maple trong hhgt

8/14/2019 Maple Trong HHGT

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> plane(P2,[d,EC],[x,y,z]);P2

> Equation(P2);

= + 288 144x 144y 48z 0 > line(d3,[P,P2]);

d3

> Equation(d3,m);[ ], ,+2 10368 m 4 31104 m

1

HNG DAN S DUNG MAPLE TRONG HNH HOCPhan I. HNH HOC GIAI TCH TRONG MAT PHANG

* Trc khi bat au lam viec vi hnh hoc giai tch trong mat phang

bang Maple, ta phai dung lenh: [> with(geometry);

1) Mot vai thao tac c bana) Nhap toa o mot iem .

e nhap toa o cua iem A(a; b) ta nhap nh sau:[> point (A, a, b);

b) Tnh khoang cach gia hai iem A(x1; y1) va B(x2; y2), ta nhap:[> point(A,x1,y1),point(B,x2,y2);

,A B

[> distance(A,B);

+( )x1 x2 2 ( )y1 y2 2

c) ng thang :e nhap phng trnh cua ng thang l : ax + by + c = 0, ta nhap

[> line(l,a*x +b*y + c = 0,[x,y]);

I. TAM GIAC VA CAC VAN E LIEN QUAN1) Khai bao mot tam giac trong Maplea) Tam giac co ten la ABC i qua ba nh A, B, C cho trc , ta nhap:

triangle[ABC, [A, B, C] );V du: Khai bao mot tam giac ABC i qua ba iem A(1; 1), B(0; 0) va

C(0; 5) ta lam nh sau:[> point(A,1,1), point(B,0,0),point(C,0,5);

, ,A B C

[> triangle(ABC,[A,B,C]);ABC

b) Tam giac co ten la T c lap bi ba ng thang l1 , l2 , l3. Ta nhap:


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triangle(T, [l1, l2, l3]);

V du: Ba canhAB,BC, CA cua tam giacABCco phng trnh lan lt la :x + 21y 22 = 0, 5x 12y + 7 = 0 , 4x 33y + 146 = 0. Khi o, ta nhap

[> line(AB, x + 21*y -22 = 0,[x,y]),line(BC,5*x - 12*y +7 =0,[x,y]),line(AC, 4*x - 33*y +146 =0,[x,y]),triangle(ABC,[AB,BC,AC]);c) Tam giac khi biet o dai ba canh.

triangle(Ten tam giac , [canh 1, canh 2, canh 3]);

V du: e nhap tam giac co o dai ba canh la 3, 4, 5. Neu tam giac nayco ten la ABC, ta nhap:[> triangle(ABC,[3,4,5]);

ABC

d) Tam giac khi biet o dai hai canh va goc xen gia hai canh o

triangle(T, [canh 1, 'angle'= goc xen gia hai canh, canh 2]) ;V du: e nhap tam giac co o dai hai canh la 2, 1 va goc xen gia haicanh la /2, ta nhap:[> triangle(T4,[2,'angle'=Pi/2,1]):

2) CAC NG AC BIET TRONG TAM GIACA. NG CAOe khai bao ng cao hA i qua nh A cua tam giac ABC, ta nhap :

altitude(hA, A, ABC);

hay altitude(hA, A, ABC, H );

v ay, H la chan ng cao.71

> a:=ParallelVector(d);:=a [ ], ,-12 0 36

> point(A,2,0,0);point(E,0,0,0);

A E

> point(B,0,4,0);B

> point(C,2,4,6);

C> line(AB,[A,B],t);

AB

> Equation(AB);[ ], ,2 2 t 4 t 0

> plane(P,[d,AB],[x,y,z]);P

> Equation(P);= 576 144x 72y 48z 0

> line(EC,[E,C]);EC

> Equation(EC,t);[ ], ,2 t 4 t 6 t

> intersection(M,EC,P);M

> coordinates(M);

, ,

4

3

8

3 4


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Use d1, d2, and d3 to define the parallelepiped pp.> parallelepiped(pp,[d1,d2,d3]);

pp

> form(pp); parallelepiped3d

> DefinedAs(pp);[ ], ,d1 d2 d3

> detail(pp);name of the object: pp

form of the object: parallelepiped3d

the 6 parallelogram faces of the object: [[[0, 0, 0], [4, 0, 0], [9, 5, 1], [5, 5, 1] \

], [[0, 2, 5], [4, 2, 5], [9, 7, 6], [5, 7, 6]], [[0, 0, 0], [4, 0, 0], [4, 2, 5], [0, 2, 5 \

]], [[4, 0, 0], [9, 5, 1], [9, 7, 6], [4, 2, 5]], [[5, 5, 1], [9, 5, 1], [9, 7, 6], [5, 7, \

6]], [[0, 0, 0], [5, 5, 1], [5, 7, 6], [0, 2, 5]]]

coordinates of the 8 vertices: [[0, 0, 0], [4, 0, 0], [5, 5, 1], [9, 5, 1], [0, 2, 5], [ \

4, 2, 5], [5, 7, 6], [9, 7, 6]]

(e d b khoi A, 2007)> plane(alpha,6*x-3*y+2*z=0,[x,y,z]);

> plane(beta,6*x+3*y+2*z-24=0,[x,y,z]);

> n1:=NormalVector(alpha);

:=n1 [ ], ,6 -3 2

> n2:=NormalVector(beta);:=n2 [ ], ,6 3 2

> line(d,[alpha,beta]);d

3

v e xem chi tiet ve ng cao hA ta dung lenh detail(hA);v Trong detail, neu khai bao theo cach 1 ta se biet c phng

trnh ng cao hA, con neu khai bao theo cach 2 ta se biet ctoa o chan ng cao H.

V du: Viet phng trnh ng cao hA cua tam giac ABC vi ba nhA(0; 0), B(2; 0) va C(1; 3). Ta lam nh sau:Cach 1[> triangle(ABC, [point(A,0,0), point(B,2,0), point(C,1,3)]):altitude(hA1,A,ABC);

hA1

[> detail(hA1);assume that the names of the horizontal and vertical axes are _x and _y,respectively

name of the object: hA1

form of the object: line2d

equation of the line: -_x+3*_y = 0

Trong detail ta co phng trnh ng cao hA1 la x + 3y = 0Cach 2[> with(geometry);[> triangle(ABC, [point(A,0,0), point(B,2,0),point(C,1,3)]):altitude(hA1,A,ABC,H);

hA1

[> detail(hA1);name of the object: hA1

form of the object: segment2d

the two ends of the segment: [[0, 0], [9/5, 3/5]]

Chu y: Trong detail [9/5,3/5] la toa o chan ng vuong goc H.

B. NG TRUNG TUYENe khai bao ng trung tuyen AM i qua nh A cua tam giac ABC, ta

nhap :


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median(AM, A, ABC);

* e xem chi tiet ve ng trung tuyen AM, ta dung lenh detail(AM);V du : Viet phng trnh ng trung tuyen AM cua tam giac ABC biet

A(5; 1), B(2; 3) va C( 6; 1).[> triangle(ABC, [point(A,5,1),point(B,2,3),point(C,-6,-1)]):median(AM,A,ABC);

AM

[> detail(AM);assume that the names of the horizontal and vertical axes are _x and _y,

respectivelyname of the object: AMform of the object: line2d

equation of the line: 7-7*_y = 0

trong detail cho biet phng trnh ng trung tuyen AM la 7 7y = 0.

C. NG PHAN GIAC TRONG CUA TAM GIAC.

e khai bao ng phan giac trong AD i qua nh A cua tam giac ABC,ta nhap :

bisector(AD, A, ABC);

* e xem chi tiet ve ng phan giac trong AD, ta dung lenhdetail(AD);V du : Viet phng trnh ng phan giac trong AD cua tam giac ABC

biet A(1; 6), B(3; 4) va C(0; 1).[>triangle(ABC,[point(A,1,6),point(B,3,4),point(C,0,1)]):bisector(AD,A,ABC);

AD

[> detail(AD);assume that the names of the horizontal and vertical axes are _x and _y,

respectively

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DescriptionA parallelepiped is a polyhedron bounded by six parallelograms. It

can be defined from three given directed segments having a

common initial point.To access the information related to a parallelepipedpp, use the

following function calls:

form(pp) returns the form of the geometric object

(that is, parallelepiped3d ifpp is a parallelepiped).

See geom3d[form].

DefinedAs(pp) returns the list of three directed segments

defining pp. See geom3d[DefinedAs].

detail(pp) returns a detailed description of the

parallelepiped pp. See geom3d[detail].

This function is part of the geom3d package, and so it can be usedin the form parallelepiped(..) only after executing thecommand with(geom3d). However, it can always be accessedthrough the long form of the command by usinggeom3d[parallelepiped](..).

Examples> with(geom3d):Define four points A, B, C, and E.> point(A,0,0,0), point(B,4,0,0), point(C,5,5,1), point(E,0,2,5):Define three directed segments d1, d2, and d3 with initial point A andend points B, C, and E respectively.

> dsegment(d1,[A,B]), dsegment(d2,[A,C]), dsegment(d3,[A,E]):


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MBD

[> Equation(MBD);

=+ + 1

2

a~ b~ x1

2

a~ b~ y

+2

+1

2

a~1

2

a~ a~ a~2 z1

2

a~2 b~ 0

[> ArePerpendicular(A1BD,MBD,'cond');FAIL

[> cond;

=+a~2 b~2 a~2

+2

+1

2a~

1

2a~ a~ a~2 0

Ta hieu la a2b2 + a2 ( 2a2 + a2) = 0 hay a2b2 a4 = 0

[> solve(a^2*b^2+a^2*(-2*(1/2*a+1/2*a)*a+a^2) = 0,{a});, , ,{ }=a~ 0 { }=a~ 0 { }=a~ b~ { }=a~ b~

Chu y :1) Cac ky hieu a ~ , b ~ ta hieu la a va b phai thoa ieu kien

ma ta a at trongassume, tc laa > 0 va b > 0.

2) Do a > 0 va b > 0, nen ta ch nhan a = b hay 1=ba .

