A Vector Space Approach to Geometry

By Melvin Hausner

The effects of geometry and linear algebra on each other receive close attention in this examination of geometry’s correlation with other branches of math and science. In-depth discussions include a review of systematic geometric motivations in vector space theory and matrix theory; the use of the center of mass in geometry, with an introduction to barycentric coordinates; axiomatic development of determinants in a chapter dealing with area and volume; and a careful consideration of the particle problem. 1965 edition.

• Language: English
• Publisher: Dover Publications
• Release date: Oct 30, 2012
• ISBN: 9780486137858

Book preview

GEOMETRY

1

THE CENTER OF MASS

1.1 INTRODUCTION

Many geometric facts may be made quite vivid and intuitive with the help of the physical notion of the center of mass. It is the purpose of this chapter to illustrate some of the techniques involved. But, lest the reader jump to hasty conclusions, we offer a note of caution. It should not be thought that the methods of this chapter merely give physical proofs. Rather, the center of mass may be used as motivation for certain parts of geometry. Furthermore, as we shall see, this notion can be made mathematically precise. To consider a rough analogy, geometry may be defined and taught without a single picture. Doing so may be considered by some to be a great advance, but it is wrong to consider geometry with pictures as merely giving artistic proofs. In any event, pictures (and physical models) are strongly suggestive and helpful.

1.2 SOME PHYSICAL ASSUMPTIONS AND CONVENTIONS

A mass-point is visualized as a particle having a certain mass m located at a point P of space. It is, of course, an idealized physical notion. We shall be concerned with bodies consisting of finitely many mass-points. These notions are used by physicists every day. Thus a ball is often idealized as a mass-point, and in astronomical contexts it is perfectly reasonable to take the earth and stars as mass-points. Furthermore, a large solid mass is approximated by cutting it up into (finitely many) small pieces and considering each part as a mass-point.

For many purposes, a solid body of total mass m may be replaced by a single mass-point of mass m, located at the center of mass of the body. We are going to make various assumptions about the center of mass. These will be made quite explicit in Section 1.5, but for the present we prefer to work intuitively.

An especially simple mass consists of two mass-points located at A and B with masses m1 and m2, respectively, as in Fig. 1.1. If we wish, we may imagine them connected by a bar, but we should think of the bar as massless in order not to introduce any new masses. We shall assume that the center of mass of this system is located on the bar at the point where the see-saw balances. This is the point P on the segment AB at which m1AP = m2PB (see Fig. 1.1). Thus,

FIGURE 1.1

FIGURE 1.2

This has the effect of pulling P closer to B when the mass of B gets larger. (If m2 increases, so does AP.) Therefore, by the device of mass-points, we shall be able to deal with points P on a segment AB which divide AB in a given ratio.

We may easily remember the rule (1) which locates P by noting that AP (distance to A) is in the numerator and so is m2 (mass of B). Diagrammatically, we merely place the mass of B on the segment opposite B, and similarly for the mass of A. This is indicated in Fig. 1.2 where we read AP/PB = 5/3 and we say that P divides AB in the ratio 5/3. The numbers 5 and 3 do not represent the distances AP and PB but give the ratio AP/PB, and any proportional numbers will do as well.

FIGURE 1.3

It is convenient to use a similar device for broken lines. In this case, the numbers along a line refer only to ratios along that line. Thus, in Fig. 1.3, we have AB = BC and CD = 2DE; but, for example, we do not necessarily have DE = BC. We also make a point of writing, for example, CD = 2DE rather than CD = 2ED. Thus we make sure that the directions CD and DE are the same.

If a mass-point is located at A and has mass m, we shall refer to the mass-point mA. Thus we may refer to a system of mass-points 3A, 4B, 9C instead of the more cumbersome "system of mass-points located at A, B, C with masses 3, 4, and 9, respectively." When we wish to replace them with a single mass-point located at their center of mass, we replace them with 16X, where X is the center of mass of this system, and 16 = 3 + 4 + 9 is the total mass. In the next few sections, we shall learn how to find such a center of mass.

1.3 PHYSICAL MOTIVATIONS IN GEOMETRY

In this section are various examples which illustrate how the notion of a center of mass may be used in geometry.

EXAMPLE 1 (The Centroid of a Triangle). Consider a triangle ABC and place a mass 1 at each vertex (see Fig. 1.4).

