1、Section 8.312Equations for Planes in SpaceA plane in space is determinedby knowing a point on the planeand its“tilt”or orientation.(,)P x y zThen M is the sets of all points for whichSuppose that plane M passes through a pointnijk.ABCand is normal(perpendicular)to the nonzero vector0000(,)P xy z0P P
2、uuu vis orthogonal to n.3Equations for Planes in Space(,)P x y zPlane M is the sets of all points0P Puuu vfor whichis orthogonal to n.Thus,the dot product0n0.P Puuu vThis equation is equivalent to000(ijk)()i()j()k0ABCxxyyzzor000()()()0.A xxB yyC zzPoint normal form(点点法式法式):scalar equation of the pla
3、ne4Equation for a planehas0000(,)P xy zThe plane throughnijkABCnormal toComponent equation simplified:0n0P PuuuuvVector equation:000()()()0A xxB yyC zzComponent equation:0,AxByCzDwhere000()DAxByCz Two planes are parallel if and only if their normals are parallel,or12nnk for some scalar k.Equations f
4、or Planes in Space5Finding an Equation for a planeperpendicular to0(3,0,7)P Example Find an equation for the plane throughn5i2j k.Solution The component equation is515270 xyz Simplifying,we obtain5(3)2(0)(1)(7)0.xyz Finish.5222.xyz 0n0P Puuuuv6Equations for Planes in SpaceGeneral Equation for a plan
5、eThe equation can be rewritten in the form0,AxByCzDwhere000()DAxByCz Therefore,the equation of any plane is a linear equation in threeConversely,any linear equation in three variables representsvariables.if A,B,C are not all 0.n(,),A B C a plane with normal vectorthe equation can be written as0,C In
6、 fact,if(0)(0)0.DA xB yC zC 7Some Planes with Special Locations (1)If a given plane passes through the origin(0,0,0),Othen0 xyzsatisfy the general equation for the plane,so that0.D Therefore,the equation of the plane through the origin is0.AxByCzOxyz8OxyzSome Planes with Special Locations(2)If a giv
7、en plane is parallel to the z-axis,andn k0.CTherefore,the equation of this plane isn(,)A B C k(0,0,1),is orthogonal to 0.AxByD0,0,ByCzDAxCzDSimilarly,the equations of planes whichare parallel to the x-axis or y-axis arerespectively.nthen the normal vector9OxyzSome Planes with Special Locations (3)If
8、 a given plane is orthogonal to the z-axis,0.ABTherefore,the equation of this plane is0CzD0,0,AxDByDSimilarly,the equations of planes whichare orthogonal to the x-axis or y-axis arerespectively.and so 0.DzzC orn0zn/kthen10Intercept Form of the Equation for a PlaneExample Review the example of findin
9、g the equation of plane throughand(0,0,1),A(2,0,0)B(0,3,0).CSolution is any point in the plane,Suppose that(,)P x y z(0,3,1)AC uuu v(,1)APx y zuuu v(2,0,1)AB uuu vSince these three vector are coplanarif and only if the point P lies in the plane,0,AP AB AC uuu v uuu v uuu v(0,3,0).C(0,0,1),A(2,0,0)Bx
10、yzO(,)P x y zthenso we have11Intercept Form of the Equation for a PlaneSolution(continued)(0,3,0).C(0,0,1),A(2,0,0)Bthat is 12010.031xyz Expanded the determinant on the leftside of above equation,we have3266,orxyzxyzO(,)P x y zFinish.Example Review the example of finding the equation of plane throug
11、hand(0,0,1),A(2,0,0)B(0,3,0).C1.231xyz12Intercept Form of the Equation for a PlaneIf the intercepts of the plane with the x-axis,y-axis and z-axis(0,0)Bb(0,0,)Cc(,0,0)A afor the plane1.yxzacband,OAa OBb are,OCc Then,just as the lastrespectively.example,we can obtain the equationxyzO(,)P x y z13Angle
12、 Between Planes(两平面的夹角两平面的夹角)1 1n2nThe angle between two intersectingplanes is defined to be the(acute)angle determined by the normalvectors as shown in the figure.Let 22222:0.A xB yC zD 11111:0,A xB yC zD 2 and1111n(,)A B C There normal vectors can be chosen asThen 2222n(,),A B C respectively.12121
13、21222222212111222|nn|cos.|n|n|A AB BC CABCABC 14If two planes are parallel or orthogonal,their normal vector alsoparallel or orthogonal.2 1212120.A AB BC C12 12nn2 111222.ABCABC12/12n/n1 2n1n1 2n1nAngle Between Planes15Position Relationships between two PlanesExample Discuss the position relationshi
14、ps between thefollowing planes:12(1):210,:310 xyzyz Solution(1)Since 22222|1 02 11 3|cos(1)2(1)13 then these two planes intersect and the angle between them is1.60 1arccos.60 16Solution(2)Since 12,1,1,n 2 4,2,2n and211,422 then these two planes are parallel.Again,since1(1,1,0)M but2(1,1,0),M these t
15、wo planes are not same.Position Relationships between two PlanesExample Discuss the position relationships between thefollowing planes:12(2):210,:42210 xyzxyz 1712(3):210,:42220 xyzxyz Solution(3)Since 211,422 these two planes are parallel.Again,since1(1,1,0)M and2(1,1,0),M these two planesare same.
