HỆ THỐNG BÀI TẬP HÌNH HỌC ÔN THI ĐẠI HỌC

8/14/2019 H THNG BI TP HNH HC N THI I HC

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Gio vin: Nguyn nh Dng - Trng THPT Nng Cng IV LTH

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Bi 1) HC 2002 K.ATrong khong gian vi he toa o ecac vuong goc Oxyz cho hai ng thang:

1 :2 0

2 2 4 0

x y z

x y z

+ =

+ + =va 2 :

1

2

1 2

x t

y t

z t

= +

= +

= +

a) Viet phng trnh mat phang (P) cha ng thang 1 va song song vi ng thang 2

b) cho iem M(2 ; 1,4). Tm toa o iem H thuoc ng thang 2 sao cho oan thang MH co o dai nhonhat.Bi 2) HC 2002 K.B

1.Trong mat phang toa o ecac vuong goc Oxy cho hnh ch nhat ABCD co tam1

;02

, phng trnh

ng thang AB la x 2y + 2 = 0 va AB = 2AD. Tm toa o cac nh A,B,C,D biet rang A co hoanh oam.

2.Cho hnh lap phng ABCDA1B1C1D1 co canh bang a.a) Tnh theo a khoang cach gia hai ng thang A1B va B1D.b) Goi M,N,P lan lt la cac trung iem cua cac can h BB1, CD, A1D1. Tnh goc gia hai ng than

MP, C1N.Bi 3) HC 2002 K.D1.Cho hnh t dien ABCD co canh AD vuong goc vi mat phang (ABC) ; AC = AD = 4cm; AB = 3cm; BC= 5cm. Tnh khoang cach t iem A ti mat phang (BCD).2.Trong khong gian vi he toa o ecac vuong goc Oxyz cho mat phang (P) : 2x y + 2 = 0

Va ng thang dm :(2 1) (1 ) 1 0

(2 1) 4 2 0

m x m y m

mx m z m

+ + + =

+ + + + =

( m la tham so ).

Xac nh m e ng thang dm song song vi mat phang (P).Bi 4) HC 2003 K.A

1) Cho hnh lap phng ABCD.ABCD. Tnh so o cua goc phang nh dien [B,AC,D].2) Trong khong gian vi he truc toa o ecac vuong goc Oxyz cho hnh hop ch nhat

ABCD.ABCD co A trunh vi goc cua he toa o, B(a; 0; 0) , D(0; a; 0), A(0; 0; b) (a>0, b>0).Goi M la trung iem canh CC.

a) tnh the tch khoi t dien BDAM theo a va b.

b) Xac nh ty soa

be hai mat phang (ABD) va (MBD) vuong goc vi nhau.

Bi 5) HC 2003 K.B1) Trong mat phang vi he toa o ecac vuong goc Oxyz cho tam giac ABC co AB = AC ,

BAD = 900. Biet M(1; -1) la trung iem canh BC va G2

;03

la trong tam tam giac ABC. Tm toa

o cac nh A, B, C.

2) Cho hnh lang tru ng ABCD.ABCD co ay ABCD la mot hnh thoi canh a, gocBAD = 600. GM la trung iem canh AA va N la trung iem canh CC. Chng minh rang bon iem B, M, D, Ncung thuoc mot mat phang. Hay tnh o dai canh AA theo a e t giac BMDN la hnh vuong.


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3) Trong khong gian vi he toa o ecac vuong goc Oxyz cho hai iem A(2; 0; 0), B(0;0;8) va iem

sao cho ACuuur

=(0; 6; 0). Tnh khoang cach t trung iem I cua BC en ng thang OA.

Bi 6) HC 2003 K.D1) Trong mat phang toa o ecac vuong goc Oxyz cho ng tron(C) : (x 1)2 + (y 2)2 = 4 va ng thang d:x y 1 = 0Viet phng trnh ng tron (C) oi xng vi ng tron (C) qua ng thang d.

Tm toa o cac giao iem cua (C) va (C).2) Trong khong gian vi he toa o ecac vuong goc Oxyz cho ng thang :

dk:3 2 0

1 0

x ky z

kx y z

+ + =

+ + =tm ke ng thang dkvuong goc vi mat phang (P) : x y 2z +5 = 0.

3) Cho hai mat phang (P) va (Q) vuong goc vi nhau, co giao tuyen la ng thang . Tren layhai iem A, B vi AB = a . trong mat phang (P) iem C , trong mat phang (Q) lay iem D sao cho

AC, BD vuong goc vi va AC = BD = AB. Tnh ban knh mat cau ngoai tiep t dien ABCD vatnh khoang cach t A en mat phang (BCD) theo a.

Bi 7)

HC2004 K.A

1) Trong mat phang toa o Oxy cho hai iem A (0; 2) va B( 3 ; 1 ). Tm toa o trc tam va toa o

tam ng tron ngoai tiep cua tam giac OAB.2) Trong khong gian vi he toa o ecac Oxyz cho hnh chop S.ABCD co ay ABCD la hnh thoi, A

cat BD tao goc toa o O. Biet A(2; 0; 0), B (0; 1; 0), S(0; 0; 2 2 ). Goi M la trung iem canh SC.a) Tnh goc va khoang cach gia hai ng thang SA, BM.b) Gia s mat phang (ABM) cat ng thang SD tai iem N. Tnh the tch khoi hnh chop A.ABMN

Bi 8) HC 2004 K.B1) trong mat phang toa o Oxy cho hai iem A(1; 1), B(4; -3). Tm iem C thuoc ng thang x 2y 1= 0 sao cho khoang cach t C en AB bang 6.

2) Cho hnh chop t giac eu S.ABCD co canh ay bang a, goc gia canh ben va mat ay bang (00 < < 900). Tnh tang cua goc gia hai mat phang (SAB) va (ABCD) theo . Tnh the tch khoi chop

S.ABCD theo a va .

