Vector Prerequisites for this Course

6jun10, 1jun12
\begin{document} \maketitle \section{Introduction} \begin{eqnarray*} r(A+B) &=& r( (x_A,y_A)+(x_B,y_B) ) \\ &=& r (x_A + x_B, y_A +y_B) \\ & =& ( r(x_A + x_B), r(y_A +y_B) ) \\ &=& ( rx_A + rx_B , ry_A + ry_B ) \ \mbox{distributive law for reals} \\ &=& ( rx_A , ry_A) + (rx_B , ry_B) \\ &=& ( r(x_A , y_A) + r(x_B , y_B) \\ &=& (rA + rB) \\ \end{eqnarray*} Do you understand what is written above? To test that you do know how to work with vectors in the plane you need to read $\S\S$ 1.1-1.6 of Tondeur's text. It's only 7 pages, but you'll need a pad of paper and a pencil to work your way through it. And, if you have any difficulty solving the six exercises you should contact your mentor immediately and consult about remediation you may require to take this course. What follows here is no substitute for the above. Nor is it a self-contained review of vector geometry in the plane. It can serve as a very brief review of the subject as it will be needed in this course. And, more importantly, it introduces a slightly different slant on this subject than in the book. If you are easily confused by studying two slightly different treatments of the same mathematical subject, then you should stick to one or the other. But, if you plan on reaching a college level understanding of advanced geometry, you will appreciate complementary expositions. And, in this case, you might also scour the web for yet different approaches. Contrary to what you might have been taught in high school, geometry is not a dry, dogmatic and mummified science. On the contrary, developing definite preferences and opinions is just as valid here as in any field that requires thought. \section{Coordinates} In high school you learned the correspondence between pairs of numbers and points in a coordinate plane, and how curves in the plane can be described by equations in the coordinates of its points. By $P=(x,y)$ or $P=(p_1,p_2)$ and much more rarely, by $P=(P_x, P_y)=(x_P,y_P)$, we mean the point located $x$ units to the right or left of the $y-axis$ and $y$ units above or below the $x-axis$; or, more precisely, $P$ is the intersection of the two lines with equations $x = p_1$ and $y=p_2$; and much more rarely, that $P_x=x_P$ is the point on the horizontal number-line that $P$ projects to vertically, and $P_y=y_P$ is the point on the vertical number-line that $P$ projects to horizontally. Whichever way an author might prefer to write this, the third idea is that "a geometric point \textit{has} two coordinate", not that "a geometric point can be found by \textit{plotting} two coordinates." You must transcend the latter idea in favor of the former. Already you see the same mathematical object, a pair of real numbers, can have several different geometrical meanings depending on how we want to think of the coordinates of a point. By the way, you do remember that the phrase ``the line $x=p_1$" refers to $\{(x,y) : x=p_1\}$. \section{Vectors} In college you learned that ordered pairs of numbers could be added, subtracted, multiplied by a scalar, and satify certain rules, such as the distributive law, to form what was called a \textit{vector field}. The geometrical interpretation of a vector also came in a, perhaps bewildering, variety of forms. \subsection{Arrows} A directed line segment, $\overline{AB}$, with its "tail" at $A$ and its "head" at $B$, is associated with the ordered pair $B-A$. For example, if $A=(1,2)$ and $B=(3,4)$ then $B-A=(2,2)$. Such an object is more precisely called a \textit{displacement vector} because it is calculated in terms of having displaced $A$ to $B$. Now, if you move $A \mapsto A' = A+D$ and $B \mapsto B'= B+D $ then the displacment vector, $\overline{A'B'}$ has exactly the same coordinates as $\overline{AB}$ because, as you can check, \begin{eqnarray*} B' - A' &=& B - A . \\ \end{eqnarray*} That is the meaning of the definition you learned in college, that a vector is an equivalence class of arrows all or which are parallel and oriented in the same way as any one of them. So now an ordered pair, say $(2,2)$ as in our example above, has two meanings: point and vector. How are these related? The Cartesian plane has a very distinguished point, called the \textit{origin}, $O=(0,0)$. If the origin is displaced by the vector $(2,2)$ then it ends up at the point $(2,2)$. And \textit{vice versa}, every point, for example $D=(-42, \pi/2)$, corresponds to the displacement $\overline{OD}$ of the origin to itself. And this vector has the same coordinates (or components, as some prefer) as the point. Some elementary expositions of this topic develop a bizarre collection of unnecessary notation in a misguided attempt to help the student keep these distinctions clearly in their head. I'm sure you've seen the little arrows over the letters, or obtrusively thick bold face type for vectors. We shall not do so. In fact, we will even give up the unnecessary effort of writing $\overline{AB}$ when $B-A$ specifies exactly the same vector. Which meaning is meant is evident from the context of its usage. If we had to write every instance of the English word "set" differently, depending on whether it is used as a verb, noun, adjective or interjection, even fewer children would learn how to spell than already do. By the way, can you use all four meanings of "set" in the same sentence? \subsection{Physics} Another definition you undoubtedly learned in college is that a vector is anything with a magnitude and direction. The geometrical interpretations of this is based on polar coordinates of the plane. To define the corresponding vector, however, requires measuring the distance between two points and the angle between two rays. Since this concept is left undefined in affine geometry, we shall not ask you to remember these concepts now. We shall not need them until a later module of the course. \subsection{Parallel Displacements} We say that two vectors that have the same direction but not necessarily the same magnitudes are \textit{parallel}. In this course we are more precise and say that two displacement vectors $B-A$ and $D-C$ are parallel if one is a scalar mutliple of the other: $D-C = r(B-A)$. In particular, we say that these two vectors are \textit{proportional} and we can take their ratio: $r = \frac{D-C}{B-A}$. In this course you may not take the ratio of two arbitrary vectors unless you know that they are proportional. In a course on Quaternions you will learn how Hamilton solved the problem of making sense of dividing one vector by another. \section{Lines} In high school analytic geometry you learned three, perhaps four formulas, each of which describes an (infinite) straight line in the plane. The \textit{point-slope formula}, such as $y=mx+b$, is the one you're most likely to remember, because it was the first you learned. Like counting on your fingers, which was still easier for you in 4th grade than using your newly learned multiplication tables, you'll fall back on this in a pinch. In algebra you learned the form $ax+by=c$ and later still, the parametric form. Here are three equivalent forms, without using coordinates. \begin{eqnarray*} \ell_{AB} &= & \{X = A + t(B-A) : t \in \mathbb{R} \} \\ & = & \{X = A(1-t) + tB : t \in \mathbb{R} \} \\ & = & \{X = aA + bB : a,b \in \mathbb{R} \ \and \ a+b=1 \} \\ \end{eqnarray*} Although they are algebraically equivalent (you should work out a proof of this assertion first on scratch paper, then enter it into your journal) they have different geometrical meaning. The first says that the points on the line through $A$ and $B$ may be reached by first putting you finger on $A$ and the moving along the displacement vector, $D=B-A$, both poisitive and negative multiples of $D$. Think of a streetcar. The second expresses every point on the line as a weighted average of the positions $A$ and $B$. The weights may be negative, but they must add up to one. The third formula says this more elegantly, expressing no prejudices. Of course $a=1-t$ and $b=t$ relates these two. Think of a teeter-totter. But you must allow a particularly light child to also have enough helium filled ballons to have a "negative" weight. The position of the fulcrum of such a teeter-totter is the position described by the pair of weights $a,b$. \section{Coordinate Free Geometry} Although there is always an underlying coordinate system, we shall (almost) never refer to it. After all, as you recall from high school, one can change the coordinate system, but the geometrical facts remain the same. In particular we do not care where the origin is located. It too can move when the coordinate system is changed. \subsection{When do we need coordinates?} We do need the coordinates (a.k.a. components) of vectores to check the rules. For instance, the \textit{distributive law} would be verified thus. To show that $ (\forall r \in \mathbb{R} \ \and \ A, B \in \mathbb{R}^2)( r(A+B) = rA + rB ) $ we begin on one side of the equation and appeal to algebra until we get to the other. You should \textbf{not} work on both sides of an equation you haven't proved yet, and then work you way to an identity, as you may have been mis-taught to do in high school. Please read the Advice on proofs elewhere in these notes. \begin{eqnarray*} r(A+B) &=& r( (x_A,y_A)+(x_B,y_B) ) \\ &=& r (x_A + x_B, y_A +y_B) \\ & =& ( r(x_A + x_B), r(y_A +y_B) ) \\ &=& ( rx_A + rx_B , ry_A + ry_B ) \ \mbox{distributive law for reals} \\ &=& ( rx_A , ry_A) + (rx_B , ry_B) \\ &=& ( r(x_A , y_A) + r(x_B , y_B) \\ &=& (rA + rB) \\ \end{eqnarray*} Note that we have identified the step where the distributive law for vectors is reduced to the distributive law for reals. All the other steps are applications of the rules for vector algebra. There will be one other time in the module on Affine Geometry where we will do something similar to verify the existence and uniqueness of \textit{barycentric coordinates} in the plane. \section{Study