16、Finish.Example Discuss the position relationships between thefollowing planes:Position Relationships between two Planes18Example Find an equation for the plane that passes throughand is parallel to the plane(1,2,0)the point13460.2xyzSolution Let the normal vector to the plane be n;thenn(1,6,8)can be
17、 taken as the normal1n/,3,42 and soorThus the equation of the plane is68110.xyzvector of.(1)6(2)8(0)0 xyzFinish.Position Relationships between two Planes19Since the two point P1 and P2 lie in the plane,we haveSolution(I)be the equation for the plane.0AxByCzDLet0.ABC0.ABCDand20,ACDExample Find an equ
18、ation for the plane that passes throughThe two pointsand is perpendicular to the plane2(1,1,1),P1(1,0,2),P 10.xyz 10,xyz Because is perpendicular to the planewe havePosition Relationships between two Planes202300,CxCyCzSolution(I)Example Find an equation for the plane that passes throughThe two poin
19、tsand is perpendicular to the plane2(1,1,1),P1(1,0,2),P 10.xyz Therefore,the equation for isPosition Relationships between two Planes2020300ACDACABCDBCABCD 230.xyzThat is21Thus,the equation of the plane is230.xyzSolution(II)Let the normal vector to the plane be n.Then 12n,P P uuuu v12(2,1,1).P P uuu
20、u vwhere by the given conditions,thenn(1,1,1)Also,(2,3,1).12n(1,1,1)P Puuuu vijk211111Finish.Position Relationships between two PlanesExample Find an equation for the plane that passes throughThe two pointsand is perpendicular to the planeP2(1,1,1)1(1,0,2),P 10.xyz 22xyzOEquations for Lines in Space
21、In the plane,a line is determined by a point and a numberAnalogously,in space a line is determined by a point andgiving the slope of the line.L0000(,)P xy za vector giving the direction of the line.v(,)P x y z23Equations for Lines in SpaceSuppose that L is a line in space passing through a pointpara
22、llel to a vector 0000(,)P xy z123vijk.vvvxyzOL0000(,)P xy zv(,)P x y zThen L(,)P x y zfor which0P Puuu vis parallel to v.is the set of all pointsThus for some scalar parameter t.0vP P tThe value of t depends on the location of the point P along the line,and the domain of t is(,).24and this last equa
23、tion can be rewrittenEquations for Lines in Space000123()i()j()k(ijk).xxyyzzvvvtasis0vP P t The expanded form of the equation000123ijkijk(ijk).xyzxyzvvvt(1)xyzOL0000(,)P xy zv(,)P x y z25000123ijkijk(ijk).xyzxyzvvvtThese equations give us the standard parametrization of the line for theparameter int
24、erval.tEquating the corresponding components of the two sides of Equation(1)Gives three scalar equations involving the parameter t:010203,.xxvyyvzzvttt010203,.xxvyyvzzv ttttParametric Equation for a Lineparallel 0000(,)P xy zThe standard parametrization of the line throughis123vijkvvvto(2)Equations
25、for Lines in Space26xyzOvL(,)P x y z0000(,)P xy z0r()r tfollowing vector form for the equation of a line in space.on the line and is the position vector of a pointIfr()t0r(,)P x y zthen we have theis the position vector of point0000(,),P xy zVector Equation for a LineA vector equation for the line L
26、 through0000(,),P xy zparallel to v is0r()rv,ttt r()tare the position vector of 0randand(,)P x y z0000(,)P xy zpoint on the line,respectively.Equations for Lines in Space27Parametrizing a Line Through a Point Parallel to a VectorSolution equation(2)become0000(,)P xy zWith equal(2,0,4)123ijkvvvequala
27、ndtoparallel to Find parametric equations for the line through(2,0,4)v2i4j2k.2i4j2k,to22,4,42.xtytzt 010203,.xxtvyytvzztvt (2)Finish.28Solution The vector(1(3)i(12)j(4(3)kPQ 4i3j7k is parallel to the line,14,13,47.xtytzt and Find parametric equations for the line through(3,2,3)P(1,1,4).Q(1,1,4)Q as