Xac nh lang truCu phap parallelepiped(pp, [d1, d2, d3]) xac nh lang tru pp vi bacanh la d1, d2, d3.

geom3d[parallelepiped] - define a parallelepipedCalling Sequence

parallelepiped(pp, [d1, d2, d3])

Parameterspp - name of the parallelepiped

d1, d2, d3 - three directed segments having a common initial point5

name of the object: ba


equation of the line: (5*8^(1/2)+2*26^(1/2))*_x+(2*26^(1/2)-8^(1/2))*_y-1

6^(1/2)+8^(1/2) = 0

D. NG PHAN GIAC NGOAI CUA TAM GIAC.e khai bao ng phan giac AE i qua nh A cua tam giac ABC, tanhap :

ExternalBisector(AE, A, ABC);

* e xem chi tiet ve ng phan giac ngoai AE, ta dung lenh detail(AE);

3) CAC IEM AC BIET TRONG TAM GIACA. TRONG TAM CUA TAM GIACGoi G la trong tam cua tam giac ABC, khi o:

a) G c khai bao bi lenh centroid(G, ABC);b) Toa o G c xac nh bi lenh coordinates(G);

V du1: Cho tam giac ABC vi A(2; 3), B(-2; 4), C( 4; 7). Tm toa o

trong tam G cua tam giac ABC.

[> triangle(ABC,[point(A,2,3),point(B,-2,4),point(C,-4,7)]);ABC

[> centroid(G,ABC);G

[> coordinates(G);

,

-4

3

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3

V du2:Cho tam giac ABC vi A(1; 2), B(2; 3), C(0; 7). Tm toa o trong

tam G cua tam giac ABC.

Ta co the lam gon hn nh sau:

[> point(A,1,2),point(B,2,3),point(C,0,7);


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, ,A B C

[> coordinates(centroid(G,triangle(ABC,[A,B,C])));[ ],1 4

B. TRC TAM CUA TAM GIACGoi H la trc tam cua tam giac ABC, khi o:

a) H c khai bao bi lenh orthorcenter(H, ABC);b) Toa o H c xac nh bi lenh coordinates(H);

V du: Cac canh AB, BC, AC cua tam giac ABC lan lt co phng trnh:

4x y 7 = 0, x + 3y 31 = 0 , x + 5y 7 = 0.

Xac nh trc tam H cua tam giac.[> line(AB,4*x- y -7 = 0,[x,y]),line(BC,x + 3*y -31 = 0,[x,y]),line(AC,x+5*y -7 =0,[x,y]),triangle(ABC,[AB,BC,AC],[x,y]);

, , ,AB BC AC ABC

[> orthocenter(H,ABC);H

[> coordinates(H); [ ],3 4

[> map(coordinates,DefinedAs(ABC));[ ], ,[ ],4 9 [ ],2 1 [ ],67 -12

Chu y: lenh map(coordinates,DefinedAs(ABC)); Cho ta xac nh ctoa o cua ba nh A, B, C.

4)TRUNG TRC CUA MOT OAN THANG.e khai bao l la trung trc cua oan thang AB, ta dunglenh

PerpenBisector(l, A, B );

Viet phng trnh trung trc l cua oan thang BC, biet B(2; 0) va C(1; 3)

67

[> assume(a>0), assume(b>0);[> dsegment(d1,[A,B]),dsegment(d2,[A,D]),dsegment(d3,[A,A1]);

, ,d1 d2 d3

[> parallelepiped(HHCN,[d1, d2, d3]);HHCN

[> detail(HHCN);* V khong u giay, nen khong in ra ay ket qua cua detail(HHCN);[> point(C1,a,a,b);

C1

[> midpoint(M,C,C1); M

[> coordinates(M);

, ,+

1

2a~

1

2a~ +

1

2a~

1

2a~

1

2b~

( ay ta hieu M

2

baa ;; )

[> gtetrahedron(BDA1M,[A,D,A1,M]);BDA1M

[> volume(BDA1M);1

6a~ b~

+

1

2a~

1

2a~

[>plane(A1BD,[A1,B,D],[x,y,z]);

A1BD

[> Equation(A1BD);=+ + a~ b~ x a~ b~ y a~2z a~2 b~ 0

[> n1:=NormalVector(A1BD);:=n1 [ ], ,a~ b~ a~ b~ a~2

[> plane(MBD,[M,B,D],[x,y,z]);


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, , ,A B D A1

[> dsegment(d1,[A,B]),dsegment(d2,[A,D]),dsegment(d3,[A,A1]);, ,d1 d2 d3

[> assume(a > 0);[> parallelepiped(ABCDA1B1C1D1,[d1, d2, d3]);

ABCDA1B1C1D1

[> detail(ABCDA1B1C1D1);[> point(D1,0,a,a), point(C,a,a,0);

,D1 C

[> plane(BA1C,[B,A1,C],[x,y,z]);BA1C

[> Equation(BA1C);= +a~2x a~2z a~3 0

[> plane(DA1C,[D,A1,C],[x,y,z]);DA1C

[> Equation(DA1C);=+ a~2y a~2z a~3 0

[> FindAngle(BA1C,DA1C);1

3

Lu y: ay la goc gia hai mat phang2) Trong khong gian vi he toa o ecac vuong goc Oxyz cho hnh

hop ch nhat ABCD.ABCDco A trung vi goc toa o,B(a; 0; 0),D(0;a; 0), A(0; 0; b) (a > 0, b > 0). GoiMla trung iem canh CC.

a) Tnh the tch khoi t dienBDM theo a va b.

b) Xac nh ty sob

a e hai mat phang (ABD) va (MBD) vuong goc.

a) [> point(A,0,0,0),point(B,a,0,0),point(D,0,a,0),point(A1,0,0,b);, , ,A B D A1

7

[> point(B,2,0), point(C,1,3);,B C

[> PerpenBisector(l,B,C);l

[> detail(l);assume that the names of the horizontal and vertical axes are _x and _y,respectivelyname of the object: l


equation of the line: -3-_x+3*_y = 0

may tra li l co phng trnh la 3 x + 3y = 0.

5) DIEN TCH CUA MOT TAM GIACe tnh dien tch cua tam giac ABC ta dung lenh area(ABC);

V du: Tnh dien tch tam giac ABC vi A(2; 3), B(3; 2) va C( 2; 5).

[> with(geometry);[>triangle(ABC,[point(A,2,-3),point(B,3,2), point(C,-2,5)]);

ABC

[> area(ABC);14

May tra lidien tch tam giac ABC la 14.

NG THANG* Khoang cach t mot iem en mot ng thange tnh khoang cach t iemMen ng thang d, ta dung lenh:

distance(M, d);

V du 1. Tnh khoang cach t iem M(2; 3) en ng thang


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d: 3x + 6y = 1

[> with(geometry);[> point(M,2,3),line(d,3*x+6*y=1,[x,y]);

,M l

[> distance(M,d);

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155

V du 2: Ba canhAB,BC, CA cua tam giacABCco phng trnh lan ltla : x + 21y 22 = 0, 5x 12y + 7 = 0 , 4x 33y + 146 = 0.

Tnh khoang cach t trong tam cua tam giac en canhBC.[> line(AB, x + 21*y -22 = 0,[x,y]),line(BC,5*x - 12*y +7 =

0,[x,y]),line(AC, 4*x - 33*y +146 =0,[x,y]),triangle(ABC,[AB,BC,AC],[x,y]);

, , ,AB BC AC ABC

[> centroid(G,ABC);G

[> coordinates(G);[ ],-2 3

[> distance(G,BC);3

*Hnh chieu cua mot iem len

mot ng thang

a) e khai bao H la hnh chieu cua iem P len ng thang l, ta dunglenh:

projection(H, P, l);

b) e tm toa o hnh chieuH, ta dung lenh: coordinates(H);V du : Tm hnh chieu Q cua iem P(2; 3) len ng thang

l : x + y + 1 = 0.[> point(P,2,3), line(l,x+y-1=0,[x,y]);

65

Bai 9: (TN, 2000, 2 iem)Trong khong gian vi he toa o Oxyz cho mat phang (P) va mat

cau (S) co cac phng trnh tng ng:(P) 2x 3y + 4z 5 = 0 va (S) x2 + y2 + z2 + 3x + 4y z + 6 = 0.

a) Xac nh toa o tamIva ban knhR cua mat cau (S).b) Tnh khoang cach t tamIen mat phang (P). T o suy ra rang

mat phang (P) cat mat cau (S) theo mot ng tron ma ta ky hieula (C). Xac nh ban knh r va toa o tam Hcua ng tron (C).

[> sphere(S,x^2+y^2+z^2+3*x+4*y-z+6=0,[x,y,z],'centername'=O);S

[> center(S);O

[> coordinates(O);

, ,

-3

2-2

1

2

[> R:=radius(S);

:=R

1

2 2

[> plane(P,2*x-3*y+4*z-5=0,[x,y,z]);P

[> d:=distance(O,P);:=d 0

Chu y : V khoang cach t tam O cua (S) en (P) bang 0 nen iem O(P). Do o, tam cua ng tron giao tuyen chnh la O va ban knh bangban knh cua (S).

Bai 10 :(H, C toan quoc, khoi A, 2003)

1) Cho hnh lap phng ABCD.ABCD . Tnh so o cua goc phangnh dien [B, AC, D].

[> point(A,0,0,0),point(B,a,0,0),point(D,0,a,0),point(A1,0,0,a);


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b) Viet phng trnh cua ng thang qua Cva vuong goc vi matphang (A, B, D).

c) Tnh khoang cach t Cti mat phang (A, B, D).a) [> point(O,0,0,0),point(A,3,0,0),point(B,0,4,0),point(C,0,0,5);

, , ,O A B C [> dsegment(d1,[O,A]),dsegment(d2,[O,B]),dsegment(d3,[O,C]);

, ,d1 d2 d3

[> parallelepiped(HH,[d1, d2, d3]);HH

[> detail(HH);[[0, 0, 0], [0, 4, 0], [0, 4, 5], [0, 0, 5]]]

coordinates of the 8 vertices: [[0, 0, 0], [3, 0, 0], [0, 4, 0], [3, 4, 0], [0, 0, 5], [ \

3, 0, 5], [0, 4, 5], [3, 4, 5]]

[> point(D,3,4,5);D

b) [> plane(ABD,[A,B,D],[x,y,z]);ABD

[> Equation(ABD);= + + 60 20x 15y 12z 0

[> n:=NormalVector(ABD);:=n [ ], ,20 15 -12

[> line(L,[C,n]); L

[> Equation(L,t);[ ], ,20 t 15 t 5 12 t

c) [> distance(C,ABD);120

769

769

9

,P l

[> projection(Q,P,l);Q

[> coordinates(Q);[ ],0 1

* iem oi xng cuamot iem qua mot ng thanga) e khai bao Q la cua iem oi xngcua iem Plen ng thang l, ta

dung lenh:reflection(Q, P, l);

b) ) e tm toa o cua Q, ta dung lenh: coordinates(Q);

V du : Tm iem M1 oi xng vi iem M2(8; 9) qua ng thang iqua hai iemA(3; 4) vaB( 1; 2).

