THREE–DIMENSIONAL GEOMETRY - Number theory

1 Chapter 8 three IntroductionIn this chapter we present a vector algebra approach to three dimensionalgeometry. The aim is to present standard properties of linesand planes,with minimum use of complicated three dimensional diagrams such as thoseinvolving similar triangles. We summarize the chapter:Pointsare defined as ordered triples of real numbers and thedistancebetween pointsP1= (x1, y1, z1) andP2= (x2, y2, z2) is defined by theformulaP1P2=p(x2 x1)2+ (y2 y1)2+ (z2 z1) line segments-ABare introduced as three dimensional columnvectors: IfA= (x1, y1, z1) andB= (x2, y2, z2), then-AB= x2 x1y2 y1z2 z1.

2 IfPis a point, we letP=-OPand callPtheposition suitable definitions oflines,parallel lines, there are important ge-ometrical interpretations of equality, addition and scalar multiplication ofvectors.(i) Equality of vectors: SupposeA, B, C, Dare distinct points such thatno three are collinear. Then-AB=-CDif and only if-ABk-CDand-ACk-BD(See Figure )149150 CHAPTER 8. three dimensional GEOMETRY -6yzxO -A CQQQQQQDQQQQQQB-AB=-CD,-AC=-BD-AB+-AC=-A DFigure : Equality and addition of vectors.(ii) Addition of vectors obeys theparallelogram law: LetA, B, Cbe non collinear.

3 Then-AB+-AC=-AD,whereDis the point such that-ABk-CDand-ACk-BD. (See Fig-ure )(iii) Scalar multiplication of vectors: Let-AP=t-AB, whereAandBaredistinct points. ThenPis on the lineAB,APAB=|t|and(a)P=Aift= 0,P=Bift= 1;(b)Pis betweenAandBif 0< t <1;(c)Bis betweenAandPif 1< t;(d)Ais betweenPandBift <0.(See Figure ) INTRODUCTION151 -6yzxO@@@@@R@@@RPAB-AP=t-AB,0< t <1 Figure : Scalar multiplication of productX Yof vectorsX= a1b1c1 andY= a2b2c2 , is definedbyX Y=a1a2+b1b2+ ||X||of a vectorXis defined by||X||= (X X)1/2and theCauchy Schwarz inequalityholds:|X Y| ||X|| ||Y||.

4 Thetriangle inequalityfor vector length now follows as a simple deduction:||X+Y|| ||X||+||Y||.Using the equationAB=||-AB||,we deduce the corresponding familiartriangle inequalityfor distance:AB AC+ 8. three dimensional GEOMETRYT heangle between two non zero vectorsXandYis then defined bycos =X Y||X|| ||Y||,0 .This definition makes sense. For by the Cauchy Schwarz inequality, 1 X Y||X|| ||Y|| said to beperpendicularororthogonalifX Y= of unit length are calledunitvectors. The vectorsi= 100 ,j= 010 ,k= 001 are unit vectors and every vector is a linear combination ofi,jandk: abc =ai+bj+ zero vectorsXandYareparallelorproportionalif the angle be-tweenXandYequals 0 or ; equivalently ifX=tYfor some real numbert.

5 VectorsXandYare then said to have the same or opposite direction,according ast >0 ort < are then led to study straight lines. IfAandBare distinct points,it is easy to show thatAP+P B=ABholds if and only if-AP=t-AB,where 0 t defined as a set consisting of all pointsPsatisfyingP=P0+tX, t Ror equivalently-P0P=tX,for some fixed pointP0and fixed non zero vectorXcalled adirection vectorfor the , in terms of coordinates,x=x0+ta, y=y0+tb, z=z0+tc,whereP0= (x0, y0, z0) and not all ofa, b, care INTRODUCTION153 There is then one and only one line passing passing through two distinctpointsAandB.

6 It consists of the pointsPsatisfying-AP=t-AB,wheretis a real productX Yprovides us with a vector which is perpendicularto bothXandY. It is defined in terms of the components ofXandY:LetX=a1i+b1j+c1kandY=a2i+b2j+c2k . ThenX Y=ai+bj+ck,wherea= b1c1b2c2 , b= a1c1a2c2 , c= a1b1a2b2 .The cross product enables us to derive elegant formulae forthe distancefrom a point to a line, the area of a triangle and the distance between twoskew we turn to the geometrical concept of a plane in three a set of pointsPsatisfying an equation of the formP=P0+sX+tY, s, t R,( )whereXandYare non zero, non parallel terms of coordinates, equation takes the formx=x0+sa1+ta2y=y0+sb1+tb2z=z0+sc1+tc2 ,whereP0= (x0, y0, z0).

7 There is then one and only one plane passing passing through threenon collinear pointsA, B, C. It consists of the pointsPsatisfying-AP=s-AB+t-AC,wheresand tare real cross product enables us to derive a concise equation for the planethrough three non collinear pointsA, B, C, namely-AP (-AB -AC) = 8. three dimensional GEOMETRYWhen expanded, this equation has the formax+by+cz=d,whereai+bj+ckis a non zero vector which is perpendicular to-P1P2forall pointsP1, P2lying in the plane. Any vector with this property is said tobe anormalto the is then easy to prove that two planes with non parallel normal vectorsmust intersect in a conclude the chapter by deriving a formula for the distance from apoint to a three dimensional spaceDEFINITION dimensional spaceis the setE3of orderedtriples (x, y, z), wherex, y, zare real numbers.

8 The triple (x, y, z) is calleda pointPinE3and we writeP= (x, y, z). The numbersx, y, zare called,respectively, thex, y, zcoordinates axesare the sets of points:{(x,0,0)}(x axis),{(0, y,0)}(y axis),{(0,0, z)}(z axis).The only point common to all three axes is theoriginO= (0,0,0).Thecoordinate planesare the sets of points:{(x, y,0)}(xy plane),{(0, y, z)}(yz plane),{(x,0, z)}(xz plane).Thepositive octantconsists of the points (x, y, z), wherex >0, y >0, z > think of the points (x, y, z) withz >0 as lying above thexy plane,and those withz <0 as lying beneath thexy plane.

9 A pointP= (x, y, z)will be represented as in Figure The point illustrated lies in the pointsP1= (x1, y1, z1)andP2= (x2, y2, z2) is defined by the formulaP1P2=p(x2 x1)2+ (y2 y1)2+ (z2 z1) example, ifP= (x, y, z),OP=px2+y2+ three dimensional SPACE155 -6yzxOQQQQQQ(x, y,0)(x,0,0) (0, y,0)QQQQQQP= (x, y, z)(0,0, z)P= (x, y, z)Figure : Representation of three - dimensional x(x2,0,0)(x1,0,0)(0, y1,0)(0, y2,0)(0,0, z1)(0,0, z2)6zbbbbbbbbbb bbbbbbbbbbTTTTT TTTTTABF igure : The 8. three dimensional GEOMETRYDEFINITION (x1, y1, z1) andB= (x2, y2, z2) we definethe symbol-ABto be the column vector-AB= x2 x1y2 y1z2 z1.

10 We letP=-OPand callPtheposition components of-ABare the coordinates ofBwhen the axes aretranslated toAas origin of think of-ABas being represented by the directed line segment fromAtoBand think of it as an arrow whose tail is atAand whose head is atB. (See Figure )Some mathematicians think of-ABas representing the translation ofspace which following simple properties of-ABare easily verified and correspondto how we intuitively think of directed line segments:(i)-AB= 0 A=B;(ii)-BA= -AB;(iii)-AB+-BC=-AC(the triangle law);(iv)-BC=-AC -AB=C B.