1、Chapter 12 Vectors and Geometry of Space 12.1 Three-Dimensional Coordinate Systems*12.2 Vectors*12.3 The Dot Product*12.4 The Cross Product*12.5 Equations of Lines and Planes 12.6 Cylinders and Quadric Surfaces*12.7 Cylindrical and Spherical CoordinatesIn this chapter we introduce vectors andcoordin
2、ate systems for three-dimensionalspace.This will be the setting for our study of the calculus of functions of two variablesin Chapter 14 because the graph of such afunction is a surface in space.In this chapter we will see that vectors provide particularly simple descriptions of lines and planes in
3、space.12.1 Three-Dimensional Coordinate Systems Coordinate axesxyzoxyzoCoordinate planesplanexyxz-planeplaneyz origin OThrough point O,three axes vertical each other,by right-hand rule,we obtain a Three-Dimensional Rectangular Coordinate SystemsxyzThree-Dimensional Rectangular Coordinate Systemso oc
4、tantsplanexyplaneyzxz-plane The Cartesian product is the set of all ordered triples of real numbers and is denoted by .We have given a one-to-one correspondence between points P in space and ordered triples(,b,c)in,|),(RzyxzyxRRR3R3Rxyzo)0,0,(aP),0,0(cR)0,(baA)0,0(bQ),(coaC),0(cbB),(cbaMrWe call a,b
5、 and c the coordinates of PDistance Formula in Three Dimensions The distance between the points and is 21PP),(1111zyxP),(2222zyxP21221221221)()()(zzyyxxPP1z2zO2xxz1xABC1P2P1y2yyEquation of a Sphere An equation of a sphere with center(h,k,l)and radius r is .In particular,if the center is the origin O
6、,then an equation of the sphere is2222)()()(rlzkyhxoxyz2222rzyx12.2 VectorsThe term vector is used by scientists to indicate a quantity that has both magnitude and direction.ABSuppose a particle moves along a linesegment from A to point B.Initial point(the tail)ATerminal point(the tip)BThe displacem
7、ent vector is denoted by v=AB=vvABvCDuu and v are equivalent u=vThe zero vertor is denoted by 0Definition of Vector Addition If u and v are vectorspositioned so the initial point of v is at the terminalpoint of u,then the sum u+v is the vector from the initial point of u to the terminal point of v.u
8、vu+vuvThe Triangle LawThe Parallelogram Lawu+vDefinition of Scalar Multiplication If c is a scalar and v is a vector,then the scalar multiple cv is thevector whose length is times the length of v andwhose direction is the same as v if and is opposite to v if If or v=0,then v=0.c0c.0c0ccuvu-vThe diff
9、rence of the two vectors u and v.u-vComponentsThe two-dimensional vector is the position vector of the point .21,aaa),(21aaP),(21aaPopoaThe three-dimensional vector is the position vector of the point .321,aaaa),(321aaaP),(321aaaPopoaAn n-dimensional vector is an ordered n-tuple:naaaa,21where are re
10、al numbers that are calledthe components of .We denote by the set ofall n-dimensional vectors.naaa,21aThe magnitued or length of the vector is denotedby the symbol or .aaa22221naaaanvwhen .1aais called a unit vectorIf ,then the unit vector that has the same direction as is0aaaauDefinition If,21naaaa
11、nbbbb,21c is scalar,thennnnnbabababbbaaaba,22112121nncacacaaaacac,2121Properties of VectorsIf a,b.and c are vectors in and c and d are scalars,then1.a+b=b+c 2.a+(b+c)=(a+b)+c3.a+0=a 4.a+(-a)=05.c(a+b)=ca+cb 6.(c+d)a=ca+cd7.(cd)a=c(da)8.1a=anvThe standard basis vectors in ,0,0,1i3v,0,1,0j1,0,0koijkoi
12、jk321,aaaakajaiaaaaa321321,We have Definition If and naaaa,21nbbbb,21,then the dot product of and is the number given bybaabnnbabababa2211The dot product is also called the scalar product(or inner product).Properties of the Dot ProductIf a,b.and c are vectors in and c is scalar,thennv0a0 5.b)(ab)(ab
13、a)(4.cabac)(ba 3.abba.2aaa.12cccTheorem If is the angle between the vectors and thenab0cosbabaProofBy the Law of Cosines,we have cos2222bababaCorollary If is the angle between the nonzerovectors and thenabbabacosExample Find the angle between the vectors kjia 2andkjib23And are orthogonal if and only