Giai[> point(M2,8,-9),point(A,3,-4),point(B,-1,-2);

, ,M2 A B [> line(AB,[A,B],[x,y]);

AB

[> Equation(AB);= 10 2x 4y 0

[> reflection(M1,M2,AB);M1

[> coordinates(M1);[ ],10 -5

Lu y: Lenh Equation(AB); cho ta phng trnh cua ng thang AB.

* NHOM LENH KIEM TRA

Sau khi anh lenh > with(geometry);


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Ta c cac lenh, trong o co cac lenh bat au bang Are hay Is.Cac lenh nay nham kiem tra tnh ung (true), sai (false) cua mot tnhchat hnh hoc nao o. Sau ay la mot so lenh c ban:

Ten lenh Cu phap Chc nangAreCollinear AreCollinear(P, Q, R, cond) Kiem tra tnh thang hang

cua ba iem P, Q, R.

AreConcurrent AreConcurrent(l1, l2, l3, cond ) Kiem tra tnh ong quycua ba ng thang l1, l2,l3.

AreParallel AreParallel(l1, l2, cond) Kiem tra tnh song songcua hai ng thang l1,

l2.

ArePerpendicular ArePerpendicular(l1, l2, cond) Kiem tra tnh vuong goccua hai ng thang l1,l2.

AreTangent AreTangent(f, g) Kiem tra s tiep xuc cuang thang f va ngtron g hay s tiep xuccua hai ng tron f va g

IsOnCircle IsOnCircle(f, c, cond) Kiem tra xem iem(hoac tap hp cac iem)f co nam tren ng tronc hay khong ?

IsOnLine IsOnLine(f, l, cond) Kiem tra xem iem(hoac tap hp cac iem)

f co nam tren ngthang l hay khong ?IsRightTriangle IsRightTriangle(ABC, cond ) Kiem tra tnh vuong goc

cua tam giac ABC.

Lu y:1) Co the bo cond hoac s dung cond trong trng hp co cha

tham so.

63

[> Equation(anpha);

= + + +1

3

1

3x

1

3y

1

3z 0

[> AreCollinear(O,B,C); true

[> sphere(S,[B,sqrt(2)],[x,y,z]);S

[> Equation(S);=+ + + x2 y2 z2 1 2x 2y 2z 0

[> R:=radius(S);:=R 2

[> d:=distance(B,anpha);

:=d2

33

[> line(AB,[A,B]);AB

[> Equation(AB,t);[ ], ,1 t t

[> projection(l,AB,anpha);l

[> Equation(l);

, ,1

2

3t

1

3t

1

3t

ay la phng trnh tham so .

Bai 8: (TN, 1999, t 2, 3 iem)Trong khong gian Oxyz cho hnh hop ch nhat co cac nhA(3; 0;

0),B(0; 4; 0), C(0; 0; 5), O(0; 0; 0) vaD la nh oi dien vi O.

a) Xac nh toa o nhD. Viet PTTQ cua mat phang (A, B, D).


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[> coordinates(center_S_1);

, ,

3

23 1

[> R:=radius(S);:=R

1

221

[> d:=distance(center_S_1,anpha);

:=d1

2+3 m

[> solve(R=d,{m}); ,{ }=m +3 21 { }=m 3 21

Bai 7: (TN, 2000 2001, 2,5 iem)Trong khong gian vi he toa o Oxyz cho ba iem

A(1; 0; 0), B(1; 1; 1),

3

1

3

1

3

1 ;;C .

a) Viet PTTQ cua mat phang ( )vuong goc ng thang OCtai C.Chng minh rang ba iem O, B, C thang hang. Xet v tr tngoi cua mat cau (S) tamB, ban knh 2 vi mat phang ( ).

b) Viet PTTQ cua ng thang g la hnh chieu vuong goc cuang thangAB tren mat phang ( ).

a) [> point(A,1,0,0), point(B,1,1,1), point(C,1/3,1/3,1/3),point(O,0,0,0);, , ,A B C O

[> line(OC,[O,C]); OC

[> v:=ParallelVector(OC);

:=v

, ,

1

3

1

3

1

3

[> plane(anpha,[C,v],[x,y,z]);anpha

11

2) Khi ket thuc cac lenh nay va nhan Enter th may tra li latrue(ung) hoacfalse (sai).

V du 1: Xet tnh thang hang cua ba iem A(1; 2), B(2; 3) va C(0; 7).Ta lam nh sau:

[ point(A,1,2),point(B,2,3),point(C,0,7);, ,A B C

[ AreCollinear(A, B, C);false

May tra lifalse, tc ba iem A, B, C khong thang hang.

V du 2 : Cho tam giac ABC vi A(1; m 2), B(2; 3 + m), C(0; 7). Tm me ABC la tam giac vuong.

Trc het ta dung lenh [> AreCollinear(A, B, C);e kiem tra tnh thang hang cuaba iem A, B, C[> AreCollinear(A, B, C);AreCollinear: "hint: could not determine if -m+14 is zero"

FAIL

may bao khong the xac nh neu m + 14 = 0, tc ba iem A, B, Cthang hang khi m + 14 = 0 hay m = 14.

Lu y : Trong mot so bai toan ta phai s dung lenh assume(gia s)m khong thoa gia tr tren th may mi thc hien tiep bai toan. Cu the,trong bai nay, ta phai gia s m 14, tc la ta phai nhap :

assume (m 14;Tuy nhien, trongbai nay th khong

[> point(A,1,m -2),point(B,2,3+m), point(C,0,7);, ,A B C

[> IsRightTriangle(ABC,cond);IsRightTriangle: "hint: one of the following conditions must be satisfied: {-76+26*m-2*m^2 = 0, -88+10*m = 0, 36-10*m = 0}"

FAIL

May thong bao : Mot trong cac ieu kien sau phai thoa man:


12/36

12

{-76+26*m-2*m^2 = 0, -88+10*m = 0, 36-10*m = 0}"

Bay gi ta dung lenh solve e tm m.

[> solve(-76+26*m-2*m^2 = 0,{m});

,{ }=m

13

2

1

2 17 { }=m +

13

2

1

2 17

[> solve(-88+10*m = 0, {m});

{ }=m44

5

[> solve(36-10*m = 0,{m});

{ }=m

18

5

Vay ta phai co :

5

18

5

4417

2

1

2

1317

2

1

2

13==+== mmmm ;;;

V du 3: Cho hai ng thang :

l1 : 2x + 5y = 1 va l2 : 5x 2y = 0.

Chng minh rang l1 vuong goc vi l2

[>line(l1,2*x+5*y=1,[x,y]),line(l2,5*x-2*y=0,[x,y]);,l1 l2

[> ArePerpendicular(l1,l2);true

May tra li true tc ta co pcm.

*Phng trnh cua ng thang qua mot iem cho trc va vuong gocvi mot ng thang cho trc.

e viet phng trnh cua ng thang lp qua iem P va vuong goc vi

mot ng thang l cho trc ta dung lenh:61

[> ArePerpendicular(AD,AC);true

[> gtetrahedron(ABCD,[A,B,C,D]);ABCD

[> volume(ABCD);4

3

c) [> plane(ABD,[A,B,D],[x,y,z]);ABD

[> Equation(ABD); =+2 2z 0

[> NormalVector(ABD);[ ], ,0 0 2

[> plane(anpha,2*x+m=0,[x,y,z]);anpha

[> sphere(S,[A,B,C,D],[x,y,z]);S

[> Equation(S);=+ + + x2 y2 z2 7 3x 6y 2z 0

[> detail(S);name of the object: S

form of the object: sphere3d

name of the center: center_S_2

coordinates of the center: [3/2, 3, 1]

radius of the sphere: 1/2*21^(1/2)

surface area of the sphere: 21*Pi

volume of the sphere: 7/2*Pi*21^(1/2)

equation of the sphere: x^2+y^2+z^2+7-3*x-6*y-2*z = 0


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60

Trong khong gian Oxyz vi he toa o Oxyz, cho bon iem A, B, C,D xac nh bi cac he thc:

A = (2; 4; 1),

+= kjiOB 4 , C = ( 2; 4; 3),

+= kjiOD 22 .

a) Chng minh rangAB AC,ACAD,AD AB. Tnh the tch cuakhoi t dienABCD.b) Viet phng trnh tham so cua ng thang vuong goc chung

cua hai ng thang AB va CD. Tnh goc gia ng thang va mat phang (ABD).

c) Viet phng trnh mat cau (S) i qua bon iem A, B, C, D. Vietphng trnh tiep dien )( cua mat cau (S) song song vi mat

phang (ABD).a) [> point(A,2,4,-1),point(B,1,4,-1),point(C,2,4,3),point(D,2,2,-1);, , ,A B C D


[> Equation(AB,t);[ ], ,2 t 4 -1

[> line(AC,[A,C]);AC

[> Equation(AC,t);[ ], ,2 4 +1 4 t

[> line(AD,[A,D]);

AD [> Equation(AD,t);

[ ], ,2 4 2 t -1

[> ArePerpendicular(AB,AC);true

[> ArePerpendicular(AB,AD);true

13

PerpendicularLine(lp, P, l; )

Viet phng trnh ng thang i qua iem A(5; 1) va vuong gocvi ng thang l : x + 4y + 5 =0.

[> point(A,5,-1),line(l,x + 4*y + 5 =0,[x,y]);,A l

[> PerpendicularLine(D,A,l);D

[> Equation(D);= + 21 4x y 0

Vay ng thang can tm co phng trnh 21 + 4x y = 0.

* Phng trnh cua ng thang qua mot iem cho trc va song song

vi mot ng thang cho trc.

e viet phng trnh cua ng thang lp qua iem P va song song vimot ng thang l cho trc ta dung lenh:

ParallelLine(lp, P, l);

Viet phng trnh ng thang i qua iem P(2; 3) va song songvi ng thang l : x + y = 1[> with(geometry):

[> point(P, 2 , 3), line(l, x + y =1, [x,y]); ,P l

[> ParallelLine(D,P,l);D

[> Equation(D);= + +5 x y 0


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14

Vay ng thang can tm co phng trnh 5 + x + y = 0.