FIGURE 1.4

FIGURE 1.5

Let M be the center of mass of this system. To find it, replace 1A and 1B with their equivalent, 2M1, located at their mid-point. Then replace 2M1 and 1C with their equivalent. The answer is 3M, the point on the median from C and two-thirds of the way from C (see Fig. 1.4). But clearly this same M might have been obtained in two equally effective ways, as in Fig. 1.5. The result is the pretty theorem about the medians of a triangle: The medians of a triangle intersect in a point two-thirds of the way along each median from each vertex. This common point of intersection has the physical significance of being the center of mass of three mass-points of equal mass located at the vertices of the triangle. It is called the centroid of the triangle. It is the planar analogue of the mid-point of a segment, which is, analogously, the center of mass of two mass-points of equal masses located at the end-points of the segment.

EXAMPLE 2. Let masses 1, 2, and 3 be placed at the vertices A, B, and C of the triangle in Fig. 1.6. As in Example 1, we find the center of mass by first replacing two of the mass-points with their equivalent mass-point and then using the other mass-point. In Fig. 1.7, 2B and 3C are replaced by 5A1 where A1 divides BC in the ratio 3/2. Then M, the center of mass, is on AA1 and divides it in the ratio 5/1. The same figure gives two other ways of proceeding. For example, M is also the midpoint of CC1 and divides BB1 in the ratio 4/2 = 2/1.

FIGURE 1.6

FIGURE 1.7

We are using a very important physical assumption here. Briefly, it is that a system of mass-points has only one center of mass, and it can be found by individually working on parts of the system. This will be made quite formal in Section 1.5.

EXAMPLE 3. In Fig. 1.8, P divides AB in the ratio 5/1 and Q divides PC in the ratio 2/3. BQ is extended until it meets AC at R. Find AR/RC and BQ/QR.

The method is to take Q as the center of mass of a suitable physical system. Give A mass 1 and B mass 5. Then P is the location of the center of mass of 1A and 5B (see Fig. 1.9). By giving C mass 4 we obtain Q as the center of mass of 6P and 4C. The previous examples point the way. By observation, AR/RC = 4/1 and BQ/QR = 5/5 = 1.

FIGURE 1.8

FIGURE 1.9

EXAMPLE 4 (Ceva’s Theorem). Let P be inside ABC. Suppose AP is extended until it meets BC at A1, and that B1 and C1 are found similarly. Then (see Fig. 1.10),

AB1 · CA1 · BC1 = B1C · A1B · C1A

The method of proof is clear from Fig. 1.11. We treat P as the center of mass of mA, nB, pC for suitable masses m, n, p. We have three ways of finding P and, according to Fig. 1.11, we have

FIGURE 1.10

FIGURE 1.11

By multiplying we obtain

Clearing of fractions, we have the result.

EXAMPLE 5. In Fig. 1.12, find the ratios AG/GE and AF/FC.

We should like A and B to have mass 2 and mass 3, respectively. Clearly, any proportional numbers can be used. Since B and C need mass 5 and mass 2, we see that B wants to have mass 5 as well as 3. Choose 15, the least common multiple for the mass of B (in order to avoid fractions), and we quickly obtain Fig. 1.13, where we read

In the exercises which follow, try to work as physically as possible, where this can be done.

FIGURE 1.12

FIGURE 1.13

Exercises

1. In Fig. 1.14 , AP = 2 PB and QC = 2 PQ . Compare AR and RC.

FIGURE 1.14

FIGURE 1.15

2. In Fig. 1.15 , find the ratios BP / PQ and AP / PR .

3. In Fig. 1.16 , the letters a , b , c , d , and e represent actual distances. Express QO in terms of them. ( Hint: First ignore e , and use only the ratios a / b and c / d . Then find the ratio QO / OT and substitute OT = e .)

FIGURE 1.16

4. Menelaus’s theorem deals with the ratios into which an arbitrary transversal divides the three sides of a triangle. In Fig. 1.17 , A 1 , B 1 , C 1 are the points of intersection of a transversal with BC , CA , AB , respectively. Prove Menelaus’s theorem:

FIGURE 1.17

5. The proof of Ceva’s theorem assumed that any point P inside a triangle ABC is the center of mass of A , B , and C with suitable masses. Show that this is so, and show further that the ratios of the masses are unique, and the weights themselves are unique, if the total mass is 1.