3) Trong khong gian vi he toa o Oxyz cho iem A(-4; -2; 4) va ng thang d :

3 2

1

1 4

x t

y t

z t

= +

=

= +

Viet

phng trnh ng thang i qua iem A, cat va vuong goc vi ng thang d.Bi 9) HC 2004 K.D

1) trong mat phang vi he toa o Oxy cho tam giac ABC co cac nh A(-1; 0); B (4; 0); C(0;m) vi m 0. tm toa o trong tam G cua tam giac ABC theo m. xac nh m e tam giac GAB vuong tai G.

2) Trong khong gian vi he toa o Oxyz cho hnh lang tru ng ABC.A1B1C1. Biet A(a; 0; 0), B(-a; 00), C(0; 1; 0), B1(-a; 0; b), a > 0, b > 0.

a) Tnh khoang cach gia hai ng thang B1C va AC1 theo a, b.b) Cho a, b thay oi nhng luon thoa man a + b = 4. Tm a,b e khoang cach gia hai ng thang B

va AC1 ln nhat.3) Trong khong gian vi he toa o Oxyz cho ba iem A(2;0;1), B(1;0;0), C(1;1;1) va mat phang (P) :

+ y + z 2 = 0. Viet phng trnh mat cau i qua ba iem A, B, C va co tam thuoc mat phang (P).


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Bi 10) HC 2005 K.A1) trong mat phang vi he toa o Oxy cho 2 ng thang

d1 : x y = 0 va d2 : 2x + y 1 = 0tm toa o cac nh hnh vuong ABCD biet rang ng A thuoc d1 , C thuoc d2 va cac nh B, D thuotruc hoanh.

2) Trong khong gian vi he truc Oxyz cho ng thang d :1 3 3

1 2 1

x y z + = =

va mat phang (P) : 2x

y 2z + 9 = 0.a) tm toa o iem I sao cho khoang canh t I en mat phang (P) bang 2.b) Tm toa o giao iem A cua ng thang d va mat phang (P). Viet phng trnh tham so cua n

thang nam trong mat phang (P), biet i qua A va vuong goc goc vi d.Bi 11) HC 2005 B

1) Trong mat phang vi he toa o Oxy cho hai iem A(2;0) va B(6;4). Viet phng trnh ng tron(C) tiep xuc vi truc hoanh tai iem A va khoang cach t tam cua (C) en iem B bang 5.

2) Trong khong gian vi he toa o Oxyz cho hnh lang tru ng ABC.A1B1C1 vi A(0;-3;0), B(4;0;0),

C(0;3;0), B1(4;0;4).a) Tm toa o cac nh A1, C1. Viet phng trnh mat cau co tam la A va tiep xuc vi mat phang(BCC1B1).b) Goi M la trung iem cua A1B1. Viet phng trnh mat phang (P) i qua hai iem A, M va song songvi BC. Mat phang (P) cat ng thang A1C1 tai iem N. Tnh o dai MN.

Bi 12) HC 2005 D1) Trong mat phang toa o Oxy cho iem C(2;0) va elp (E) :

2 2

14 4

x y+ = . Tm toa o cac iem A, B

thuoc (E), biet rang hai iem A,B oi xng vi nhau qua truc hoanh va tam giac ABC la tam giaeu.

2) Trong khong gian vi he toa o Oxyz cho hai ng thang

d1 :1 2 1

3 1 2

x y z + += =

va d2 :

2 0

3 12 0

x y z

x y

+ =

+ =

a) chng minh rang d1 , d2 song song vi nhau. Viet phng trnh mat phang (P) cha ca haing thang d1 va d2.

b) Mat phang toa o Oxz cat hai ng thang d1, d2 lan lt tai cac iem A,B. Tnh dien tchtam giac OAB ( O la goc toa o).

Bi 13) HC 2006 ATrong khong gian vi he toa o Oxyz, cho hnh lap phng ABCD.ABCD vi A(0;0;0), B(1;0;0),D(0;1;0) , A(0;0;1). Goi M va N lan lt la trung iem cua AB va CD.

1. Tnh khoang cach gia hai ng thang AC va MN.

2. Viet phng trng mat phang AC va tao vi mat phang Oxy mot goc biet cos=1

6.

Bi 14) HC 2006 ATrong khong gian vi he toa o Oxyz, cho iem A(0;1;2) va hai ng thang :

d1 :1 1

2 1 1

x y z += =

, d2 :

1

1 2

2

x t

y t

z t

= +

=

= +


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Bi 22. (Cc bi ton tm hnh chiu)

1. Cho im ( )2; 3;1M v mt phng (P): 3 2 0 x y z+ + = . Tm hnh chiu H ca M trn (P).

2. Cho im ( )2; 1;1M v ng thng

1 2

: 1

2

x t

d y t

z t

= +

=

=

. Tm hnh chiu H ca M trn d.

3. Trong khng gian vi h ta Oxyz, cho ng thng 2 2 0:2 4 0

x y zdx y

=

+ =

Tm hnh chiu ca dtrn mt phng (P): 2 2 3 0 x y z + = .Bi 23. (Cc bi ton v khong cch)

1. Trn trc Oy tm im cch u hai mt phng ( ) : 1 0P x y z+ + = v ( ) : 5 0Q x y z + = .

2. Gi s (P) l mt phng c phng trnh ( ) : 2 3 7 0P x y z+ + = v ( )2; 4; 6A ; ( )4;0; 2B l h

im cho trc.Bi 24. (Bi ton vng vung gc chung)

Cho hai

ng thng 11 2

: 2 1 1

x y z

d

+= =

; 2

1 2

: 13

x t

d y tz

= +

= +

=

1. Chng minh d1, d2 l hai ng thng cho nhau.2. Vit phng trnh ng vung gc chung ca d1 v d2.

Bi 25. Cho ng thng ( )1 1

:3 2 1

x y zd

+= =

v hai im ( )3;0;2A , ( )1;2;1B . K AA, BB vung g

vi ng thng (d). Tnh di on thng AB.