GOC

* Goc tao bi hai ng thang.e tnh goc cua hai ng thang d1 va d2 ta dung lenh :

FindAngle(d1, d2);V du : Xac nh goc gia hai ng thang:

1) 5x y + 7 = 0, 3x + 2y = 0;

2) 3x + 2y 1 = 0, 5x 2y + 3 = 0.1)[> line(D1,5*x-+7=0,[x,y]),line(D2,3*x+2*y=0,[x,y]);

,D1 D2

[> FindAngle(D1,D2);1

4

2) [> line(D1,3*x+2*y-1=0,[x,y]),line(D2,5*x-2*y+3=0,[x,y]);,D1 D2

[> FindAngle(D1,D2);

arctan

16

11

NG TRON59

= +56 28x 14z 0

Bai 5: (TN, 1999, t 1, 4 iem)Trong khong gian Oxyz cho iemD( 3; 1; 2) va mat phang i

quaA(1; 0; 11),B(0; 1; 10) va C(1; 1; 8).a) Viet phng trnh ng thangAC.b) Viet PTTQ cua mat phang .c) Viet phng trnh mat cau tam D, ban knh R = 5. Chng minh

rang mat cau nay cat mat phang .a) [> point(A,1,0,11),point(B,0,1,10),point(C,1,1,8),point(D,-3,1,2);

, , ,A B C D

[> line(AC,[A,C]); AC

[> Equation(AC,t);[ ], ,1 t 11 3 t

b) [> plane(anpha,[A,B,C],[x,y,z]);anpha

[> Equation(anpha);= 13 2x 3y z 0

[> sphere(S,[D,5],[x,y,z]);S

c) [> Equation(S);=+ + + x2 y2 z2 11 6x 2y 4z 0

[> distance(D,anpha);14

[> radius(S);5

e y : distance(D,anpha)< radius(S)

Bai 6: (TN, 2002 2003, 2,5 iem).


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58

a) [> point(A,1,0,-2),point(B,0,-4,-4), plane(anpha,3*x-2*y+6*z+2=0,[x,y,z]);

, ,A B anpha

[> sphere(S,[A,anpha],[x,y,z]); S

[> Equation(S);=+ + + +x2 y2 z2 4 2x 4z 0

[> line(L,[A,B]);L

[> Equation(L,t); [ ], ,1 t 4 t 2 2 t

[> intersection(giaodiem,L,anpha);giaodiem

[> coordinates(giaodiem);[ ], ,2 4 0

b) [> a:=ParallelVector(L);:=a [ ], ,-1 -4 -2

[> n:=NormalVector(anpha);:=n [ ], ,3 -2 6

[> with(linalg);Warning, the name inverse has been redefined

[> crossprod(a,n); [ ], ,-28 0 14

[> v:=vector([-28, 0, 14]);:=v [ ], ,-28 0 14

[> plane(P,[point(A,1,0,-2),v],[x,y,z]);P

[> Equation(P);15

1) Khai bao phng trnh mot ng tron

Neu ng tron C, co phng trnh

x2 + y2 2ax 2by + c = 0

Trong Maple ta nhap:

circle(C,x^2 + y^2 2*a*x 2*b*y + c = 0,[x, y]);

2)Thiet lap phng trnh ng tron .

Maple cho phep lap phng trnh ng tron thoa mot trong cac . K sau:

a)ng tron i qua ba iem A, B, C cho trc vi cu phap nh sau:

circle(ten ng tron, [A, B, C], [x, y]);

VD: Lap phng trnh ng tron i qua ba iem A(5; 0), B(0; 1), C(3; 3)[> point(A,5,0),point(B,0,1),point(C,3,3);

, ,A B C

> circle(ABC,[A,B,C],[x,y]);ABC

[> Equation(ABC);

=+ x2 y2 5x y 0

[> coordinates(center(ABC));

,

5

2

1

2

[> radius(ABC);1

213 2

Chu y:a) Lenh coordinates(center(ABC));cho ta toa o cua tam ng

tron ABC.b) Lenh radius(ABC); cho ta ban knh cua ng tron ABC.


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16

c) Lenh Equation(ABC);cho ta phng trnh cua ng tron ABC.d) Neu khong dung cac lenh nay, ta co the xem chi tiet ve ng

tron ABC bang lenh detail(ABC);[> detail(ABC);name of the object: ABC

form of the object: circle2d

name of the center: center_ABC

coordinates of the center: [5/2, 1/2]

radius of the circle: 1/2*13^(1/2)*2^(1/2)

equation of the circle: x^2+y^2-5*x-y = 0

b) ng tron co tam A cho trc va ban knh R cho trcvi cu phap nh sau:

circle(ten ng tron, [A, R], [x, y]);

V du 1 : Viet phng trnh ng tron co tam A(4; 8) va ban knh la 5.

[> point(A,4,-8);

A [> Equation(circle(C,[A,5],[x,y]));

=+ + +x2 55 y2 8x 16y 0

V du 2: Viet phng trnh ng tron co tam C(1; 1) va tiep xuc ving thang 5x 12y + 9 = 0.[> point(C,1,-1),line(D,5*x-12*y+9=0,[x,y]);

,C D

[> R:= distance(C,D);

:=R 2

[> circle(T,[C,R],[x,y]);T

[> Equation(T);

= + +x2

2 y2

2x 2y 0 57

, , ,A B C O

[> sphere(S,[A,B,C,O],[x,y,z]);S

[> Equation(S);=+ + x2 y2 z2 2x 4y 4z 0

[> center(S);center_S_1

[> coordinates(center_S_1);[ ], ,1 2 2

[> R:=radius(S);:=R 3

b) [> plane(P,[A,B,C],[x,y,z]);P

[> Equation(P);= + + +32 16x 8y 8z 0

[> NormalVector(P);[ ], ,16 8 8

[> line(D,[point(center_S_1,1,2,2),[16, 8, 8]]);D

[> Equation(D,t);[ ], ,+1 16 t +2 8 t +2 8 t

Bai 4: (TN, 1998, t 2, 2 iem)

Trong khong gian Oxyz cho hai iemA(1; 0; 2),B(0; 4; 4) vamat phang ( ) co phng trnh 3x 2y + 6z + 2 = 0.

a) Viet phng trnh mat cau tiep xuc vi mat phang ( ) va nhanAlam tam. Tm toa o giao iem cua AB va mat phang ( ).

b) Viet phng trnh mat phang cha ng thangAB va vuong gocvi mat phang ( ).


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56

obj

[> coordinates(obj);[ ], ,4 0 1

Bai 2.( TN, 1997, t 2)Trong khong gian Oxyz choA(1; 4; 0),B(0; 2; 1), C(1; 0; 4).a) Viet PTTS cua ng thang AB.b) Viet PTTQ cua mat phang qua Cva vuong goc vi AB. Xac

nh toa o giao iem cuaAB va .a) [> point(A,1,4,0),point(B,0,2,1), point(C,1,0,4),line(L,[A,B]);

, , ,A B C L

[> Equation(L,t);[ ], ,1 t 4 2 t t

[> v:=ParallelVector(L);:=v [ ], ,-1 -2 1

b)[> plane(anpha,[C,v],[x,y,z]);anpha

[> Equation(anpha);= +3 x 2y z 0

[> intersection(H,L,anpha);H

[> coordinates(H);[ ], ,-1 0 2

Bai 3 : (TN, 1998, t 1, 2 iem)Trong khong gian Oxyz cho cac iemA(2; 0; 0),B(0; 4; 0), C(0; 0; 4).

a) Viet phng trnh mat cau i qua bon iem O, A, B, C. Xac nhtoa o tamIva o dai ban knhR cua mat cau o.

b) Viet phng trnh mat phang (ABC). Viet PTTS cua ng thangi quaIva vuong goc vi mat phang (ABC).

a) [> point(A,2,0,0),point(B,0,4,0), point(C,0,0,4),point(O,0,0,0);17

3) Phng tch cua mot iem oi vi mot ng tron.e tnh phng tch cua iem P oi vi ng tron C, ta dung lenh:

powerpc(P, C);

V du : Cho ng tron co phng trnh x2+ y2 2x + 4y 8 = 0va cac iem A(1; 5), B(6; 1) va C( m; m). Hay xet xem iem A, Bnam trong hay nam ngoai ng tron. Tm m e C thuoc ng tron.[> point(A,1,-5), point(B,6,1), circle(DTRON,x^2+y^2-2*x+4*y-8=0,[x,y]),point(C,-m,m);

, , ,A B DTRON C

[> powerpc(A,DTRON);-4

[> powerpc(B,DTRON);21

[> powerpc(C,DTRON);

+ ( ) m 12

( )+m 22

13 [> simplify((-m-1)^2+(m+2)^2-13);

+ 2 m2 6 m 8

[> solve(2*m^2+6*m-8 =0,{m});,{ }=m 1 { }=m -4

Lu y:a) T ap so ta thay: A trong ng tron va B ngoai ng tron.b) Co the dung lenh IsOnCircle e tm m e C thuoc ng tron.

4) Lenh intersection : Tm giao iem cua hai ng thang, mot ngthang va mot ng tron, hoac hai ng tron.

(find the intersections between two lines, a line and a circle, or twocircles).

Cu phap :


18/36

18

intersection(obj,f,g);hay intersection(ten,f,g,[M, N]);

obj - (mot ten ) a name

f, g - (ng thang hoac ng tron ) the lines or circles404. Trong cac trng hp sau xac nh xem ng thang cat, tiep xuchoac khong co iem chung vi ng tron :

1) y = 2x 3, x2 + y2 3x + 2y 3 = 0;

2) y =2

1 x 2

1 , x2 + y2 8x + 2y + 12 = 0;

3) y = x + 10, x2+ y2 1 = 0.1) [> circle(C,x^2+y^2-3*x+2*y-3=0,[x,y]),line(D,y=2*x-3,[x,y]);

,C D

[> intersection(H,D,C,[M,N]);[ ],M N

[> coordinates(M);

,

11

5

7

5

[> coordinates(N);

May bao ng thang cat ng tron tai 2 iem M );(; 305

7

5

11

Nva

2) [> circle(C,x^2+y^2-8*x+2*y+12=0,[x,y]),line(D,y=1/2*x-1/2,[x,y]);,C D

[> AreTangent(C,D);true

[> intersection(TX,C,D,[M,N]);TX

[> coordinates(TX);

[ ],0 -3

55

a) [> point(A,3,-2,-2),point(B,3,2,0),point(C,0,2,1),point(D,-1,1,2);, , ,A B C D

[> plane(BCD,[B,C,D],[x,y,z]);BCD

[> Equation(BCD);= + + +7 x 2y 3z 0

[> IsOnObject(A,BCD);false

b) [> sphere(S,[A,BCD],[x,y,z]);

S [> Equation(S);

=+ + + + +x2 y2 z2 3 6x 4y 4z 0

[> detail(S);name of the object: S

form of the object: sphere3d

name of the center: Acoordinates of the center: [3, -2, -2]

radius of the sphere: 14^(1/2)

surface area of the sphere: 56*Pi

volume of the sphere: 56/3*Pi*14^(1/2)

equation of the sphere: x^2+y^2+z^2+3-6*x+4*y+4*z = 0

[> n:=NormalVector(BCD); :=n [ ], ,1 2 3

[> line(L,[point(A,3,-2,-2),[1, 2, 3]]);L

[> Equation(L,t);[ ], ,+3 t +2 2 t +2 3 t

[> intersection(obj,L,BCD);


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54

gd1

[> coordinates(gd1);[ ], ,3 7 -6

[> plane(P,2*x + 4*y - 2 = 0,[x,y,z]);P

[> coordinates(intersection(gd2,l1,P));;[ ], ,-1 1 2

[> sphere(S,x^2+y^2+z^2-3*z=25,[x,y,z]);S

[> intersection(gd3,l1,S);Error, (in intersection) wrong number/type of arguments

[> detail(gd3);name of the object: l1_intersect1_S[

form of the object: point3d

coordinates of the point: [-25/29, 35/29, 50/29] name of the object: l1_intersec \,t2_S

form of the object: point3d

coordinates of the point: [-1, 1, 2] ]

Qua detail(gd3) ta co hai giao iem cua l1 va S la :( 25/ 29; 35/29 ; 50/ 29)va ( 1; 1; 2).