28、the“base point”and written We could have chosen34,23,37.xtytzt These equations serve as well as the first;different point on the line for a given value of t.give000(,)(3,2,3)xy z and equation(2)withthey simply place you at a010203,.xxtvyytvzztvt (2)Finish.Parametrizing a Line Through a Point Paralle
29、l to a Vector29If we eliminate the parameter t in the equations,we obtain theequivalent formscalled the symmetric form equations of L.000123.xxyyzzvvv00023,.yyzzxxvvlies on the line L if and only if the(,)P x y zObviously,a point satisfy the equations,x y zcoordinatesPof010203,.xxvyyvzzvtttIn this c
30、ase,we write10,v If0.xx 00 xxthis impliesorthe direction numbers of LEquations for Lines in Space30Lines of IntersectionTwo planes that are not parallel intersect in a line.Suppose that the equations of two planes are22222:0.A xB yC zD and11111:0A xB yC zD 1 2 Then the equation for the line of inter
31、sectioncan be represented by the systemof equations111122220.0A xB yC zDA xB yC zD This is called the generalform of the equations of the line.L311 2 LFinding a Vector Parallel to the Line of Intersection of Two Planes Find a vector parallel to the line of intersection of the planes and225.xyz36215x
32、yz1n2n12nn Solution As in the right figure,the required vector is12ijknn362212 14i2j15k.Finish.32Parametrizing the Line of Intersection of Two Planes Find parametric equations for the line in which the planes225xyzintersect.36215xyzandSolution We find a vector parallel to the line and a point on the
33、 lineand use equation(2).The last example identifiesv14i2j15kas a vector parallel to the line.To find a point on the line,we can takeany point common to the two planes.Substituting0z in the planeequations and solving for x and y simultaneously identifies one of thesepoints as(3,1,0).The line is3 14,
34、12,15.xtytzt 010203,.xxtvyytvzztvt (2)Finish.331 2 LFinding the Equation for a Plane Find the equation of a plane that passes through the andline L of intersection of the two planes1:25340 xyz 2:310 xyz and is perpendicular to the plane 2.Solution(I)12ann2n1nnIt is easy to see that0(7,2,0)P is on L
35、and the directionvector of L isa(4,1,1).By the assumptions,we know na.Notice that2nn,thenn can be chosen as34Finding the Equation for a PlaneSolution(I)(continued)1 2 2n12annn2ijkna n411(2,5,13).131Sincelies in the plane ,0(7,2,0)P the equation of is2513240.xyzor2(7)5(2)130,xyzLFinish.351 2 LFinding
36、 the Equation for a PlaneSolution(II)The equation of the pencil of planesthrough L is(2)(53)(3)(4)0.t xt yt zt Since is perpendicular to the plane 2,we haveExample Find the equation of a plane that passes through the andline L of intersection of the two planes1:25340 xyz 2:310 xyz and is perpendicul
37、ar to the plane 2.361 2 LFinding the Equation for a PlaneSolution(II)(continued)(2)(1)(53)(3)(3)(1)0.ttt Solving this equation,we obtain that20,11t and therefore the equation of the plane is2513240.xyzCompare this result with the previous result,it is easy to see that this method serves just as the
38、previous method.(2)(53)(3)(4)0.t xt yt zt 2nn2:310 xyz Finish.371L2L1L2LPosition Relationships Between Two LinesThe included angle between two lines is defined as the acute angle 1v2v1v2v 1v2v1v2vbetween their direction vectors.Thus,two lines are parallel(ororthogonal)iff their direction vectors are