Bi 26. Cho hai im ( )1; 3; 2A , ( )9;4;9B v mt phng (P): 2 1 0 x y z + + = . Tm im K trn m

phng (P) ao cho AK BK + nh nht.

Bi 27. Lp phng trnh mt phng cha ng thng

2 3

1 5

x t

y t

z t

= +

=

=

v c khong cch n im ( 1; 1;A

bng 1.

Bi 28. Cho hai ng thng: 1

1

:

x t

d y t

z t

=

=

=

v 2

2

: 1

x t

d y t

z t

=

=

=

1. Chng minh d1 v d2 l hai ng thng cho nhau.2. Vit phng trnh cc mt phng (P), (Q) sao cho (P) cha d1, (Q) cha d2 v (P)//(Q).

Bi 29. Vit phng trnh hnh chiu ca( )1

7 3 9:

1 2 1

x y z = =

theo phng( )2

3 1:

7 2 3

x y z = =

ln mt phng (): 3 0 x y z+ + + = .

Bi 30. Lp phng trnh ng thng () i qua ( )4; 5;3M , ct ( )11 3 2

:1 2 1

x y zd

+ + = =

v c

( )22 1 1

:2 3 5

x y zd

+ = =

.


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Bi 31. Vit phng trnh ng thng () i qua ( )3; 2; 4A song song vi mt ph

( ) : 3 2 3 7 0P x y z = , ng thi ct ng thng ( )2 4 1

:3 2 2

x y zd

+ = =

Bi 32. Cho hai ng thng:

1

2

:

4

x t

d y t

z

=

= =

v 2 3 0:4 4 3 12 0x yd

x y z

+ =

+ + =

1. Chng minh d1 v d2 cho nhau.2. Lp phng trnh mt cu (S) nhn on vung gc chung ca d1 v d2 lm ng knh.

Bi 33. Cho ng thng d:1 2

3 1 1

x y z += = v mt phng (P): 2 2 2 0 x y z+ + =

1. Vit phng trnh mt cu (S) c tm nm trn d, tip xc vi (P) v c bn knh bng 1.2. Gi M l giao im ca (d) vi (P), T l tip im ca (S) vi (P). Tnh MT.

Bi 34. Lp phng trnh mt cu c tm ti im ( )2;3; 1I v ct ng thng (d) c phng trn

11 2

25 2

x ty t

z t

= +

=

=

ti hai im AB sao cho AB = 16.

Bi 35. Trong khng gian vi h ta Oxyz cho hai im ( )0;0;4A ; ( )2;0;0B . Vit phng trnh mt c

qua O, A, B v tip xc vi mt phng (P): 2 5 0 x y z+ = .

Bi 36. Trong khng gian vi h ta Oxyz, cho ng thng ( )2 0

:2 6 0

x yd

x y

=

=v mt cu (S

2 2 2 2 2 2 1 0 x y z x y z+ + + + = . Vit phng trnh mt phng (P) cha (d) sao cho giao tuyn ca mphng (P) v mt cu (S) l ng trn c bn knh r = 1.

Bi 37. Trong khng gian vi h ta Oxyz, cho bn im ( )1; 1; 2A , ( )1;3;2B , ( )4;3;2C v ( 4; 1; 2D

Gi A l hnh chiu ca A ln mt phng Oxy. Vit phng trnh mt cu (S) qua A, B, C, D.Vit phng trnh tip din vi mt cu (S) ti im A.Bi 38. Trong khng gian vi h ta Oxyz cho mt cu (S): 2 2 2 2 2 4 3 0 x y z x y z+ + + + = v h

ng thng ( )12 2 0

:2 0

x y

x z

+ =

=, ( )2

1:

1 1 1

x y z = =

Vit phng trnh tip din vi mt cu (S), bit n song song vi (1) v (2).

Bi 39. Lp phng trnh mt cu (S) c tm ( )1;0;3I v ct ng thng:1 1 1

: 2 1 2

x y z +

= =

Ti hai im A, B sao cho tam gic IAB vung.

Bi 40. Lp phng trnh mt cu (S) c tm ( )4;1;1I v ct mt phng ( ) : 2 2 1 0 x y z + + = theo gi

tuyn l mt ng trn c bn knh bng 2 2 .

Bi 41. Lp phng trnh mt cu c tm thuc ng thng d:1 0

2 0

x z

y

+ =

=v ct mt phng (P) theo th

din l ng trn ln c bn knh bng 4, y (P): 0y z = .


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Bi 55:Lp phng trnh tham s, chnh tc ca ng thng (d) i qua im A(1;2;3) v song song vi

ng thng ( ) cho bi : ( )

2 2

: 3 t

3

x t

y t R

z t

= +

=

= +

.

Bi56: Xt v tr tng i ca ng thng (d) v mt phng (P) ,bit:

a) ( ) Rt,

2

3

1

:

+=

=

+=

tz

ty

tx

d (P): x-y+z+3=0 b) ( ) Rt,

1

9

412

:

+=

+=

+=

tz

ty

tx

d (P): y+4z+17=0

Bi 57: (HNN_TH-98): Cho mt phng (P) v ng thng (d) c phng trnh (P): 2x+y+z=0 v

( )3

2

12

1:

+==

zyxd .

a) Tm to giao im A ca (d) v (P) .b) Lp phng trnh ng thng (d1) qua A vung gc vi (d) v nm trong mt phng (P) .

Bi 58: Cho hai ng thng (d1),(d2) c phng trnh cho bi :

( )1

12

11

2:1== zyxd ( ) ( )t

31

2

21

:2 R

tz

ty

tx

d

+=

+=

+=

a) CMR hai ng thng ct nhau.Xc nh to giao im ca n.b) Vit phng trnh tng qut ca mt phng (P) cha (d1),(d2).