Guide} This course is an upper level college math course, and by now you have surely learned how study mathematics. Nevertheless, since this is the "remediation lesson" let me remind you of some simple rules of thumb. Just as every contact hour of a college course requires an average of two hours of study time, so every lesson in these notes requires an average of two hours of concetrated study time. Sometimes less, sometimes more. And by \it{concentrated study time} I do not mean reading the words and looking at the pictures. \subsection{How to read a mathematical passage} Three times. First time through you skim it to see what's there. You may recognize some items you already know, and other items that make no sense at all. Mentally tag each sentence. Congratulate yourself on those that are "obvious" to you, and skip over those that make no sense, .... yet! The second time you work slowly with pencil and paper at your side. When sentence takes a moment of thought, write it out on paper. When a sentence makes an assertion you don't immediately understand why, then work it out on paper. When you come to one you cannot understand, read it more slowly. If you cannot fill in the details of assertion, play with some numerical or graphical examples. You may have to lay aside a particularly diffiult difficult passage. You may want to come back to it later, or ask someone else. But don't go surfing the web. Chances are you won't find anything, and chances are you'll get distracted by something more interesting. At all events, write down carefully (in your journal, for example) exactly what you don't understand, but also what of the question to do understand. \subsection{How to ask someone else for help} This too has to be learned. Above all, \textbf{do not} preface your question by useless phrases like "I am so confused...", "I am probably stupid for not understanding this but ... ", "You are so clever, I bet you know the answer right away ...", etc. This just wastes your helper's time. (S)he already suspects all that, so no need to remind! Instead, take a dispassionate, diagnostic attitude. Objectify your problem with the problem. Help your helper to help you! \subsection{How to ask by email or on a bulleting board.} Since most of your consulting will be done electronically in this course it is time to learn that discipline too. Here, avoid such unhelpful questions like "I don't understand problem 4.2. How can I solve it?" That kind of a questions is particularly annoying because the only logical response to it is to solve it for you. In which case the second question is moot. No need for you to solve it, the solution is already in your hands. You've learned nothing, and your helper did you no favor. In this geometry course there are very few good exercises, and a great deal of text to master before attempting to solve any one. Therefore, seeing the solution won't help you solve the next exercise because there rarely are two exercises are "solved exactly the same way." So, the \textit{drill} is in working out the lesson texts. Solving an actual exercise is a confirmation that have understand! So, when asking a question be sure to state the question in your own words. Don't just copy the phrase from the lesson or exercise set verbatim. By weritin out your own rephrasing, your astute helper can often tell what your difficulty really is. Clearing up the misundertanding, and maybe dropping a little hint, will give you the help you need to solve the problem yourself. \section{What's with the figure at the top?} Whenever a figure appears at the top of a lesson, it often is meant to intrigue and elicit questions. The figure concerns yet another idea that you acquired in the last geometry course you've taken, which was most probably a course in analytic geometry in high school. You learned to \textit{graph} functions like $ y = 3(1-x)x$, by locating the locus of points that satisfy this equation. So far so good. But then you were taught to read the function $y=f(x)$ by locating $x$ on the x-axis, rising/falling to the graph, and then moving right/left to hit $y$ on the y-axis. All along you were reminded that $f(x) = 3(1-x)x $ is a function takes real numbers to real numbers. Which real numbers? The one's on the x-axis or the reals on the y-axis? A very simple device resolves this mystery, and opens up a tool for investigating interesting phenomena in the sciences. You place the real numbers on the diagonal line, and forget about the x-axis and the y-axis after you've plotted the graph of the function. Locate x on the diagnonal, draw a vertical to the graph, then a horizontal back to the diagonal real line, and presto, you have a new way of representing functions. The payoff is that \textit{feedback}, where the output of the function is used as the input of the next interation of the function, is easy to visualize. The moral of this story? Don't be afraid of new ways of looking at old things! \end{document}