6. State and prove the converse of Ceva’s theorem.

7. A heavy wire is bent in the form of a triangle. Where is its center of mass? ( Note: It is not at the centroid of the triangle.)

8. Problems such as Exercises 1 and 2 above often distress students because the given information does not serve to determine the figure. Comment on this.

1.4 FURTHER PHYSICAL MOTIVATIONS

The methods of the previous section may also be applied to solid geometry. Furthermore, even in the plane, some algebra (rather than the arithmetic of the last section) is often convenient. As in the previous section, we illustrate the techniques by examples.

EXAMPLE 1 (The Centroid of a Tetrahedron). Let ABCD be a tetrahedron (not necessarily a regular tetrahedron). Assign the mass 1 to each vertex and let M be the center of mass of the resulting configuration. There are quite a few ways to find M. As in Fig. 1.18, we may replace three mass-points 1A, 1B, 1C with 3M1, where M1 is the centroid of ABC. Then M is found along M1D by dividing M1D in the ratio 1/3. In Fig. 1.19, 1A and 1B are replaced by 2X1 (where X1 is the mid-point of AB). Similarly, X2 is found along CD, and M is then the mid-point of X1X2. (Note again that it is strongly assumed that there is only one center of mass M.) Hence we have physically proved the following theorem:

FIGURE 1.18

FIGURE 1.19

The lines connecting a vertex of a tetrahedron to the centroid of the opposite face all meet in a common point M, called the centroid of a tetrahedron. It is located three-fourths of the way from each vertex to the centroid of the opposite face. Furthermore, M bisects each segment which joins the mid-points of opposite ‘edges of the tetrahedron.

EXAMPLE 2. In Fig. 1.20, EF/FC = 4/3 and AF/FD = 7. Show how F may be realized as a center of mass involving the vertices A, B, and C, and in this manner compute BD/DC.

FIGURE 1.20

FIGURE 1.21

We proceed as in the last section. The mass at D (which is the replacement of the masses at B and C) is 7 times that of A. Let x = mass at A, 7x = mass at D. Similarly, the masses at E and C are in the ratio 3/4, so we let 3y = mass at E, 4y = mass at C. (Here x and y are proportionality factors.) Now, letting z = mass at B, we read from Fig. 1.21:

Eliminating z by subtracting, and then simplifying, we obtain

Now the values of x, y, z are determined only up to a proportionality factor. This is physically clear by considering what would happen if all masses were, for example, doubled.† Equation (3) gives the ratio of x to y. For simplicity, we choose x = 7 and y = 8. Then by using (1), we obtain z = 3y – x = 24 – 7 = 17. Therefore, the mass system giving Fig. 1.20 is given in Fig. 1.22. Thus, BD/DC = 32/17, which is the answer to the problem.

FIGURE 1.22

EXAMPLE 3. Transversals may be obtained by breaking up a mass into parts. Thus, in Fig. 1.23, the mass-point 3A is split into two parts, 1A and 2A. 1A and 1C may be replaced by 2D, while 2A and 3B may be replaced by 5E, and in this way the center of mass M may be obtained. By considering Fig. 1.24 we see, for example, that DE and the median CF meet at M and divide each other in the indicated ratios. (Recall that ratios may be read only along a line. Thus we do not say, for example, that MF/ME = 1/2.) The next example exploits this method of treating transversals.

FIGURE 1.23

FIGURE 1.24

EXAMPLE 4. In Fig. 1.25, AD/BD = 1/2, AE = EC and DF/FE = 2. Find BG/GC and AF/FG.

We proceed algebraically, keeping in mind the previous example. The transversal DE is considered to come from a split of the mass at A, combining some with B (yielding D) and some with C (yielding E). Hence, let x = mass at C = part of mass at A combining to give E. Similarly, a mass y at B and 2y at A may be replaced by the mass 3y at D (see Fig. 1.26). Comparing Figs. 1.25 and 1.26 with respect to DF/FE, we have

FIGURE 1.25

FIGURE 1.26

Hence, 6y = 2x. We choose x = 3, y = 1. (As usual, any proportional numbers can be used.) The mass at B is 1 and at C it is 3. Therefore, BG/GC = 3. Furthermore, the mass at A is x + 2y = 3 + 2 = 5. We leave to the reader the problem of finding AF/FG from these data.