14、 ifab0baCorollary:gnl Direction Angles and Direction CosinsThe direction angles of a nonzero vector are the angles and ,that makes with the positive a,a,yxand z-axes.321,aaaao,0,cos,cosand are called the directio cosin of cosaWe have aaaaaa321cos,cos,coscos,cos,cosThe vector is a unit vector inthe d
15、irection of aProjectionsThe vector with representation is called the vectorprojection of onto and is denoted by.projababpspsbasbapThe number is called the scalar projection of onto (also called the componen of along )and is denoted by pabWe have The scalar projection of ontob:abaabababbbcoscompaThe
16、vector projection of ontob:aaabaaabaab2a)(projExample Find the scalar projection and the vectorprojection of onto.2kjiakjib23SolutionThe scalar projection isababacompThe vector projection isaabbaacompproj12.4 The Cross Product(The Vector Product)Definition If and 321,aaaa,21nbbbbthen the cross produ
17、ct of and is the vector ab122131132332,babababababababaNote 1is defined only when and are three-abdimensional vectors.kbbaajbbaaibbaabbbaaakjiba212131313232321321Note 2A determinant of order 222112121bababbaaA determinant of order 3TheorembaThe vector is orthogonal to both anda.bProofSince0)(3212123
18、13113232abbaaabbaaabbaaabaA similar computation shows that.0)(bbaTherefore,The vector is orthogonal to both andbaa.bTheorem If is the angle between the vectors and ,thenab0sinbabaProperties of the cross productProof From the definitions of the cross productand length of a vector,we have 212212311322
19、3322)()()(bababababababa22223322112322212322212122123113223322)()()()()()(babababababbbaaabababababababaCorollary Two nonzero vectors and areparallel if and only if ab0baThe length of the cross product is equal tothe area of the parallelogram determined by and aba.bExample Find a vector perpendicula
20、r to the planethat passes through the points),4,3,1(),5,1,2(QP).6,0,3(RandProperties of the cross productprlelgrm Solution The vector perpendicular to bothPQPRPRPQAnd and therefore perpendicular to theplane through p,Q,and R.Example Find the area of the triangle with vertices),4,3,1(),5,1,2(QP).6,0,
21、3(RandSolutionThe area of the parallelogram withadjacent sides and is the length of the crossPRPQproduct:PQPRThe area of the triangle PQR is half the area of thisparallelogram.Properties of the cross ProductIf a,b,and c are vectors and c is scalars,thenb)ca(c)ba(c)(ba 6.cb)a(c)(ba 5.cbcacb)a (4.caba
22、c)(ba 3.b)(ab)(aba)(.2abba.1cccThe product is called the scarlar triple product of the vectors and,a,b.c)(cba321321321)(cccbbbaaacbaWe haveThe volume of the parallelepiped determined by thevector and is the magnitude of their scalartriple product:,a,bc)(vcbaExample Use the scalar triple product to s
23、how thatthe vectorsand,4,1,2,7,4,1ba18,9,0care coplanar.12.5 Equations of Lines and PlanesA line L in three-dimensional space:We know a point on L,and v be a vector of parallel to L,Let be anarbitrary point on L,then a vector equation of L is ),(0000zyxP),(zyxPRtv trr,0where zyxr,0000,zyxrand be the
24、position vectors of and.0PPIf ,we have three scalar equation:),(cbav Rtctzzbtyyatxx,000these equation are called parametric equation of the line L.The numbers are called cbaand,direction numbers of LIf ,and none of a,b,or c is 0,then),(cbav czzbyyaxx000czzbyyxx000,these equation are called symmetric
25、 equation of the line L.If one of a,b,or c is 0,we can still eliminate t.For instance,If a=0,we could write the equation of L asThis means that the line L lies in the vertical plane.0 xx Example Find a vector equation,parametric equa-tions,and symmetric equations of the line that passes through the