MOT SO E THIBai 1. (TN, 1997, t 1)

Trong khong gian Oxyz cho bon iemA(3; 2; 2),B(3; 2; 0), C(0;2; 1) vaD( 1; 1; 2).

a) Viet phng trnh mat phang (BCD). Suy ra ABCD la mot tdien.

b) Viet phng trnh mat cau tam A tiep xuc vi mat phang (BCD).Tm toa o tiep iem.

19

[ ],3 1

May bao ng thang tiep xuc vi ng tron tai iem TX(3; 1)

3) [> circle(C,x^2+y^2-1=0,[x,y]),line(D,y=x+10,[x,y]);,C D

[> AreTangent(C,D);false

[> intersection(H,C,D,[M,N]);intersection: "there is no point of intersection"

May bao ng thang va ng tron khong co iem chung

434.T iem M(4; 4), ve cac tiep tuyen ti ng tron

x2 + y2 6x + 2y + 5 = 0.

Tnh o dai day cung noi cac tiep iem.(S. 10 )

[> circle(C,x^2+y^2-6*x+2*y+5=0,[x,y]);C

[> point(M,4,-4); M

[> TangentLine(TT,M,C,[L1,L2]);[ ],L1 L2

[> Equation(L1);= 4 x 2y 0

[> Equation(L2);= +12 2x y 0

[> intersection(TX1,L1,C,[M,N]);TX1

[> A:=coordinates(TX1);:=A [ ],2 -3


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20

[> intersection(TX2,L2,C,[M,N]);TX2

[> B:=coordinates(TX2);:=B [ ],5 -2

[> with(student);Warning, the names distance, midpoint and slope have been redefined

[> distance(A,B);10

* H, C toan quoc, khoi D, 2003

Trong mat phang vi he toa o Oxy cho ng tron

(C) : (x 1)2 + (y 1)2 = 4 va ng thang d: x y 1 = 0.

Viet phng trnh cua ng tron (C) oi xng vi ng tron (C) quang thang d. Tm toa o giao iem cua (C) va (C).

[> circle(C,(x-1)^2+(y-2)^2=4,[x,y]);C

[> line(d,x-y-1=0,[x,y]);d

[> reflection(C1,C,d);C1

[> Equation(C1);=+ + y2 5 x2 6x 0

[> intersection(giaodiem,C,C1,[M,N]);[ ],M N

[> coordinates(M);[ ],3 2

53

[> IsTangent(Q,S,'cond');IsTangent: "unable to determine if 2*77^(1/2)-1/77*abs(-51+m)*77^(1/2) is zero"

FAIL

[> solve(2*77^(1/2)-1/77*abs(-51+m)*77^(1/2)=0,{m});,{ }=m~ 205 { }=m~ -103

lenhintersection

Cu phap Chc nangintersection(obj, l1, l2) Tm toa o giao iem cua hai

ng thang l1 va l2.

intersection(obj, l1, p1) Tm toa o giao iem cua ngthang l1 va mat phang P1.intersection(obj, l1, S) Tm toa o giao iem cua ng

thang l1 va mat cau S.intersection(obj, p1, p2) Tm giao tuyen cua hai mat phang

p1 va p2.intersection(obj, p1, p2, p3 ) Tm toa o giao iem cua ba mat

phang p1, p2, p3.V du : Cho hai ng thang :

l1 :5,

4 1,4.

x t

y t

z t

= +

= =

l2 :5,

4 1,4.

x t

y t

z t

= +

= =

Cho mat phang P : 2x + 4y 2 = 0 va mat cau S : x2 + y2 + z2 = 25.Tm toa o giao iem cua:

a) l1 va l2;b) l1 va P;c) l1 va S.

[> line(l1,[2*t-3,3*t-2,-4*t+6],t),line(l2,[t+5,-4*t-1,t-4],t);,l1 l2

[> intersection(gd1,l1,l2);


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[> plane(P,4*x+3*z-17=0,[x,y,z]),sphere(S,(x-3)^2+(y+2)^2+(z-1)^2=25,[x,y,z]);

,P S

[> assume( m -17);[> plane(P1,4*x+3*z+m=0,[x,y,z]);

P1

[> IsTangent(P1,S,'cond');IsTangent: "unable to determine if 5-1/5*abs(15+m) is zero"

FAIL

[> solve(5-1/5*abs(15+m)=0,{m});,{ }=m~ 10 { }=m~ -40

1117. Viet phng trnh cua cac mat phang tiep xuc vi mat caux2 + y2 + z2 10x + 2y + 26z 113 = 0

va song song vi cac ng thang

0

8

2

1

3

7

2

13

3

1

2

5 =

+=

++=

=

+ zyxzyx , .

S. 4x + 6y + 5z 103 = 0, 4x + 6y + 5z + 205 = 0.[> line(L1,[point(M1,-5,1,-13),[2,-3,2]]),line(L2,[point(M2,-7,-1,8),[3,-2,0]]);

,L1 L2

[> sphere(S,x^2+y^2+z^2-10*x+2*y+26*z-113,[x,y,z]);S

[> plane(P,[L1, L2]); P

[> Equation(P);=+ + +79 4x 6y 5z 0

[> plane(Q,4*x+6*y+5*z+m=0,[x,y,z]);Q

21

[> coordinates(N);[ ],1 0

ay, (C1) la (C) cua e bai.

412. Viet phng trnh cua ng tron i qua iem A(1; 1) va qua cacgiao iem cua hai ng tron

x2 + y2 + 2x 2y 23 = 0, x2 + y2 6x + 12y 35 = 0.Write the equation of the circle which passes through the point A(1;

1) and through the points of intersection of the two circlesx2 + y2 + 2x 2y 23 = 0, x2 + y2 6x + 12y 35 = 0.

S. x2 + y2 + 6x 9y 17 = 0.[> circle(C1,x^2+y^2+2*x-2*y-23=0,[x,y]),circle(C2,x^2+y^2-6*x+12*y-35=0,[x,y]),point(A,1,-1);

, ,C1 C2 A

[> intersection(GIAODIEM,C1,C2,[M,N]);[ ],M N

[> coordinates(M);

,

47

13

38

13

[> coordinates(N);[ ],-5 -2

[> circle(C,[M,N,A],[x,y]);C

[> Equation(C);= + + x2 17 y2 6x 9y 0

4) TIEP TUYEN CUA NG TRON*Loai 1 : Tiep tuyen tai mot iem thuoc ng tron.

e viet phng trnh tiep tuyen cua ng tron C tai iem P, tadung lenh

tangentpc(l, P, c );


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22

l -the name of the tangent line (ten tiep tuyen ).

418. Viet phng trnh tiep tuyen vi ng tron (x + 2) 2 + (y 3)2 =25tai iem A(5; 7)

Write the equation of the line tangent to the circle(x + 2)2

+ (y 3)2

=25at the point A(5;7). (S. 3x 4y + 43= 0).[> point(A,-5,7),circle(C,(x+2)^2+(y-3)^2=25,[x,y]);

,A C

[> tangentpc(l,A,C);l

[> Equation(l); = +43 3x 4y 0

* Loai 2 : Tiep tuyen vi ng tron i qua mot iem cho trc.e viet phng trnh tiep tuyen cua ng tron C i qua iem P

cho trcTangentLine( TEN , P, C, [x, y] );

427. Viet phng trnh cac tiep tuyen vi ng tron x2 + y2 = 5 ke t

iem

3

5

3

5 ;A .

From the point

3

5

3

5 ;A , tangent lines are drawn to the circle

x2 + y2 = 5. Find their equations. (S. x 2y 5 = 0 va 2x y 5 = 0).

[> point(A,5/3,-5/3),circle(C,x^2+y^2=5,[x,y]);,A C

[> TangentLine(Tieptuyen,A,C,[L1,L2]);[ ],L1 L2

[> TangentLine(Tieptuyen,A,C,[L1,L2]);[ ],L1 L2

51

V du1: Viet phng trnh tiep dien cua mat cau x2 + y2 + z2= 49 tai iemM(6; 3; 2).[>sphere(S,x^2+y^2+z^2=49,[x,y,z]),point(M,6,-3,-2);

,S M

[> TangentPlane(P,M,S);P

[> Equation(P,[x,y,z]);= + +49 6x 3y 2z 0

V du 2: Tm phng trnh cua mat phang tiep xuc vi mat cau

(x 3)2

+ (y 1)2

+ ( z + 2)2

= 24 tai iemM1( 1; 3; 0).Find the equation of the tangent plane to the sphere

(x 3)2 + (y 1)2 + ( z + 2)2 = 24 at the point M1( 1; 3; 0).(S. 2x y z + 5 = 0 ).

[> sphere(S,(x-3)^2+(y-1)^2+(z+2)^2=24,[x,y,z]), point(M1,-1,3,0);,S M1

[> TangentPlane(P,M1,S);P

[> Equation(P,[x,y,z]);=+ 10 4x 2y 2z 0

* Loai 2 : Tiep dien song song hoac vuong goc vi mot mat phang chotrc

1116. Viet phng trnh cua cac mat phang tiep xuc vi mat cau

(x 3)2

+ (y+ 2)2

+ ( z 1)2

= 25va song song vi mat phang 4x + 3z 17 = 0.

Write the equations of the tangent planes to the sphere

(x 3)2 + (y+ 2)2 + ( z 1)2 = 25which are parallel to the plane 4x + 3z 17 = 0.