FIGURE 1.27

FIGURE 1.28

EXAMPLE 5. In Fig. 1.27, ABCD is a tetrahedron and M is the mid-point of EF. Construct a line through M which meets both AB and CD.

As usual, M is conceived of as a center of mass. Let 2x and x be the masses of A and C (so that 2xA and xC may be replaced by 3xF), and let 3y and 2y be the masses of B and D (see Fig. 1.28). M is the center of mass of the system if 3x = 5y. Choose x = 5, y = 3 to obtain Fig. 1.29. By first replacing 10A and 9B with 19G, and then 5C and 6D with 11H, we see that M is on the line GH where G and H are on AB and CD and divide them, respectively, in the ratios 9/10 and 6/5. This is the solution of the problem.

FIGURE 1.29

Exercises

1. In Fig. 1.30 , determine masses at A , B , and C in such a way that G is the center of the mass of the resulting system.

2. In Fig. 1.31 , find PT / TQ.

FIGURE 1.30

FIGURE 1.31

3. In Fig. 1.32 , find BI / IC .

FIGURE 1.32

FIGURE 1.33

4. In Fig. 1.33 , a , b , c , d , e , are given lengths. Find x.

5. In Fig. 1.34 , C ′ divides AB in the ratio 1/2, and similarly for A ′ and B ′. By assigning suitable masses to A , B , and C , and by suitably breaking them up, show that the centroid of A ′ B ′ C ′ is also the centroid of ABC . Generalize and state a converse.

FIGURE 1.34

FIGURE 1.35

6. Let A , B , C , D be points in a plane. Prove that the three lines formed by joining the mid-point of any two of these points with the mid-point of the other two, all meet in a point. State the relationship between this theorem and the centroid of a tetrahedron.

7. In Fig. 1.35 , what masses, if any, must be placed at E , F , and G so that the resulting system has its center of mass located at the centroid of ABC ? ( Hint: E comes from mass 2 x at A and x at B , etc. The total masses at A , B , and C must be equal.)

8. In Fig. 1.36 , where does the line through P and the centroid M of ABCD meet the face of the tetrahedron?

9. Prove that any point in the interior of a tetrahedron is the center of mass of suitable mass-points located at its vertices. Show also that the ratio of the masses are unique and that masses themselves are unique, provided the total mass is 1.

FIGURE 1.36

FIGURE 1.37

10. A point in the interior of a tetrahedron may be characterized as the center of mass of mass-points located at all of the vertices. Similarly characterize:

(a) points on the face of a tetrahedron (and not on an edge),

(b) points on an edge (not a vertex), and

(c) vertices.

11. In the tetrahedron of Fig. 1.37 , determine G and H on BC and AD , respectively, so that GH passes through the mid-point of EF.

12. In the tetrahedron of Fig. 1.38 , determine H so that GH meets EF . (Note that for this H , the points E , F , G , H will lie in a plane, so that we are finding where the plane through E , F , and G meets the line AD. )

FIGURE 1.38

FIGURE 1.39

13. In Fig. 1.39 , determine conditions on the ratios a / b , . . . , g / h in order that the lines EF and GH intersect. ABCD is a tetrahedron in space.

1.5 AN AXIOMATIC CHARACTERIZATION OF CENTER OF MASS

We now investigate and formalize some of the basic assumptions which we have been making about the center of mass. In doing this, a neat notation will be introduced, as well as some algebraic tools which can replace physical reasoning. We shall always use capital letters A, B, . . . , P, Q, . . . to designate points and lower case letters a, b, . . . , m, n, . . . to designate positive numbers (the masses). A mass-point (mass m located at point P) will continue to be designated by mP. A mass-point is, from the present point of view, a positive number m and a point P.*

Suppose we have k mass-points m1P1, . . . , mkPk. We have assumed that they uniquely determine a new mass-point mP, where m = m1 + · · · + mk and where P is their center of mass. We shall write

mP = m1P1 + · · · + mkPk

Thus m1P1 + · · · + mkPk is a shorthand way of writing, "The mass-point obtained when all of the masses of m1P1, . . . , mkPk are concentrated at their center of mass." We have seen in the examples that the center of mass can be obtained by taking the centers of two points at a time, and repeating the operation. Hence it is necessary to concern ourselves only with the center of mass of two mass-points.

