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Lncidence Geometry of the Sphere 52

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Lncidence Geometry of the Sphere 52 , X - P : (X- P,IY)/V+ (X - P,v'lv+ (X- P,wlw, - - which showsthat X P lies in [u, w] if and only if (X P, N) : 6. - Thus,iXl(.X P, N) : 0) isa plane. n lncidence geometry of the sphere The sphere 52 on whose geometry we will be concentrating is determined by the familiar condition 52: {x e E3llxl : 11. If onebegins at a pointof 52and travels straight ahead on thesurface, one 88 will traceout a greatcircle. Viewed as a setin E3,this is theintersection of with a planethrough the origin. However,from the point of view of our Incidencegeometry on s2 it is more appropriateto call this path a line. This motivatesthe ofthe sphere ins definition. Let I be a unit vector. Then 4 : {x. Stl(€,x) : 0} called the line with pole (. We also call ( the polar line o/ {' Remark: Sphericalgeometry is non-Euclidean.This meansthat whenever we representa figure by a diagram,distortions are inevitable.Diagrams that faithfully representone aspect(e.g., straightnessof lines) will distort someother aspect(e.g., lengthsand angles).You are cautionedagainst basingarguments on a diagram,but you are encouragedto use them to suggestfacts that can then be verified rigorously.often it is desirableto havemore than one diagramof the samesituation, each providing insight, {, yet containingsome misleading information. Figures 4.3 and 4.4 showtwo Figure 4.3 A point ( and its polar line first view. ways of thinking about a point and its polar line' : -Q' Two pointsP andQof 52are saidto be antipodalif P Linesof 52 cannotbe parallel, and two lines intersectnot in just one point but in a pair of antipodalpoints. We assertthe followingfacts that you may verify as exercises(Exercise 5)' Theorem5. -l' i. If t is a pole of (, so is its antipode ii. If P lies on (, so does its antipode -P. However, oncethese facts are noticed,there are no further anomalies, the Euclideanresults. and we get the following analoguesof Figure4.4 A point { and its polar line {, secondview. Theorem6. Let P and Q be distinct points of 32 that are not antipoM. Then there is a unique line containing P and Q, which we denoteby PQ' Proof: In order to determinea candidatebr FD, we need a pole (. This mustbe a unit vectororthogonal to both P andQ. BecauseP andQ arenot antipodal,we may choose( equal to (P x qllP x Ql. Clearly,the line with pole ( passesthrough P and Q' We now consideruniqueness. If 1 is a pole of any line throughP andQ, we must have ("r,P) : (1, O) = o. Thus, by the triple product formula, in Theorem 1' 89 I i Geometryon the sphere rx(PxQ):0, Appl and, hence,q is a multiple of the nonzerovector P x Q. Becauselll : 1, we musthave 1 : t€. Thus,Pp is uniquelydetermined. n and I Theorem7. Let ( and nz be distinct lines of 52. Then ( qnd m have exactly two points of intersection,and thesepoints are antipodal. (SeeFigures 4.5 and 4.6.) Thus Proof: Suppose{ and 11are poles of ( and4r4, respectively. Because { and rz are distinct, * *1, and, hence,€ x + 0. But clearly,both points | "l t(€ x tl)/l€ x rtl lie in the intersection.Any third point, however,could lie and, on at mostone of / afidzzzby the uniquenesspart of the previoustheorem. tr Beca Figure4.5 Two intersectinglines 4 and Corollary. No two lines of 32 can be parallel. and, rn, firstview. q< Remak: Even linesthat havea commonperpendiiular will intersect.See w Figures4.7 and 4.8 for two views of this situation. as re Distanceand the triangle inequality Coro The distancebetween two pointsP and of 52is definedby the equation Q Proo d(P,Q): cos-t(P,O). ineq 2" Assr This definitionreflects the idea that the measureof the anglesubtended at colli ln the centerof the sphereby the arcPQ shouldbe numericallyequal to the henr Figure4.6 Twointersectinglines{ and lengthof the arc. SeeFigures 4.9 and4.10. The following theoremshould zn,secondview. be comparedwith Theorem 5 of Chapter 1. Rem Par Theorem8. If P, Q, and R are points of 52, then resu i. d(P, Q) > o. chal ii. d(P, Q) : 0 if and only if P : 9. iii. d(P, Q) : d(Q, P). iv. d(P, Q) + d(Q, R) > d(P, R) (the triangleinequality). Par Proof: Properties(i)-(iii) follow from the Cauchy-Schwarzinequality andthe propertiesof the cos-1function. (See Appendix F.) The detailsare Just Figure4.7 Even lineswith a common left to the readeras exercises. We concentrateour attentionon the triangle Sup perpendicularare not parallel, first view. inequality. iso Let r : d(P, Q), p : d(Q, R), and q : d(P, R). By the Cauchy- 90 Schwarzinequality we have.
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