Bi 59: (HNN-96): cho hai ng thng (d1),(d2) c phng trnh cho bi :

( )

34

24

37

:1

+=

=

+=

tz

ty

tx

d ( ) ( )R

tz

ty

tx

d

=

+=

+=

1

1

1

1

2 tt,

12

29

1

:

a) Chng t rng hai ng thng (d1),(d2) cho nhau.b) Vit phng trnh ng thng vung gc chung ca (d1),(d2) .

Bi 60: Cho 3 ng thng (d1),(d2), (d3) c phng trnh :

( )1

1

4

2

3

2:1

=

+=

zyxd , ( )

1

9

2

3

1

7:2

=

=

zyxd , ( )

1

2

2

3

3

1:3

=

+=

+ zyxd

a) Lp phng trnh ng thng (d) ct c hai ng thng (d1),(d2) v song song vi ng thng(d3).

b) Gi s ( ) ( ) { }Add = 1 , ( ) ( ) { }Bdd = 2 .Lp phng trnh mt cu ng knh AB.

Bi 61 Cho 2 ng thng (d1),(d2) c phng trnh : ( ) Rtz

ty

tx

d

==

+=

t21

2

:1 , ( ) 1

9

2

3

1

7

:2

=

=

zyx

d

a) CMR (d1) v (d2) cho nhau.b) Vit phng trnh ng vung gc chung ca (d1) v (d2).c) Lp phng trnh mt cu (S) c ng knh l on vung gc chung ca (d1) v (d2).d) Vit phng trnh tng qut ca mt phng cch u (d1) v (d2).

Bi 62: Vit phng trnh mt cu (S) bit :a) Tm I(1;2;-2) v tip xc vi mt phng (P):6x-3y+2z-11=0.b) (CGTVT-2000): Tm I(1;4;-7) v tip xc vi mt phng (P) :6x+6y-7z+42=0.


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c) Bn knh R = 9 v tip xc vi (P): x+2y+2z+3=0 ti im M(1;1;-3).Bi 63:(H Hu-96): Trong khng gian vi h to 0xyz ,cho bn im A(1;0;1), B(2;1;2),C(1;-1;1),D(4;5;-5

a) Vit phng trnh tham s ca ng thng i qua D v vung gc vi mt phng (ABC).b) Vit phng trnh mt cu ngoi tip t din ABCD.

Bi 64: Cho bn im O(0;0;0),A(6;3;0), B(-2;9;1), S(0;5;8)a) (HKT-99): CMR SB vung gc SA.b) (HKT-99): CMR hnh chiu ca cnh SB ln mt phng (0AB) vung gc vi cnh 0A. Gi K l

giao im ca hnh chiu vi 0A. Hy xc nh to d ca K.c) Vit phng trnh mt cu ngoi tip t din ABCD.d) (HKT-99): Gi P,Q ln lt l im gia ca cc cnh S0,AB . Tm to ca im M trn SB sa

cho PQ v KM ct nhau.Bi 65: Trong khng gian vi h to 0xyz ,cho bn im A(4;4;4), B(3;3;1), C(1;5;5), D(1;1;1).

a) (HVKTQS-98): Tm hnh chiu vung gc ca D ln (ABC) v tnh th tch t din ABCD.b) (HVKTQS-98): Vit phng trnh tham s ng thng vung gc chung ca AC v BD.c) Vit phng trnh mt cu ngoi tip t din ABCD.d) Tnh th tch t din ABCD.

Bi 66: Cho bn im A(-1;3;2), B(4;0;-3), C(5;-1;4), D(0;6;1).a) (HVNHTPHCM-99):Vit phng trnh tham s ca ng thng BC .H AH vung gc BC .Tm t

ca im H.b) (HVNHTPHCM-99):Vit phng trnh tng qut ca (BCD) .Tm khong cch t A n mt phng

(BCD).c) Vit phng trnh mt cu ngoi tip t din ABCD.

Bi 67: Trong khng gian 0xyz, cho hnh chp .bit to bn nh S(5;5;6), A(1;3;0), B(-1;1;4), C(1;-1;4),D(3;1;0).

a) Lp phng trnh cc mt ca hnh chp. b) Lp phng trnh mt cu (S) ngoi tip hnhchp .

c) Tnh th tch hnh chp SABCDBi 68: (HVKTMM-97) Cho bn im A(1;2;2), B(-1;2;-1), C(1;6;-1), D(-1;6;2).

a) CMR t din ABCD c cp cnh i din bng nhau . b) Xc nh to trng tm G ca tdin.

c) Vit phng trnh mt cu ngoi tip ,ni tip t din ABCD.Bi 2:Lp phng trnh mt phng i qua im M(2;1;-1) v qua hai giao tuyn ca hai mt phng (P1) v (Pc phng trnh : (P1): x - y + z - 4 = 0 v (P2) 3x y + z 1 = 0

Bi 3: Lp phng trnh mt phng cha ng thng ( )

=

=+

02

0323:

zx

zyxd v song song vi mt phng

(Q) c phng trnh: 11x - 2y - 15z 6 = 0.Bi 4: Lp phng trnh mt phng qua giao tuyn ca (P1): y + 2z 4 = 0 v (P2) : x + y z 3 = 0 v songsong vi mt phng (Q): - 2 0 x y z+ + = .

Bi 5: Lp phng trnh mt phng cha ng thng ( )

=

=+

02

0323:

zx

zyxd v vung gc vi (Q) c

phng trnh:a) (HNNI-95): (Q): - 2 5 0 x y z+ + = . b) ( ) : 3 1 0Q x y z+ + =

Bi 6: Lp phng trnh ca mt phng qua hai giao tuyn ca hai mt phng (P1): 3 - - 2 0 x y z+ = v (P2):

4 -5 0x y+ = v vung gc vi mt phng : 2 - 7 0x z + = .