S. 4x + 3z 40 = 0, 4x + 3z + 10 = 0.


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50

III. Phng tch cua mot iem oi vi mat cau.

e tnh phng tch cua iem P oi vi mat cau S (The power ofpoint P with respect to sphere S ), ta dung lenh :

Powerps(P, S);

V du : Cho mat cau S : x2 + y2 + z2 2x 4y + 6z 25 = 0 va cac iem :A(1; 2; 0), B(n; n 3; 4), C( m; 2; 4)

a) Tm phng tch cua iem A oi vi mat cau S;b) Tm n e iem B trong mat cau S;c) Chng minh rang iem C luon ngoai mat cau mR.

a [> powerps(A,S);-30

b) [> point(B,n, n-3,-4);B

[> powerps(B,S); +12 12 n 2 n2

[> solve(-12-12*n+2*n^2 Equation(L1);

= + 5

3

1

3x

2

3y 0

[> Equation(L2);

= +5

3

2

3x

1

3y 0

*Loai 3. tiep tuyen cua ng tron song song hoac vuong goc vi motng thang cho trc

436. Viet phng trnh cac tiep tuyen vi ng tron

x2

+ y2

+ 10x 2y + 6 = 0,biet tiep tuyen song song vi ng thang 2x + y 7 = 0.Find the equations of the lines tangent to the circle

x2 + y2 + 10x 2y + 6 = 0 and parallel to the line 2x + y 7= 0.(S. 2x + y 1 = 0, 2x + y + 19 = 0).

[> line(D,2*x+y+m=0,[x,y]),circle(C,x^2+y^2+10*x-2*y+6=0,[x,y]);,D C

[> AreTangent(D,C,'cond');AreTangent: "hint: unable to determine if -19-18*m+m^2 is zero"

FAIL

[> solve(m^2-18*m-19=0,{m});,{ }=m 19 { }=m -1

a) dong lenh 1, ta goi ng thang song song vi ng thang 2x+ y 7 = 0 co dang D : 2x + y + m = 0 ( m 7).

b) dong lenh 2, ta dung lenh AreTangent(D,C,'cond'); e tmieu kien e D tiep xuc vi ng tron C. T o, ta tm c m.

* ng tron noi tiep tam giac

e viet phng trnh ng tron noi tiep tam giac ABC, ta thc hien:

v Bc 1: Khai bao tam giac ABC. Gia s tam giac co ten la T.


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24

v Bc 2 : Dung lenh incircle(ten ng tron , T); e khai baong tron.

v Bc 3 : Dung lenh Equation(ten ng tron) e viet phngtrnh ng tron .

* Co the xem chi tiet ng tron bang lenh detailV du: Viet phng trnh ng tron noi tiep tam giac ABO, biet A(0; 1),B(1; 0) va O(0; 0)[> point(A,0,1),point(B,1,0),point(O,0,0),triangle(T,[A,B,O]);

, , ,A B O T

[> incircle(noitiep,T);

noitiep [> Equation(noitiep);enter name of the horizontal axis > x;enter name of the vertical axis > y;

=+ + x2 y22x

+2 2

2y

+2 2

2

( )+2 2 21

2

12

+2 2

2

0

e cho ap so gon lai, ban s dung lenh factor(%);nh sau:[> factor(%);

=+ + + + x2 y2 2x x 2 2y y 23

22 0

Cac phep bien oi trong hnh hoc phang phan trc, chung ta a xet phep oi xng cua mot iem qua

mot ng thang, trong phan nay ta xet tat ca cac phep bien oi tronghnh hoc phang.1) Phep tnh tien

geometry[translation] - find the translation of a geometric object withrespect to a directed segmentCalling Sequence

translation(Q, obj, AB)

Parameters49

a)[>point(A,3,-2,-2),point(B,3,2,0),point(C,0,2,1),point(D,-1,1,2);, , ,A B C D

[> Equation(sphere(S,[A,B,C,D],[x,y,z]));

=+ + + x2 y2 z2 724

7x

15

7y

16

7z 0

b) [> plane(BCD,[B,C,D],[x,y,z]);BCD

[> Equation(sphere(S,[A,BCD],[x,y,z]));=+ + + + +x2 y2 z2 3 6x 4y 4z 0

c) [> Equation(sphere(S,[A,B],[x,y,z]));

=+ + + +x2 y2 z2 5 6x 2z 0

d) [> line(BC,[B,C]);BC

[> R:=distance(D,BC);

:=R1

1014 10

[> Equation(sphere(S,[D,R],[x,y,z]));

=+ + + + x2 y2 z223

52x 2y 4z 0

Chu y: Trong cau d)v dong lenh th nhat, ta khai bao BC la ng thang qua hai

iem B va C.v dong lenh th hai, ta gan R la khoang cach t iem D en

ng thang BC.


25/36


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26

k - number which is the ratio of the dilatationO - point which is the center of the dilatation

[>point(A,2,3),point(B,-1,-1),line(d,2*x+3*y-1=0,[x,y]),circle(C,x^2+y^2-2*x+4*y-4=0,[x,y]);

, , ,A B d C [> coordinates(reflection(P,A,d));

,

-22

13

-33

13

[> rotation(Q,P,Pi/2,counterclockwise,B);Q

[> coordinates(Q);

,

7

13

-22

13

[> dilatation(E,A,1/3,B);E

[> coordinates(E);

,0

1

3

[> translation(F,E,dsegment(BA,B,A));F

[> coordinates(F);

,3

133

[> rotation(C1,C,Pi/3,clockwise,A);C1

[> Equation(C1);=+ + + + x2 y2 ( )5 3 3 x ( ) 1 3 y 13 7 3 0

47

[> solve(2-4*m = 0,{m});

{ }=m1

2

[> assume(m1/2):gtetrahedron(ABCD,[A,B,C,D]);ABCD

[> v:=volume(ABCD);

:=v +1

3

2

3m~

[> solve(abs(-1/3+2/3*m)=5,{m});,{ }=m~ 8 { }=m~ -7

* Lu y:

a) Neu ta goi D(0, m, 0) la toa o cua iem D, th trc het, ta phai tm

ieu kien e cho bon iem A, B, C, D la bon nh cua mot t dien.

b) Lenh AreCoplanar(A,B,C,D);khong kiem tra c tnh ong phang

cua bon iem A, B, C, D.

MAT CAU

I) Cach khai bao mat cau trong Maple1) Neu phng trnh mat cau S, tam I(a; b; c) co dang :

x2 + y2 + z2 2ax 2by 2cz + d = 0

th ta khai bao:[>sphere(S, x^2 + y^2 + z^2 2*a*x 2*b*y 2*c*z + d = 0, [x, y, z],centername= I);2) Neu phng trnh mat cau S, tam I co dang :

(x a)2+ ( y b)2 + ( z c)2 = R2th ta khai bao:[>sphere(S, (x a)^2 + ( y b)^ 2 + (z - c)^2 = R^2, [x, y, z],

centername= I);


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46

a) [> point(A,-2,4,5),point(B,0,1,-1),point(C,1,3,-6),point(D,0,-1,4),triangle(ABC,[A,B,C]);

, , , ,A B C D ABC

[> area(ABC);1

2794

b) [> gtetrahedron(ABCD,[A,B,C,D]);ABCD

[> volume(ABCD);9

2

VD 2. Mot t dien co the tch la v = 5, co ba nh la cac iem A(2; 1; 1), B(3; 0 ; 1), C(2; 1; 3); nh th t D nam tren truc Oy. Tm toa onh D.

A tetrahedron of volume v = 5 has three of its vertices A(2; 1; 1),B(3; 0 ; 1), C(2; 1; 3); the fourth vertex D lies on the axis Oy. Find thecoordinates of D. S.D1( 0; 8; 0),D2( 0; 7; 0).

[> point(A,2,1,-1), point(B,3,0,1), point(C,2,-1,3), point(D,0,m,0);, , ,A B C D

[> AreCoplanar(A,B,C,D);FAIL

[> plane(P,[A,B,C],[x,y,z]); P

[> IsOnObject(D,P,'cond');geom3d/onobjps: "hint: unable to determine if 2-4*m is zero"

FAIL

[> cond;=2 4 m 0

27

[> Equation(dilatation(C2,C,1/3,B));=+ + + +9x2 6x 8 9y2 24y 0

Tm tam v t trong, tam v t ngoai cua hai ng tron khong ong tamgeometry[similitude] - find the insimilitude and outsimilitude of twocirclesCalling Sequence

similitude(obj, c1, c2)Parameters

obj - name which is assigned the internal and external centers ofsimilitude

c1,c2 - two circles

V du:[> circle(c1,(x-4)^2+(y-3)^2=4,[x,y]);

c1

[> circle(c2,(x-1)^2+(y-3)^2=1,[x,y]);c2

[> similitude(obj,c1,c2);[ ],in_similitude_of_c1_c2 ex_similitude_of_c1_c2

[> map(coordinates,obj);[ ],[ ],2 3 [ ],-2 3

Ban chu y tam v t trong la (2; 3) va tam v t ngoai la ( 2; 3).

PHAN IIHNH GIAI TCH TRONG KHONG GIAN

Truc khi lam viec vi hnh giai tch trong khong gian ta phai bat

au bang lenh with(geom3d);I. VAI CACH NHAP THONG DUNG


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28

1) Nhap mot iem.e nhap iem M(x; y; z), ta nhap nh sau: point(M, x, y, z);

2) Nhap mat phange nhap phng trnh mat phang P : Ax + By + Cz + D = 0, ta nhap :

Plane(P, A*x + B*y + C*z + D = 0, [x, y, z]);3) Nhap mot ng thang .a) Neu phng trnh ng thang dco dang tham so

+=

+=

+=

.,,

30

20

10

tazz

tayy

taxx

Khi nhap vao maple, ta lam nh sau:

line(d, [x0 + t*a1,y0 + t*a2 , z0 + t*a3], t );b) Neu phng trnh ng thang dco dang chnh tac

3

0

2

0

1

0

a

zz

a

yy

a

xx =

=

Gia s d i qua iem M(x0; y0; z0) va co vect ch phng la

);;( 321 aaaa =

, khi nhap vao maple, ta nhap nh sau:

line(d,[point(M,x0; y0; z0),[a1,a2,a3]],t);c) Neu phng trnh ng thang dco dang tong quat :

=+++

=+++

.,0

0

2222

1111

dzcybxa

dzcybxa

dla giao tuyen cua hai mat phang:P1 : a1x + b1y + c1z + d1 = 0 va P2 : a2x + b2y + c2z + d2 = 0

khi nhap vao maple, ta nhap nh sau:

[ > plane(P1,[a1*x + b1*y + c1*z + d1 = 0, [x, y, z]), plane(P2,[a2*x +b2*y + c2*z + d2 = 0, [x, y, z]), line(d,[P1, p2];4) Khai bao mot vect

e nhap vect

u = (x; y; z), ta nhap :u:=([x, y, z]);5) Tch vo hng va tch co hng cua hai vect.

e tnh tch vo hng va tch co hng cua hai vect

u va

v .