It is convenient to introduce the operation of multiplying a mass-point mP by a positive number n. The result is the mass-point still located at P but whose mass is multiplied by n. In a formula:

n(mP) = (nm)P

We are going to state axioms for the center of mass and then briefly discuss their physical and geometric significance. The reader should not be surprised that the axioms seem so algebraic. First, one of the important aspects of algebra (and mathematics in general) is its ability to present complicated ideas concisely, and in great generality. Secondly, all the statements and rules are quite familiar to the reader, though perhaps in different contexts. Finally, this is not an algebra book, so that we shall not (except for a few exercises) concentrate on the formal processes of proofs and justifications which are consequences of these axioms. This will be left to the algebra texts, since a rigorous exposition of these methods would take us too far afield.

Axioms for the Center of Mass

I (Closure Law). There is a set of points P, Q, . . . , and an operation + on mass-points. If mP and nQ are two mass-points, then mP + nQ is a uniquely determined mass-point of mass (m + n): mP + nQ = (m + n)R. R is called the center of mass of mP and nQ.

II (Commutative Law).

mP + nQ = nQ + mP*

III (Associative Law).

mP + (nQ + rR) = (mP + nQ) + rR

IV (Idempotent Law).

mP + nP = (m + n)P

V (Homogeneity Law).

k(mP + nQ) = kmP + knQ

VI (Subtraction Law). If m > n, the equation

mP = nQ + xX

may be solved for the unknown mass-point xX. There is only one solution.

This concludes the statement of the axioms. In our interpretation, the closure law (I) merely states that two mass-points have a single center of mass. The commutative law (II) says that the center of mass depends on the mass-points and not on the order in which they are mentioned. The associative law (III) is the formal statement of our assumption that, in computing the center of mass, we may replace any subsystem with its equivalent, namely, the total mass concentrated at the center of mass. Thus, in computing the center of mass of mP, nQ, and rR, we may first work with mP and nQ and then combine with rR, or first with nQ and rR and then combine with nP. Laws I, II, and III justify the statement that any finite system of mass-points m1P1, . . . , mkPk have only one center of mass, namely, the P of the equation

(m1 + · · · + mk)P = m1P1 + · · · + mkPk

For it is an algebraic fact that, in the light of I, II, and III, the right-hand sum may be computed by arbitrarily rearranging the terms and by arbitrarily grouping them (inserting parentheses). This is known as the generalized commutative and associative law, and is the formal expression of previous remarks that there is only one center of mass.

The idempotent law (IV) was implicit in our treatment of transversals, while the homogeneity law (V) states that the center of mass depends only on the ratios of the masses involved. (Thus, if all masses were tripled, the total mass would triple, but the center of mass would be unchanged.) Finally, the subtraction law (VI) states (physically) that a mass located at a given point Q may be balanced with a fulcrum at given point P by a given mass at a suitable point X. In Fig. 1.40, we wish to solve the equation

FIGURE 1.40

mP = nQ + xX

where m, n, P, Q are given (m > n). By comparing masses we have m = n + x, and hence, x = m – n > 0. The unknown point X is somewhere on the ray from P directed away from Q. Physically, we take a mass (m – n) and slide it on the see-saw until it balances the mass-point nQ. Also, there is but one such point. Thus, geometrically, Axiom VI (subtraction) permits the extension of a line segment for an arbitrary distance in any direction.

The reader should be cautioned against assuming that Axioms I to VI constitute a proof of some geometric facts. Taken by themselves, they constitute assumptions about an abstract set of points. On the other hand, we have been guided in the previous sections by the geometric constructions of mP + nQ: the center of mass of mP and nQ is the point R on the segment PQ such that PR/RQ = n/m. Thus it is necessary to verify that this definition of mP + nQ actually satisfies these axioms.* As might be supposed, the only difficulty is the associative law which turns out to be basically a statement of Menelaus’s theorem. A proof, using vector methods, is given in Section 2.7. For the present, we assume that Axioms I to VI are true for the points of the plane, or space, with our usual definition of center of mass.