Bi 7: Lp phng trnh cha mt phng ng thng : ( )

=

=+

02

0323:

zx

zyxd v song song vi ng thn

(d) c phng trnh :


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a) ( )

=++

=+

0323

0723:

zyx

zyxd b) ( )

5

5

4

3

2

2:

+=

=

zyxd

Bi 8:Lp phng trnh cha mt phng ng thng : ( )

=+

=

0323

02:

zyx

yxd v vung gc ng thng (d

c phng trnh :

a) ( )

=++

=+

0323

0723: zyx

zyx

d b) ( ) 55

4

3

2

2:

+

=

=

zyxd

Bi 9: Lp phng trnh cha mt phng ng thng v vi mt phng (Q) mt gc 60 bit:

( )

=

=+

02

0323:

zx

zyxd v (Q):3x+4y-6=0

Bi 10: Lp phng trnh mt phng (P) cha ng thng ( )

=+

=

015

023:

zy

zxd v c khong cch t im

A(1;-1; 0) ti (P) bng 1.

Bi 11: Cho ng thng (d) v hai mt phng ( )

=+

=

01

02:

zy

zxd v (P1): 5x+5y-3z-2=0 v (P2):2x-y+z-6=0

Lp phng trnh mt phng (P) cha ng thng (d) sao cho: ( ) ( )1PP v ( ) ( )2PP l hai ng vunggc.Bi 12: (HKT-93): cho hai ng thng (d1) v (d2) c phng trnh :

( ) ,014

0238:1

=+

=+

zy

zxd ( )

=++

=

022

032:2

zy

zxd .

a) Vit phng trnh cc mt phng ( )1P , ( )2P song song vi nhau v ln lt cha ( )1d ( )2d

b) Tnh khong cch gia ( )1d , ( )2d

c) Lp phng trnh ng thng (D) song song vi trc Oz v ct c 2 ng thng ( )1d , ( )2d Bi ton 4. Khong cch t mt im ti mt phngBi 1:Tnh khong cch t im M(2;2;1) n mt phng (P) trong cc trng hp sau:

a) ( ) : 2 -3 3 0P x y z+ + = b) ( ) : 2 3 1P x y z + =

Bi 2:Trong khng gian vi h to Oxyz , cho t din c 4 nh A(5;1;3) B(1;6;2) C(5;0;4) D(4;0;6)a) Lp phng trnh tng qut mt phng (ABC)b) Tnh chiu di ng thng cao h t nh D ca t din, t suy ra th tch ca t din

Bi 3:Trong khng gian vi h to Oxyz , cho t din c 4 nh A(1;1;1) B(-2;0;2) C(0;1;-3) D(4;-1;0)a) (H Lut 1996) Tnh chiu di ng thng cao h t nh D ca t dinb) Vit phng trnh mt phng phn gic ca 2 mt (ABC) v (BCD) ct on AD

Bi 3: (HNN_TH-98): Cho mt phng (P) v ng thng (d) c phng trnh (P): 2x+y+z=0 v

( )3

2

12

1:

+==

zyxd .

a) Tm to giao im A ca (d) v (P) .b) Lp phng trnh ng thng (d1) qua A vung gc vi (d) v nm trong mt phng (P) .

Bi 4: (H Khi A-2002): Trong khng gian 0xyz ,cho mt phng (P) v ng thng (d m) c phng trnh

( ) : 2 - 2 0P x y + = , ( ) 024)12(

01)1()12(:

=++++

=+++

mzmmx

mymxmdm xc nh m (dm)//(P)

Bi 3: Cho hai ng thng (d1),(d2) c phng trnh cho bi:

( )4

9

1

5

3

7:1

=

=

+ zyxd , ( )

4

18

1

4

3:2

+=

+=

zyxd

a) Chng t rng hai ng thng (d1),(d2) song song vi nhau .


11/24


- 11 -

b) Vit phng trnh ng thng (d) song song ,cch u (d1),(d2) v thuc mt phng cha (d1),(d2).Bi 4: Trong khng gian 0xyz ,cho hai ng thng (d1),(d2) c phng trnh cho bi :

( ) Rt

46

2

23

:1

+=

+=

+=

tz

ty

tx

d , ( ) 015

0194:2

=+

=+

zx

yxd

a) Chng t rng hai ng thng (d1),(d2) ct nhau .

b) Vit phng trnh ng phn gic ca (d1),(d2)Bi5: Trong khng gian 0xyz ,cho hai ng thng (d1),(d2) c phng trnh cho bi :

( )3

4

1

2

2

1:1

=

+=

zyxd ( ) ( )t

32

1

:2 R

tz

ty

tx

d

+=

=

+=

a) Chng t rng hai ng thng (d1),(d2) ct nhau.b) Vit phng trnh ng phn gic ca (d1),(d2)

Bi 6: Trong khng gian 0xyz ,cho hai ng thng (d1),(d2) c phng trnh cho bi :

( )

1

1

:1

=

=

=

z

ty

tx

d , ( ) ( )R

tz

ty

tx

d

=

+=

=

1

1

1

1

2 tt,1

2

:

a) Chng t rng hai ng thng (d1),(d2) cho nhau.b) Vit phng trnhmt phng(P) song song ,cch u (d1),(d2) .

Bi 7: Trong khng gian 0xyz ,cho hai ng thng (d1),(d2) c phng trnh cho bi :

( )

=+

=++

0104z-y

0238zx:d1 , ( )

022

032:2

=++

=

zy

zxd

a) Chng t rng hai ng thng (d1),(d2) cho nhau.b) Vit phng trnhmt phng(P) song song, cch u (d1),(d2) .

Bi8: Trong khng gian 0xyz ,cho hai ng thng (d1),(d2) c phng trnh cho bi :

( ) 3

3

2

2

1

1:1

=

=

zyxd ( ) 0532

02

:2

=+

=+

zyx

zyx

d

a) Chng t rng hai ng thng (d1),(d2) cho nhau.b) Vit phng trnh mt phng(P) song song, cch u (d 1),(d2) .

