Proposition 20}.
Oriented Orthonormal Basis.

Let $\vi, \vj, \vk$ be an orthonormal basis so that $\vi\times\vj=\vk$.
Then suppose
$$
\aligned
& \vA=A_1 \vi+A_2 \vj+A_3 \vk\text{ \ \  and}\\
& \vB=B_1 \vi+B_2\vj+B_3\vk\text{ \ \  and }\\
& \vC=C_1\vi+C_2\vj+C_3\vk
\endaligned
$$
$$
\align
& \vA\times \vB=\det \vmatrix \vi & \vj & \vk\\ A_1 & A_2 & A_3\\ B_1 
& B_2 & B_3\endvmatrix=(A_2 B_3-A_3 B_2)\vi-(A_1 B_3-A_3 B_1)\vj+(A_1 
B_2-A_2 B_1)\vk\tag"(a)"\\
& (\vA\times \vB)\cdot \vC=\det \vmatrix A_1 & A_2 & A_3\\ B_1 & B_2 
& B_3\\ C_1 & C_2 & C_3\endvmatrix\tag"(b)"
\endalign
$$

\noindent
{\bf Proposition 21}.
Vector Identities.

\item{a)}$(\vA\times \vB)\cdot \vC=\vC\cdot (\vA\times \vB)=\vA\cdot 
(\vB\times \vC)$

\item{b)}\line{$\vA\times (\vB\times \vC)=(\vA\cdot \vC)\vB-(\vA\cdot 
\vB)\vC$\hfill (replaces associativity)}

\item{c)}$(\vA\times \vB)\times (\vC\times \vD)=[(\vA\times \vC)\cdot 
\vD]\vB-[(\vB\times \vC)\cdot \vD]\vA$

\item{d)}$(\vA\times \vB)\cdot (\vC\times \vD)=(\vA\cdot 
\vC)(\vB\cdot \vD)-(\vA\cdot \vD)(\vB\cdot \vC)$

\bigskip\noindent
{\bf VIII.\ Level $-3$}

In level $-2$ we have all we need to solve any problem or prove any 
proposition in Euclidean Geometry (with the exception of 
constructions with ruler and compass).
So in level $-2$ we have algebracized Geometry.
In the next Levels we will arithmetize Geometry.
That is we will introduce notation which gives names to every point 
in $\bE$, and in such a way that the vector operations of $+, \cdot$, 
and $\times$ become arithmetic calculations.

In this level we choose a point $x\in \bE$ and call it $\v0$, the origin.
With the origin fixed, we can identify the points of $\bE$ with 
unique vectors, and thus we can do all the vector operations directly 
on the points of $\bE$.

\vfill\eject
\noindent
{\bf IX.\ Level $-6$}

We choose a basis $\vb, \vec\beta, \vB$ for $\bE$.
Then every point $x$ can be written as a unique linear combination of 
this basis.
For example $\vx=3\vb-4\vB+2\vec\beta$.
Thus $x$ can be completely described by giving its $\vec\beta$ number 
2, its $\vB$ number $-4$, and its $\vb$ number 3.

\bigskip\noindent
{\bf X.\ Level $-8$}

We introduce an ordering to the basis.
Let us choose $\vB$ to be the first vector, $\vb$ to be the second 
vector, and $\vec\beta$ to be the third vector.
Let us change our notation to reflect this by denoting $\vB$ as 
$\vB_1$, $\vb$ as $\vB_2$, $\vec\beta$ as $\vB_3$.

Then $\vx=-4\vB_1+3\vB_2+2\vB_3$.
We can condense this into $\vx=(-4,3,2)$.
In this notation
$$
\align
\vx+\vy &=(x_1,x_2,x_3)+(y_1,y_2,y_3)=(x_1+y_1,\ x_2+y_2,\ x_3+y_3)\tag8\\
a\cdot \vx&=(ax_1, ax_2, ax_3).\tag9
\endalign
$$

As a surprise extra benefit, linear transformations are described
by means of matrices.
This follows from Proposition 10.

Let
$$
\aligned
T(\vB_1)&=\vB_1+2\vB_2+3\vB_3\\
T(\vB_2)&=4\vB_1+5\vB_2+6\vB_3\\
T(\vB_3)&=7\vB_1+8\vB_2+9\vB_3\\
\endaligned\tag10
$$

Then $T$ is completely described by the matrix
$$
\pmatrix 1&2&3\\ 4&5&6\\ 7&8&9\endpmatrix\tag11
$$
and $T(\vx)$ is given by
$$
(-4,3,2)\ \ \pmatrix 1&2&3\\ 4&5&6\\ 7&8&9\endpmatrix=(22,23,24)\tag12
$$

So matrix arithmetic is a slave of Level $-8$!!