26、points and)3,1,6(A).5,4,2(BSolution We take the point as)3,1,6(A.p0Because is parallel to the line and ABv8,3,453,41,26vThus direction number are a=4,b=-3 and c=-8.So the vector equation isHere,3,1,60rThe line segment from to is given by the vectorequation 10,)1(10tr trtr1r0rPlanesA plane in space i
27、s determined by a point),(0000zyxP),(zyxPand a vector n that id orthogonal to the plane.this orthogonal vector n is called a normal vector.Let be an arbitrary point in the plane,then a vector equation of the plane is 0)(0rrnwhere zyxr,0000,zyxrand be theposition vectors of and.0PPIf ,),(cban 0)()()(
28、000zzcyybxxa0dczbyaxthen the scalar equation of the plane is)(000czbyaxdThe linear equation iswhereExample Find an equation of the plane that pass through the points),6,1,3(),2,3,1(QP).0,2,5(RandSolutionLet.2,1,4PR,4,4,2PQbaBecause is orthogonal to the plane and can betaken as the normal vector n.th
29、us bakjikjiban142012214442With the point and the normal vector n an eqution of the plane is )2,3,1(P0)2(14)3(20)1(12zyxor50142012zyxThe distance D from a point to the plane is ),(1111zyxP0dczbyax222111cbadczbyaxDSolutionLet be any point in the),(0000zyxPgiven plane and let .,000zzyyxxbThen nbnbDn co
30、mp222010101)()()(cbazzcyybxxanbnbDn comp222000111)(cbaczbyaxczbyax222111cbadczbyaxIn three-dimensional analytic geometry,an equation in x,y,and z represents a surface in .For example:1czbyax)0,(cbaPozyxRQ3RThe equation of a plane00axbyczdabc12.6 Cylinders and Quadric SurfacesnA cylinder is a surface
31、 that consists of all lines that are parallel to a given line and pass through a given plane curve.MCL1.The parabolic Cylinder24xcyThis surface is formed by all lines that pass throughthe parabolic and are parallel to the z-axis.24xcyyxzoparabolic1KK:DJ:2.The Elliptic Cylinder22221xyab3.The Hyperbol
32、ic Cylinder22221xyabxyzoiliptik haipblik nQuadric Surfaces A quardic surface is the graph of a second-degree equation in three variables x,y,and z.The most general such equation is nilipsid 1.ellipsoid0222JIzHyGxFxzEyzDxyCzByAxoxyz2222221xyzabc2.Elliptic paraboloidiliptik prblid 224yxzzxyn3.Hyperbol
33、ic paraboloid2222yxzba-505-10010-4-2024-505-100104.The Quadric Cone22222xyzab5.The Hyperboloid of One Sheet6.The Hyperboloid of Two Sheet2222221xyzabc2222221xyzabc kun 12.7 Cylindrical and Spherical CoordinatesIn the cylindrical coodinates,a point P in three dimensional space is represented by the o
34、rdertriple .),(zr),(zrP),(1rPrzTo convert from cylindrical to rectangular coodinates,we use the equationszzryrx，sin,coswhereas to convert from rectangular to cylindrical coodinates,we usezzxyyxr,tan222，Example(a)Plot the point with cylindrical coordinates and find its rectangularcoordinates.)1,32,2(
35、b)Find cylindrical coordinates of the point withrectangular coordinates).7,3,3(a)The point with cylindrical coordinates Solution)1,32,2(is plotted in Figure.Its rectangular coordinates are1332sin2sin132cos2coszryrx(b)We have71tan3322zxyyxrTherefore,one set of cylindrical coordinate is).7,47,23(Anoth
36、er is)7,4,23(Spherical CoordinatesIn the spherical coodinates,a point P in three dimensional space is represented by the ordertriple .),(),(PTo convert from spherical to rectangular coodinates,we use the equationscossinsin,cossinzyx，whereas to convert from recrangular to spherical coodinates,we use2
37、222zyxExample(a)The point is given in spherical coordinates.Plot the point and find its rectangularcoordinates.(b)The point is given in rectangular coordinates.Find its spherical coordinates.)2,32,0(a)The point with spherical coordinates Solutionis plotted in Figure.)3,4,2()3,4,2(Its rectangular coordinates are13cos2cos234sin3sin2sinsin234cos3sin2cossinzyxThus,the point is rectangularcoordinates.)3,4,2()1,23,23(b)We have3221cos20sincos4222zxzyxTherefore,spherical coordinates of the given point are)32,2,4(