Bi ton 5. Hai ng thng ng phng v bi tp lin quanBi 1: (HBK-TPHCM-93): Vit phng trnh mt phng (P) cha (d1),(d2) ,bit:

( )2

3

2

1

3

1:1

=

=

+ zyxd ( )

2

3

1

1

1:2

=

=

zyxd

Bi 2: (HSPII-2000): Cho im A(1;-1;1) v hai ng thng (d1),(d2) c phng trnh cho bi :

( )

=+

=+

01y-2x

03z-y-3x:d1 ( ) ( )t

3

21:2 R

tz

ty

tx

d

=

=

=

CMR (d1),(d2) v im A cng thuc mt phng.

Bi 3: Cho hai ng thng (d1),(d2) c phng trnh cho bi : ( )

=+

=++

01y-x

01y2x:d1

z( )

012

033:2

=

=++

yx

zyxd

a) CMR hai ng thng ct nhau.b) Vit phng trnh tng qut ca mt phng (P) cha (d1), (d2).c) Vit phng trnh ng phn gic ca(d1), (d2)

Bi 4: Cho hai ng thng (d1),(d2) c phng trnh cho bi :


12/24


- 12 -

( )1

1

2

1

1

2:1

=

=

zyxd ( ) ( )t

31

2

21

:2 R

tz

ty

tx

d

+=

+=

+=

a) CMR hai ng thng ct nhau.Xc nh to giao im ca n.b) Vit phng trnh tng qut ca mt phng (P) cha (d1),(d2).c) Vit phng trnh ng phn gic ca(d1),(d2)

Bi5: cho hai ng thng (d1),(d2) c phng trnh cho bi :

( )3

2

4

1

1

3:1

=

+=

zyxd , ( )

03

024:2

=

=

zx

yxd

a) Chng t rng hai ng thng (d1),(d2) song song vi nhau.b) Vit phng trnh tng qut ca mt phng (P) cha (d1),(d2).c) Vit phng trnh ng thng (d) trong (P) song song cch u (d1),(d2) .

Bi ton 6. Hai ng thng cho nhau v bi tp lin quanBi 1: (HNN-96): cho hai ng thng (d1),(d2) c phng trnh cho bi :

( )

34

24

37

:1

+=

=

+=

tz

ty

tx

d ( ) ( )R

tz

ty

tx

d

=

+=

+=

1

1

1

1

2 tt,

12

29

1

:


Bi 2: (HTCKT-96): Trong khng gian 0xyz , cho hai ng thng (d1),(d2) c phng trnh cho bi :

1( ): - 1 -1d x y z= + = , 2( ) : - 1 -1d x y z+ = = . Tm to im A1 thuc (d1) v to im A2 thuc (d2)

ng thng A1A2 vung gc vi (d1) v vung gc vi (d2) .Bi 3: (H L 1996) Cho hai ng thng (d1),(d2) c phng trnh cho bi :

( )

1

1

:1

=

=

=

z

ty

tx

d , ( ) ( )R

tz

ty

tx

d

=

+=

=

1

1

1

1

2 tt,1

2

:

a) Chng t rng hai ng thng (d1),(d2) cho nhau.Vit phng trnh mt phng (P),(Q) song songvi nhau v ln lt cha (d1),(d2)

b) Tnh khong cch gia (d1),(d2) .Bi 4: (HTS-96): Cho hai ng thng (d1),(d2) c phng trnh cho bi :

( ) ( )Rt

12

23

31

:1

=

+=

+=

z

ty

tx

d ( ) 01225

0823:2

=+

=

zx

yxd

a) Chng t rng hai ng thng (d1),(d2) cho nhau. Tnh khong cch gia (d1),(d2)b) Vit phng trnh ng thng vung gc chung ca (d1),(d2) .

Bi 5: : (PVBC 99) Cho hai ng thng (d1

),(d2

) ,bit:

( )1

2

3

1

2

1:1

=

=

+ zyxd ; ( )

25

2

2

2:2

=

+=

zyxd


Bi 6: (HSPQui Nhn-D-96): cho hai ng thng (d1),(d2) ,bit:


13/24


- 13 -

( )

=+

=+

04y-x

0yx:d1

z ( ) ( )t

2

31

:2 R

tz

ty

tx

d

+=

=

+=

a) Chng t rng hai ng thng (d1),(d2) cho nhau. b) Tnh khong cch gia (d1),(d2Bi 7: : cho hai ng thng (d1),(d2) ,bit:

( ) 1

9

2

3

1

7

:1

=

=

zyx

d ( ) 3

1

2

1

7

3

:2

=

=

zyx

d a) Chng t rng hai ng thng (d1),(d2) cho nhau.b) Vit phng trnh ng thng vung gc chung ca (d1),(d2) .

Bi 8: (H Hu 1998) Cho hai ng thng (d1),(d2) c phng trnh cho bi :

( )

1

1

22

: 1

1

1

=

+=

+=

z

ty

tx

d , ( ) ( )R

tz

ty

x

d

=

+=

=

21

2

22 t,t

3

1

1

:

a) Chng t rng hai ng thng (d1),(d2) cho nhau.b) Vit phng trnh mt phng (P) cha (d1) v song song vi (d2) .c) Tnh khong cch gia (d1),(d2) .