39、 parallel(or orthogonal).38Position Relationships Between Two LinesLet 2222222:xxyyzzLlmnand1111111:xxyyzzLlmnbe two given lines.Then their direction vectors can be chosen as2222a(,),l m n respectively.1111a(,)l m n andBy the formula ofinclude angle between two vectors,we have1212|aa|cos|a|a|1 21212
40、222222111222|.l lm mn nlmnlmn By this formula,we can easily find the angle between two lines.391L2LPosition Relationships Between Two LinesThen,by the necessary and sufficient condition for two vectors tobe parallel and orthogonal,we obtain:1a2a12LL 12aa1 212120l lm mn n1L2L1a2a12/LL12a/a111222:lmnl
41、mn40Solution The direction vectors of L1 and L2 areExample Discuss the position relationships between the two lines2524:211xyzL1221:123xyzL If they intersect,find the point of intersection.If they are coplanar,find the plane in which they lie.andand2a(2,1,1)1a(1,2,3)respectively.Then the cosine of t
42、he angle between these two line is222222|(1)(2)(2)(1)(3)(1)|(1)23211 1 21212222222111222|cosl lm mn nlmnlmn 3.2 21 Then these two lines neither parallel nor orthogonal.Position Relationships Between Two Lines411L2LSolution(continued)It is easy to see that pointis1(2,2,1)P2(5,2,4)Pis on on the linean
43、d1L2.Lshould coplanar.12a,a12P P andSince1212345,a,a123211P P 2Lare coplanar,the vector Then,if and1LTherefore,these two line intersect.1P2P0,12P P 2a1aAre they coplanar?2Lare coplanar.1LThus and2524:211xyzL1221:123xyzL Do they intersect?Position Relationships Between Two Lines42Solution(continued)w
44、e haveTherefore,the point of intersection of these two linesParametrizing 2,L1Land252:2.4xsLyszs 12:2213xtLytzt andSolving the equations252,222,134,tststs we find1,2.ts is(1,0,2).Where is the point of intersection?2524:211xyzL1221:123xyzL Position Relationships Between Two Lines43Solution(continued)
45、Since are coplanar,we will find the equation of the plane.2L1LandThen the equation of the plane is(1)(2)(7)(2)(5)(1)0,xyz or75110.xyzThe normal vector of the plane is12naaijk123211 (1,7,5).What is the equation of the plane?2524:211xyzL1221:123xyzL Finish.Position Relationships Between Two Lines44Ang
46、le between a Line and a Plane L nThe included angle between the line Land the plane is defined as the acuteangle between L and its projectionvector in the plane.be a plane and:0AxByCzD LetThen the include angle between000:xxyyzzLlmnbe a line.of L and the normal vector n of is a(,)l m n the direction
47、 vector.2 2 oran 45Position Relationships Between a Line and a Plane L nasincos2 cos(a,n)222222|.AlBmCnABClmn Then,it is easy to see that/L an0,AlBmCnL a/n:.l m nA B CTherefore n:0AxByCzD 000:xxyyzzLlmn46Position Relationships Between a Line and a Plane1,22,2,.xtytztt Solution The parametric equatio
48、ns of L areExample Find the point of intersection of the line L and the plane,and the include angle between L and,where12:,:410.122xyzLxyz is a point of intersection of L and .(,)P x y zSuppose thatThen thecoordinates of P must satisfy both the equations of L and of .Substituting the expression for
49、x,y and z from the parametric equationsinto the equation of,we obtain47Position Relationships Between a Line and a PlaneSolution(continued)Substituting this value into the parametric equations of L,124,.333xyz we find2.3t Solving the equation,we obtainTherefore,the point of intersection is124,.333(1
50、)4(22)210.ttt 48Position Relationships Between a Line and a PlaneSolution(continued)222222|1 14(2)(1)2|arcsin14(1)1(2)2 the included angle between L and isn(1,4,1),to isand the normal vector a(1,2,2)Since the direction vector of L is1arcsin.42 Finish.49The Distance from a Point to a Plane We choose