Trc het, ta phai m goi[ > with(linalg); Sau o, ta dung lenh :45

V du 3 : Tnh goc tao bi ng thang D2 1,3 2,

5.

x t

y t

z t

=

= + = +

va mat phang P : 4x + y 7z 1 = 0.

[> line(D,[2*t - 1,3*t+2,-t+5],t),plane(P,4*x+y-7*z-1=0,[x,y,z]);,D P

[> FindAngle(D,P);

arcsin3

77231

V du 4 . Cho tam giac ABC vi A(1; 2; 4), B( 3; 4; 0), C( 7; 6; 3).Tnh so o goc trong cua goc A.[>point(A,1,2,-4),point(B,-3,-4,0),point(C,-7,6,3),triangle(ABC,[A,B,C]);

, , ,A B C ABC

[> FindAngle(A,ABC);

arccos

6

731

17 129

DIEN TCH CUA TAM GIAC THE TCH T DIEN

Cu phap Chc nang Chu yarea(ABC) Tnh dien tch

cua tam giacABC.

Trc het phai khai bao tam giacABC bang lenh:

triangle (ABC, [A, B, C])

volume(ABCD) Tnh the tch tdien ABCD.

Trc het phai khai bao t dienABCD bang lenh :gtetrahedron(ABCD, [A, B, C, D])

V du 1: Cho cac iem :A( 2; 4; 5), B(0; 1; 1), C(1; 3; 6), va D(0; 1; 4).

a) Tnh dien tch cua tam giac ABC;b) Tnh the tch t dien ABCD.


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44

V du 1 : Tm goc tao bi hai ng thang:( Find the acute angle between the lines: )

.,2

5

1

3

1

2

21

2

1

3 +=

=

+=

+=

zyxzyx (S. 600)

[> with(geom3d);[> line(D1,[point(M,3,-2,0),[1,-1,sqrt(2)]]);

D1

[> line(D2,[point(N,-2,3,-5),[1,1,sqrt(2)]]);D2

[> FindAngle(D1,D2);13

V du 2 : Xac nh cosin cua goc gia cac ng thang:( Determine the cosine of the angle between the lines)

=++

=+

=+

=

.,

;,

01922

0266

0422

054

zyx

zyx

zyx

zyx

(S. cos21

4= )

[> plane(p1,x-y-4*z-5=0,[x,y,z]),plane(p2,2*x+y-2*z-4=0,[x,y,z]),line(L1,[p1,p2]);

, ,p1 p2 L1

[> plane(p3,x-6*y-6*z+2=0,[x,y,z]),plane(p4,2*x+2*y+9*z-

1=0,[x,y,z]),line(L2,[p3,p4]); , ,p3 p4 L2

[> FindAngle(L1,L2);

arccos

4

21

29

crossprod(u,v); e tnh tch co hng va lenh dotprod(u,v); e tnh tchvo hng.

V du : Cho cac vect

u = (1; 2; 3) va

v = (3; 5; 7).

Tm

u .

v va [

u ,

v ][> u:=([1,2,3]),v:=([3,5,7]);:=u [ ], ,1 2 3 :=v [ ], ,3 5 7

[> with(linalg);[> crossprod(u,v);

[ ], ,-1 2 -1

[>dotprod(u,v);

34

6) Mot so lenh kiem tra

Ten lenh Cu phap Chc nangAreCollinear AreCollinear(P, Q, R, cond) Kiem tra tnh thang hang

cua ba iem P, Q, R.

AreConcurrent AreConcurrent(l1, l2, l3, cond ) Kiem tra tnh ong quycua ba ng thang l1, l2,l3.

AreCoplanar

*AreCoplanar(A, B, C, D )

*AreCoplanar(l1, l2 )

* Kiem tra tnh ongphang cua bon iem A,B, C, D.* Kiem tra tnh ongphang cua hai ng

thang l1 va l2.

AreParallel

* AreParallel(l1, l2, cond)

* AreParallel(l1, p1, cond)

* Kiem tra tnh songsong cua hai ngthang l1, l2.* Kiem tra tnh songsong cua ng thang l1va mat phang P1.* Kiem tra tnh song

song cua hai mat phang


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30

* AreParallel(p1, p2, cond) p1 va p2.

ArePerpendicular

* ArePerpendicular(l1, l2,

cond)

*ArePerpendicular(l1, p1,cond)

* ArePerpendicular(p1, p2,cond)

* Kiem tra tnh vuong

goc cua hai ng thangl1, l2.* Kiem tra tnh vuonggoc cua ng thang l1va mat phang p1* Kiem tra tnh vuonggoc cua hai mat phangp1 va p2 .

IsEquilateral IsEquilateral(ABC, cond ) Xet xem tam giac ABCco eu hay khong ?

IsOnObject IsOnObject(f, obj, cond)

Kiem tra xem iem hoac

tap hp iem f co thuocobj hay khong ? Trongo, obj co the la ngthang, mat phang haymat cau.

IsRightTriangle IsRightTriangle(ABC, cond ) Kiem tra tnh vuong goccua tam giac ABC.

MAT PHANG

Mot mat phang trong Maple co the c khai bao vi cu phap va chcnang nh sau:

Cu phap Chc nangplane(P, [A, v] ) Khai bao P la mat phang i qua iem A va co

phap vect la v.43

name of the object: anpha

form of the object: plane3d

equation of the plane: -182+14*x-14*y-56*z = 0

V du10 . Tm iem Q oi xng vi iem P(3; 2 ; 5) qua mat phang iqua cac ng thang

=++

=+++

=+

=+

.,

;,

05235

0733

0342

0532

zyx

zyx

zyx

zyx

[> point(P,-3,2,5),plane(P1,x-2*y+3*z-5=0,[x,y,z]),plane(P2,x-2*y-4*z+3=0,[x,y,z]),plane(P3,3*x+y+3*z+7=0,[x,y,z]),plane(P4,5*x-3*y+2*z+5=0,[x,y,z]),line(L1,[P1,P2]),line(L2,[P3,P4]);

, , , , , ,P P1 P2 P3 P4 L1 L2

[> plane(anpha,[L1,L2],[x,y,z]);anpha

[> Equation(anpha);= + +98 98x 196y 49z 0

[> reflection(Q,P,anpha);Q

[> coordinates(Q);[ ], ,1 -6 3

GOC

Cu phap Chc nangFindAngle(l1, l2) Tm goc cua hai ng thang l1 va l2.FindAngle(p1, p2) Tm goc cua hai mat phang p1 va p2.FindAngle(l1, p1) Tm goc cua ng thang l1 va mat phang

p1.FindAngle(A, T) Tm so o goc trong nh A cua tam giac T.


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42

[>point(P,2,-5,7),point(M1,5,4,6),point(M2,-2,-17,-8),line(M1M2,[M1,M2]);

, , ,P M1 M2 M1M2

[> reflection(Q,P,l);Q


V du 8. Tm iem Q oi xng vi iem P(1; 3; 4) qua mat phang

3x + y 2z = 0.

[> point(P,1,3,-4),plane(anpha,3*x+y-2*z=0,[x,y,z]);,P anpha

[> reflection(Q,P,anpha);Q

[> coordinates(Q);[ ], ,-5 1 0

V du 9. Tm iem Q oi xng vi iem P(3; 4; 6) qua mat phang iqua cac iemM1( 6; 1; 5),M2(7; 2; 1),M3(10; 7; 1).

[> point(P,3,-4,-6),point(M1,-6,1,-5),point(M2,7,-2,-1),point(M3,10,-7,1),plane(anpha,[M1,M2,M3],[x,y,z]);

, , , ,P M1 M2 M3 anpha

[> reflection(Q,P,anpha); Q


[> detail(anpha);

31

plane(P, [l1, l2] ) Khai bao P la mat phang i qua hai ng thangl1 va l2.

plane(P, [A, B, C] ) Khai bao P la mat phang i qua ba iem A, B,C.

plane(P, [A, l1, l2] ) Khai bao P la mat phang i qua iem A va songsong vi hai ng thang l1 va l2.Plane(P,a*x + b*y

+c*z + d = 0,[x, y, z]Khai bao P la mat phang co phng trnh a*x +b*y +c*z + d = 0.

Parallel(P, M, alpha) Khai bao P la mat phang i qua iem M va songsong vi mat phang alpha.

Parallel(P, M, l) Khai bao P la mat phang i qua iem M va song

song vi ng thang l.Parallel(P, l1, l2) Khai bao P la mat phang cha ng thang l1va song song vi ng thang l2.

parallel(w, u, v)

Parametersw - name of the object to be createdu - point or a linev - line or plane; v can be a plane only ifu is a point

Description Ifu is a point and v is a line (or plane), the parallel(w, u, v)

function defines w as the line (or plane) that passes throughuand is parallel to v.

Ifu is a line, and v is a line, the parallel(w, u, v) functiondefines w as the plane that contains u and is parallel to v.V du 1. Viet phng trnh cua mat phang i qua iem M0(1; 2; 1) va

vuong goc vi ng thang

=++

=+

.,

02

032

zyx

zyx


32/36

32

Write the equation of the plane passing through the pointM0(1; 2;

1) and perpendicular to the straight line

=++

=+

.,

02

032

zyx

zyx

S. x + 2y + 3z = 0.

[> plane(P1,x-2*y+z-3=0,[x,y,z]),plane(P2,x+y-z+2=0,[x,y,z]),point(M0,1,-2,1);

, ,P1 P2 M0

[> line(D,[P1,P2]);D

[> v:=ParallelVector(D);:=v [ ], ,1 2 3

[> Equation(plane(P,[M0,v],[x,y,z]));=+ +x 2y 3z 0

* Chu y : Lenh [>ParallelVector(D);e xac nh vect ch phng cuang thang D.

V du 2: Viet phng trnh cua mat phang i qua ba iemA(3; 1; 2), B(4; 1; 1) va C(2; 0; 2).[> point(A,3,-1,2),point(B,4,-1,-1),point(C,2,0,2);

, ,A B C

[> plane(ABC,[A,B,C],[x,y,z]);ABC

[> Equation(ABC);= + + +8 3x 3y z 0

V du 3 :Viet phng trnh cua mat phang i qua ng thang

=

+=

+=

2

32

13

tz

ty

tx

,,

va song song vi ng thang

=+

=+

.