Bi 9: (HNN-97): Cho hai ng thng (d1),(d2) c phng trnh cho bi :

( )

=++

=++

01y-x

02zyx:d1

z ( ) ( )t

2

5

22

:2 R

tz

ty

tx

d

+=

=

+=

a) Chng t rng hai ng thng (d1),(d2) cho nhau. b) Tnh khong cgia (d1),(d2) .

c) Vit phng trnh ng thng (d) i qua M(1,1,1) v ct ng thi (d1),(d2) .Bi 10: (HKT-98): Cho t din SABC vi cc nh S(-2;2;4), A(-2;2;0) ,B(-5;2;0) ,C(-2;1;1). Tnh khongcch gia hai cnh i SA v SB.V. im, ng thng v Mt Phng

Bi ton1: ng thng i qua mt im ct c hai ng thng cho trc.Bi 1: Vit phng trnh ng thng i qua A(1;2;3) v ct c hai ng thnga) ( )

=+

=++

0104z-y

0328zx:d1 ( )

022

032:2

=++

=

zy

zxd

b) ( )3

3

2

2

1

1:1

=

=

zyxd ( )

0532

02:2

=+

=+

zyx

zyxd

Bi 2: Vit phng trnh ng thng i qua gc to v ct c hai ng thng:

( ) R

tz

ty

tx

d

+=

+=

+=

t

33

2

21

:1 , ( )

13

23

2

:2

+=

+=

+=

uz

uy

ux

d

Bi 3: Vit phng trnh ng thng (d) song song vi ng thng () v ct c hai ng thng:

( ) 01

02:

=++

=++

zyx

zyx ( ) R

tz

ty

tx

d

=

=

+=

t

2

1

2

:1 ( ) 03

022:2

=

=+

y

zxd

Bi 4: (HDL-97): Vit phng trnh ng thng i qua A(1;-1;0) v ct c hai ng

thng: ( )2

1

1

1

1:1

=

+=

zyxd ( )

121

1:2

zyxd ==

+


14/24


- 14 -

Bi 5: (HTS-99): Vit phng trnh ng thng i qua A(1;-1;0) v ct c hai ng thng:

( )

=+

=

012-2z5x

08-2y-3x:d1 ( ) ( )t

2

23

31

:2 R

tz

ty

tx

d

=

=

+=

Bi 6: Vit phng trnh ng thng (d) vung gc vi (P) :x+y+z-2=0 v ct c hai ng thng (d 1) v (d2

( ) R

tz

ty

tx

d

=

=

+=

t

2

1

2

:1 ( ) 03

022:2

=

=+

yzxd

Bi 7: Vit phng trnh ng thng (d) i qua gc to v ct c 2 ng thng (d1) v (d2):

( ) R

tz

ty

tx

d

=

+=

+=

t

33

2

12

:1 ( )

0313

23

2

:2

=+=

+=

+=

uz

uy

ux

d

Bi ton 2: ng thng i qua mt im vung gc vi c hai ng thng cho trc.Bi 1: Vit phng trnh ng thng i qua A(1;2;3) v ct c hai ng thng (d1) ,(d2):

a) ( )

=+

=++

0104z-y

0328zx

:d1 ( ) 022

032

:2

=++

=

zy

zx

d b)

( ) 01225

0823:1

=+

=

zx

yxd ( ) ( )t

2

23

31

:2 R

tz

ty

tx

d

=

=

+=

Bi 2: (HTCKT 1999) Vit phng trnh ng thng (d) i qua A(1;1;-2) song song vi mt phng (P) v

vung gc vi ng thng (d):1 1 2

, ( ) : - - -1 02 1 3

x y zP x y z

+ = = =

Bi ton 3: ng thng i qua mt im vung gc vi mt ng v ct mt ng thng khcBi 1: (HSP TPHCM-95): Vit phng trnh ng thng i qua A(0;1;1) v vung gc vi ng thng

(d1) v ct (d2) ,bit: ( )11

23

1:1zyxd =+= ( )

0102:2

=+

=++

xzyxd

Bi 2: Vit phng trnh ng thng i qua A(1;1;1) v vung gc vi ng thng (d1) v ct (d2) ,bit :

( )

=+

=++

01-zy

03-zyx:d1 ( )

01

0922:2

=+

=+

zy

zyxd

Bi 3: Vit phng trnh ng thng ct c ba ng thng (d1) (d2) , (d3)v vung gc vi vect ( )1;2;3ur

,

bit: ( )

=+

=+

01z

01y-x:d1 ( )

0

01:2

=

=+

z

yxd ( )

1

01:3

=

=

z

yxd

Bi 4: Tm tt c cc ng thng ct (d1), (d2) di cng mt gc, bit: ( )

=

=

az

0y-mx

:d1 ( ) 0

:2

=

=+

az

ymx

d

Bi 5: (HTL-97):Vit phng trnh ng thng i qua A(3;-2;-4) song song vi mt phng (P) :3x-2y-3z-

7=0 v ct ng thng (d) bit: ( )2

1

2

4

3

2:

=

+=

zyxd

Bi ton 4: Hnh chiu vung gc caim ln mt phngBi 1: Tm to im i xng ca A(-2;1;3) qua (P) cho bi: 2x+y-z-3=0.Bi 2: (HKTCN-97): Cho im A(1;2;3) v mt phng (P) c phng trnh :2x-y+2z-3=0

a) Lp phng trnh mt phng qua A v song song vi (P).b) Gi H l hnh chiu vung gc ca A ln (P). Xc nh to ca H


15/24


- 15 -

Bi3: (HGTVTTPHCM-99): Cho ba im A(1;1;2), B(-2;1;-1), C(2;-2;-1) .Xc nh to hnh chiu vungc ca im O ln mt phng (ABC).Bi 4: (HTCKT-2000): Cho im A(2;3;5) v mt phng (P) c phng trnh: 2x+3y+z-17=0

a) Lp phng trnh ng thng (d) qua A v vung gcvi (P).b) CMR ng thng (d) ct trc 0z , tm giao im M ca chng.c) Xc nh to im A1 i xng vi A qua (P).