In the next section, we shall show how the methods of the previous sections look algebraically. But let us first consider the more formal and manipulative aspects of these axioms. As previously stated, none of these axioms states anything which should algebraically shock the reader. In fact, a good guide to manipulation of equations involving mass-points is, If it comes naturally, and makes sense, then it’s very likely true. For example, Axiom VI says that the equation mP = nQ + xX may be solved for xX. This justifies the following definition.

Definition. If m > n, the unique solution of the equation

is called the difference of mP and nQ and is denoted

Thus Eqs. (1) and (2) say the same thing and each is a consequence of the other. In short, transposing equations such as (1) and (2) is meaningful and is justified.

For another example of our guide to manipulation, suppose

Acting naturally, we have

which also turns out to be correct. Such algebraic manipulation will be assumed in what follows, since its justification, while not especially difficult, is somewhat time-consuming and properly belongs in an algebra text. However, it is partially justified in Section 2.7 using vector methods.

Even if we wish to work as abstractly as possible, it is convenient to use geometric language. Thus we may define betweenness as follows:

Definition. P is between A and B if there are positive numbers m, n such that

mA + nB = (m + n)P

Similarly, we define the notion of insidedness in the following manner:

Definition. P is inside A, B, C if there are positive numbers m, n, p such that

mA + nB + pC = (m + n + p)P

Similar definitions appear in the exercises.

As our sole example of a rigorous proof, we state and prove the following theorem.

Theorem (The Cancellation Law). If mA + nB = mA + pC, then nB = pC (i.e., n = p and B = C). Thus mA may be cancelled from both sides of the equation.

Proof. Let

mA + nB = qD

By hypothesis,

mA + pC = qD

By Axiom VI, the equation

mA + xX = qD

has a unique solution. Since nB and pC are both solutions, we have

xX = nB = pC

This completes the proof. It is clear from this proof that this theorem is simply a restatement of the uniqueness part of Axiom VI.

Unless otherwise stated, any algebra needed in the following exercises may be justified by the general guide to manipulation.

Exercises

In Exercises 1 to 5 below, express each of the geometric statements in algebraic language. No proofs are required. Many of your answers will be somewhat clumsy, but this will be improved in Chapters 2 and 3.

1. P is inside the tetrahedron QRST.

intersect.

3. A , B , and C are collinear (i.e., all on a line); D , E , and F are not collinear.

4. P , Q , R , and S are coplanar (i.e., all on a plane). (See Exercise 2 above and recall the definition of inside.)

5. P divides QR in the ratio a / b.

6. Prove that, if S and T intersect. Illustrate.

7. By algebra alone, prove that if A ≠ B , and if mA + nB = m ′ A + n ′ B , then m = m ′ and n = n ′. ( Hint: Assume m ′ > m and write m ′ = m + p , p > 0. Also recall that the masses agree, so that m + n = m ′ + n ′.)

8. If A , B , and C are given points in the plane, show how P and Q may be constructed:

7P = 2A + 5B,      6Q = 8A – (1B + 1C)

9. Show how P may be constructed by each of the following equivalent equations:

P = (3A – 1B) – 1C,      P = 3A – (1B + 1C)

10. Solve for X and Y :

2X + 3Y = 5A

X + Y = 2B

Restate the problem and the solution in geometric terms.

1.6 AN ALGEBRAIC ATTACK ON GEOMETRY

In this section, we consider the techniques of Sections 1.3 and 1.4 in light of the formal apparatus of Section 1.5. We start with an example typical of Section 1.3.

EXAMPLE 1. In Fig. 1.41, find BE/EC.

FIGURE 1.41

FIGURE 1.42

We first state this problem algebraically and then give the solution, keeping in mind Section 1.3. We are given:

Equation (1) locates F, and (2) locates D. Furthermore, P . Finally (locating E), P is on segment AE and E is on segment BC. The problem is to "express E in terms of B and C."

If we proceed in a characteristically nonalgebraic way, we "pull P out of the sky" by defining

(See Fig. 1.42 for the genesis of this definition.) Comparing (3) and (1), we have

Multiplying (2) by 2, we obtain

and by comparing (5) and (3), we have

Equations (4) and (6) show that P is on segments CD and BF.

Now define E by the equation

(See Eq. (3) for an explanation of this definition.) Comparing (7) and (3), we obtain

Equations (7) and (8) show that E is on segment