,

052

032

zyx

zyx

41

Giai[> point(M2,8,-9),point(A,3,-4),point(B,-1,-2);

, ,M2 A B

[> line(AB,[A,B],[x,y]);AB

[> Equation(AB);= 10 2x 4y 0

[> reflection(M1,M2,AB);M1

[> coordinates(M1); [ ],10 -5

V du 6.Tm iem Q oi xng vi iem P( 4; 1; 6) qua ng thang

=++

=+

.,

0322

0124

zyx

zyx

[> with(geom3d);

[>point(P,4,1,6),plane(P1,x-y-4*z+12=0,[x,y,z]),plane(P2,2*x+y-2*z+3=0,[x,y,z]);, ,P P1 P2

[> line(l,[P1,P2]);l

[> reflection(Q,P,l);Q


V du 7. Tm iem Q oi xng vi iem P( 2; 5; 7) qua ng thang iqua hai iemM1( 5; 4; 6) vaM2( 2; 17; 8).

5


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40

V du 3 . Tm hnh chieu cua iem C(3; 4; 2) tren mat phang i qua

hai ng thang song song4

3

1

3

13

2

4

3

1

6

13

5

+=

=

+=

=

zyxzyx ,

[> point(C,3,-4,-2),line(D1,[point(M,5,6,-3),[13,1,-

4]]),line(D2,[point(N,2,3,-3),[13,1,-4]]),plane(P,[D1,D2],[x,y,z]);, , ,C D1 D2 P

[> Equation(P);=+ +120 12x 12y 36z 0

[> projection(H,C,P);H

[> coordinates(H);[ ], ,2 -3 -5

V du 4 . Tm hnh chieu cua ng thang

=+

=

022

,05245

zx

zyx

len mat phang 2x y + z 1 = 0.Giai :[> plane(p1,5*x-4*y-2*z-5=0,[x,y,z]), plane(p2,x+2*z-2=0,[x,y,z]),line(l,[p1,p2]);

, ,p1 p2 l

[> plane(Q,2*x-y+z-1=0,[x,y,z]);Q

[> projection(R,l,Q);R

[> Equation(R,t);

, ,

17

128 t

37

2412 t +

7

244 t

V du 5 .Tm iemM1 oi xng vi iemM2(8; 9) qua ng thang iqua hai iemA(3; 4) vaB( 1; 2).

33

S. 13x 14y + 11z + 51 = 0.[> line(L1,[3*t+1,2*t+3,-t-2],t),plane(P1,2*x-y+z-3=0,[x,y,z]),plane(P2,x+2*y-z-5=0,[x,y,z]),line(L2,[P1,P2]);

, , ,L1 P1 P2 L2

[> parallel(P,L1,L2);P

[> Equation(P);V du 4. Viet phng trnh cua mat phang i qua iem M1(1; 2; 3) va

song song vi cac ng thang1

3

2

2

3

5

3

7

3

1

2

1

+=

=

+=

+=

zyxzyx ,

S. 9x + 11y + 5z 16 = 0.[> line(D1,[point(A,1,-1,7),[2,-3,3]]),line(D2,[point(A,-5,2,-3),[3,-2,-1]]),point(M1,1,2,-3);

, ,D1 D2 M1

[> plane(P,[M1,D1,D2]);P

[> Equation(P,[x,y,z]);= + + +16 9x 11y 5z 0

V du 5: Chng minh rang bon iemA(1; 2; 1),B(0; 1; 5), C(1; 2 ; 1),D(2; 1; 3) nam tren cung mat phang.

Prove that the four points A(1; 2; 1), B(0; 1; 5), C(1; 2 ; 1), D(2;

1; 3) lies in the same plane.* Cach 1:

[> point(A,1,2,-1), point(B,0,1,5), point(C,-1,2,1), point(D,2,1,3);, , ,A B C D

[> AreCoplanar(A,B,C,D);true

* C h 2 l h P


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34

* Cach 2:[> point(A,1,2,-1), point(B,0,1,5), point(C,-1,2,1), point(D,2,1,3);

, , ,A B C D

[> plane(P,[A,B,C],[x,y,z]);P

[> Equation(P);= 20 2x 10y 2z 0

[> IsOnObject(D,P);true

Lenh IsOnObject(D, P) ; e kiem tra xem iem P co nam tren mat phang

P hay khong ?

V du 5 : Xac nh cac gia tr cua l va m de hai mat phang co phngtrnh sau la song song nhau:

mx + 3y 2z 1 = 0, 2x 5y lz = 0.[> plane(P1,m*x+3*y-2*z-1=0,[x,y,z]),plane(P2,2*x-5*y-l*z=0,[x,y,z]);

,P1 P2

[> AreParallel(P1,P2,'cond');

FAIL

[> cond;( )&and , ,= 3 l 10 0 = +4 m l 0 = 5 m 6 0

[> solve({-3*l-10 = 0,-4+m*l = 0,-5*m-6 = 0},{m,l});

{ },=m -65

=l -103

NG THANG

Maple cho phep khai bao ng thang theo cac cach sau:

Cu phap Chc nang

line(l, [A, B] ) Khai bao ng thang l i qua hai iem A va B. 39

len mat phang P.

OI

XNG

reflection(Q, P, c )

Q - the name of the object to becreated

P - a geometric object

c - a point, a line, or a plane

V du 1 : Tm hnh chieu Q cua iem P(2; 1; 3) len ng thang

D:

=

+=

=

.,

,

tz

ty

tx

57

3

[> with(geom3d);

[> point(P,2,-1,3),line(D,[3*t,-7+5*t,2+2*t],t);,P D

[> projection(Q,P,D);Q

[> coordinates(Q);

[ ], ,3 -2 4 V du 2 .Tm hnh chieu H cua iem P(5; 2; 1) len mat phang

Q: 2x y + 3z + 23 = 0[> with(geom3d);

[> point(P,5,2,-1),plane(Q,2*x- y+3*z+23=0,[x,y,z]);

,P Q

[> projection(H,P,Q);H

[> coordinates(H);[ ], ,1 4 -7

[> distance(A P); line(l [A ] ) Kh i b th l i i A


35/36

38

[> distance(A,P);9

4794

[> distance(B,l);926

17 26


[> Equation(AB,t);[ ], ,1 2 t +2 2 t 3 10 t

[> distance(AB,l);27

7474

[> distance(P,Q);4

4794

[> distance(AB, Q);Error,(in geom3d/distancelp)the line and plane intersect

Lu y : ng thang AB va mat phang Q cat nhau.

HNH CHIEU VA OI XNGVan e Cu phap Chc nang

projection(Q, A, l ) Tm hnh chieu Q cua iem A lenng thang l.

projection(Q, A, P) Tm hnh chieu Q cua iem A lenmat phang P

HNHCHIEU

projection(Q, l, P) Tm hnh chieu Q cua ng thang l

35

line(l, [A, u] ) Khai bao ng thang l i qua iem A va co

VTCP la

u .line(l, [A, p1] ) Khai bao ng thang l i qua iem A va

vuong goc vi mat phang p1.line(l, [p1, p2] ) Khai bao l la giao tuyen cua hai mat phang p1

va p2.line(l, [a1+b1*t,a2+b2*t, a3+b3*t ],t)

Khai bao ng thang l la ng thang cophng trnh tham so x = a1+b1*t, y =a2+b2*t, z = a3+b3*t

parallel(l, A, d) Khai bao ng thang l i qua iem A song

song vi ng thang d.V du 1 : Viet phng trnh ng thang i qua hai iem A(3; 1; 2) vaB(4; 1; 1).[> point(A,3,-1,2),point(B,4,-1,-1),line(l,[A,B]);

, ,A B l

[> Equation(l,t);

[ ], ,+3 t -1 2 3 t

Chu y: ap so cho phng trnh tham so cua ng thang l la

=

=

+=

.,

,

tz

y

tx

32

1

3

V du 2: Viet phng trnh ng thang i qua M(5; 2; 3) va vuong gocvi mat phang 2x 3y + z 1 = 0

[> point(M,5,-2,3),plane(P,2*x-3*y+z-1=0,[x,y,z]);,M P

[> Equation(line(l,[M,P]));enter name of the independent variable > t;

[ ], ,+5 2 t 2 3 t +3 t

V du 3: Viet phng trnh chnh tac cua ng thang : Cu phap Chc nang


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36

V du 3: Viet phng trnh chnh tac cua ng thang :2 3 4 5 0,5 4 4 5 0x y z

x y z

+ =

+ + =

[> plane(P1,2*x-3*y+4*z-5=0,[x,y,z]),plane(P2,5*x+4*y-

4*z+5=0,[x,y,z]), line(D,[P1,P2]);, ,P1 P2 D

[> Equation(D);enter name of the independent variable > t;

, ,

5

234 t +

35

2328 t 23 t

[> FixedPoint(M,D);M

[> coordinates(M);

, ,

5

23

-35

230

Chu y: Lenh FixedPoint(M,D); cho ta mot iem M co nh thuoc ng

thang a cho.

V du 4 : Viet phng trnh ng cao AH cua tam giac ABC vi A(2; 3; 4), B(3; 2; 7) va C( 2; 5; 5).[> with(geom3d);[> triangle(ABC,[point(A,2,-3,4),point(B,3,2,7), point(C,-2,5,5)]),altitude(AH,A,ABC);

,ABC AH[> Equation(AH,t);

, ,+2

29

19t +3

89

19t +4

61

19t

KHOANG CACHTrong Maple cho phep tnh cac khoang cach sau:

37

Cu phap Chc nangdistance(A, B) Tnh khoang cach gia hai iem A

va B.distance(l1, l2) Tnh khoang cach gia hai ng

thang l1 va l2.distance(p1, p2) Tnh khoang cach gia hai matphang p1 va p2.

distance(A, p1) Tnh khoang cach t iem A enmat phang p1.

distance(A, l1) Tnh khoang cach t iem A enng thang l1.

distance(l1, p1) Tnh khoang cach gia ngthang l1 va mat phang p1.

V du : Cho cac iem A(1; 2; 3), B( 1; 4; -7);cac mat phang P : 2x + 3y 9z + 1 = 0 va Q : 2x + 3y 9z + 9 = 0

va ng thang l :3 1,4 6,

.

x t

y t

z t

=

= +

=

Tnh :1) Khoang cach gia hai iem A va B.2) Khoang cach t iem A en mat phang P.3) Khoang cach t iem B en ng thang l.4) Khoang cach gia hai ng thang AB va l.5) Khoang cach gia hai mat phang P va Q.6)

Khoang cach gia ng thang AB va mat phang Q.[> point(A,1,2,3), point(B,-1,4,-7),plane(P,2*x+3*y-

9*z+1=0,[x,y,z]),line(l,[3*t-1,4*t+6,-t],t),plane(Q,2*x+3*y-9*z+9=0,[x,y,z]);

, , , ,A B P l Q

[> distance(A,B);6 3

maple trong hhgt

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