\bigskip\noindent
{\bf XI.\ Level $-9$}

Let us choose our ordered basis to be an orthonormal basis.
Let us denote it by $\vi, \vj, \vk$.
Then Proposition 16a allows us to express the dot product arithmetically by
$$
(x,y,z)\cdot (x',y',z')=xx'+yy'+zz'.\tag13
$$

\bigskip\noindent
{\bf XII.\ Level $-10$}

We require $\vi, \vj, \vk$ to be an {\it oriented} basis.
That means that
$$
\vi\times \vj=\vk.\tag14
$$

Then Proposition 20 allows us to express the cross product arithmetically
$$
(x,y,z)\times (x',y',z')=(yz'-zy',\ zx'-xz', xy'-yx')
\tag15
$$

\bigskip\noindent
{\bf XIII.\ Level $-13$\ The Standard}

We choose symbols $x,y,z$ to represent a variable vector 
$\vX=x\vi+y\vj+z\vk=(x,y,z)$.
We call the coefficient of $\vi$ {\it the $x$ coordinate}, the 
coefficient of $\vj$ {\it the $y$ coordinate}, and the coefficient of 
$\vk$ {\it the $z$ coordinate}.
The lines given by $t\vi$ and $t\vj$ and $t\vk$ are called the $x,y$, 
and $z$ axes.
We can represent the $x$ axis as the solution of two equations:\ 
$y=0$ and $z=0$.

In fact, that is the power of level $-13$.
Many curves and surfaces can be expressed by equations.
Since everybody knows the standard Level $-13$, information can be 
quickly and efficiently exchanged.

For example:\ the intersection of the cylinder $x^2+y^2=1$ with the 
sphere $x^2+y^2+z^2=1$.
The smallest distance between $(1,2,3)$ and the plane $x+2y+3z=4$.

\bigskip\noindent
{\bf XIV. Calculus and Vectors}

We assume that a vector $\vX$  is a function of time  $t$. Then as $t$ varies
we see that $\vX$ travels over a path in space or on a plane or maybe on a
line. We define the velocity vector as the derivative with respect to $t$
of $\vX$ and we denote it by $\vX'$.

$$ \vX' = \lim_{\Delta t\to 0}\frac{1}{\Delta t} (\vX(t + \Delta t) - \vX(t))
\tag16 $$

\noindent
{\bf Proposition 22}.
Differentiation of Vectors.

\item{a)}If $\vX =  x\vi + y\vj + z\vk$ then $\vX' =  x'\vi + y'\vj + z'\vk$

\item{b)}$(\vX \times \vY )' = \vX' \times \vY + \vX \times \vY'$

\item{c)}$(\vX \cdot \vY )' = \vX' \cdot \vY + \vX \cdot \vY'$

\item{d)}$(a\vX + b\vY )' = a\vX ' + b\vY '$


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\noindent
{\bf Vector Problems}

\item{1.}Give a vector which bisects the angle between $\vA$ and $\vB$.

\item{2.}A 100 lb.~weight is resting on an icy inclined plane of angle 45$^o$.
What tension in the rope keeps the weight motionless?

\item{3.}If the wind $\vv_1$ is blowing at 30 miles/hour from the 
west, and the plane $\vv_2$ is flying NW at 100 mph.~with respect to 
the air, find the resulting velocity $\vv_1+\vv_2$ with respect to 
the ground.

\item{4.}
\vskip-.28truein
\itemitem{a)}A smuggler's ship is 5 miles due west of the lighthouse 
and the coast guard ship is 5 miles south west of the lighthouse.
What is the vector $\vA$ from the coast guard to the smuggler?

\itemitem{b)}What direction should the coast guard go to intercept 
the smuggler if the smuggler is travelling due north at 15 miles per 
hour and the coast guard can go at 20 mph?

\itemitem{c)}How long will it take the coast guard to catch the smuggler?

\itemitem{d)}What is the time and direction if the smuggler is 
heading due south instead of north?

\item{5.}Find the length of $2\vi+\vj+2\vk$ and of $3\vi-4\vk$ and 
the angle between them.

\item{6.}Do the lines $t(\vi+\vj+\vk)$ and $s(\vi+\vj)+4\vk$ intersect?

\item{7.}Prove the sum of the squares of the diagonals of a 
parallelogram equals the sum of the squares of its sides.

\item{8.}Prove the diagonals of a rectangle are perpendicular if and 
only if the rectangle is a square.

\item{9.}Prove the sum of the squares of the sides of any quadrangle 
minus the sum of the squares of the two diagonals, equals four times 
the square of the distance between the midpoints of the diagonals.

\item{10.}Show that the midpoints of a quadrilateral are the vertices 
of a parallelogram whose area is one half that of the quadrilateral.

\item{11.}What is the locus of points $\vx$ so that $(\vx-\va)\cdot 
(\vx-\va)=0$?

\item{12.}What is the locus of points $\vx$ so that $(\vx+\va)\cdot 
(\vx-\va)=0$?
\bye