Bi 5: Cho mt phng (P) v ng thng (d) c phng trnh:

(P): 2x+5y+z+17=0 v ( ) 0736

02743:

=++

=+

zyxzyxd

a) Xc nh to giao im A ca (d) v (P).b) Lp phng trnh ng thng (d1) i xng vi (d) qua (P)

Bi 6: Cho mt phng (P) v ng thng (d) c phng trnh :

( ) : 2 4 0P x y z+ + + = v ( ) 0723

032:

=

=+

zx

yxd

a) Xc nh to giao im A ca (d) v (P).b) Lp phng trnh ng thng (d1) i xng vi (d) qua (P)

Bi 7: (HQG 1998) Cho cc im A(a;0;0); B(0;b;0); C(0;0;c) (a,b,c dng ). Dng hnh hp ch nht nh

O,A,B,C lm 4 nh v gi D l nh i din vi nh O ca hnh hp a) Tnh khong cch t C n mt phng (ABD)b) Tnh to hnh chiu vung gc ca C xung mt phng (ABD). Tm iu kin i vi a,b,c

hnh chiu nm trong mt phng (xOy)Bi ton 5: Hnh chiu vung gc ca ng thng ln mt phngBi 1: (HQG TPHCM 1998) Trong khng gian vi h trc to trc chun 0xyz ,cho ng thng (d) v

mt phng (P) c phng trnh: (P):x+y+z-3=0 v ( ) 032

03:

=

=+

zy

zxd Lp phng trnh hnh chiu vung g

ca ng thng (d) ln (Q).Bi 2: Lp phng trnh hnh chiu vung gc ca giao tuyn (d) ca hai mt phng 3x-y+z-2=0 v x+4y-5=ln mt phng 2x-z+7=0.

Bi 3: (HMC-98) :Trong khng gian vi h to trc chun 0xyz cho ng thng (d) v mt phng (P)c phng trnh: ( )

2

1

3

4

4:

+=

=

zyxd v (P): x-y+3z+8=0. Hy vit phng trnh chnh tc hnh chiu

vung gc ca (d) ln (P) .Bi 4: Trong khng gian 0xyz cho ng thng (d) v mt phng (Q) c phng trnh :

( )

=

=+

02z-x

03-z2y-3x:d ( ) ( )R

ttz

tty

ttx

Q

+=

+=

++=

21

21

21

21

t,t

5

24

34

: . Lp phng trnh hnh chiu vung gc ca

ng thng (d) ln (Q) .

Bi 5: Cho ng thng (d) v mt phng (Q) c phng trnh: ( )

=+

=++

03-z-2yx

01zy-2x

:d (Q): x-y+z+10=0

Hy vit phng trnh chnh tc hnh chiu vung gc (d1) ca (d) ln (P) .Bi 6: (H Cn Th 1998) Trong khng gian vi h to vung gc 0xyz cho ng thng (d) v mt phn

(P) c phng trnh: ( )3

1

2

2

1

1:

=

=

zyxd v (P): x+y+z+1=0. Hy vit phng trnh chnh tc hnh chi

vung gc (d1) ca (d) ln (P) .Bi 7: (HVQY-95): Trong khng gian vi h to vung gc 0xyz cho ng thng (d) v mt phng (P) c

phng trnh : ( )3

1

2

2

1

1:

=

=

zyxd v (P): x+y+z+1=0.


16/24


- 16 -

a) Hy vit phng trnh chnh tc hnh chiu vung gc (d1) ca (d) ln (Oxy) .b) CMR khi m thay i ng thng (d1) lun tip xc vi mt ng trn c nh trong mt phng 0xy. Bi 8:(HQG-98): Trong khng gian vi h to vung gc 0xyz cho mt phng (P) v hai ng thn

(d1) v (d2) c phng trnh: (P):x+y-z+1=0, ( )

=+

=+

02yx

01z-2y:d1 , ( )

02

0123:2

=+

=+

zx

zyd

a) Hy vit phng trnh hnh chiu vung gc (1), (2) ca (d1), (d2) ln (P). Tm to giao im I

ca (d1), (d2).b) Vit phng trnh mt phng ( )1P cha (d1) v vung gc vi (P).Bi ton 6: Hnh chiu vung gc ca im ln ng thng

Bi 1: cho im A(1;2;3) v ng thng (d) c phng trnh : ( ) 01

0922:

=+

=+

zy

zyxd . Xc nh to

hnh chiu vung gc ca A ln (d) .T tm to im A1 i xng vi A qua (d) .

Bi 2: cho im A(1;2;-1) v ng thng (d) c phng trnh : ( ) R

tz

ty

tx

d

=

+=

+=

t

33

2

12

: .Xc nh to hn

chiu vung gc ca A ln (d) .T tm to im A1 i xng vi A qua (d) .

Bi 3: cho im A(2;1;-3) v ng thng (d) c phng trnh : ( )1

3

2

2

1

1:

+=

=

zyxd .Xc nh to

hnh chiu vung gc ca A ln (d) .T tm to im A1 i xng vi A qua (d) .Bi 4: (Hhu /A,B phn ban 98): Trong khng gian 0xyz cho im A(2;-1;1) v ng thng (d) c phng

trnh : ( ) 022

04:

=+

=+

zyx

zyd

a) Vit phng trnh mt phng (P) i qua A v vung gc (d) .b) Xc nh to im B i xng vi A qua (d) .

Bi 5: ( 60-Va): Lp phng trnh ng thng qua A(3;2;1) v vung gc vi ng thng

(d)1

3

42:

+==

zyxv ct vi ng thng .

Bi 6: (HTM-2000): Lp phng trnh ng thng qua A(2;-1;0) v vung gc vi ng thng

( ) 012

025:

=++

=+++

zyx

zyxd v ct vi ng thng .

Bi7: (HV BCVT-2000): Cho 2 ng thng () v (d) c phng trnh :

( )3

1

2

1

7

3:

=

=

zyx ( )

1

9

2

3

1

7:

=

=

zyxd

Lp phng trnh ng thng (d1) i xng vi (d) qua ()Bi 8: (HHH-1999): Trong khng gian cho 2 ng thng (d1),(d2) :

( ) Rt

54

21:)(d01

012: 21

+=

+=

=

=+

=++

tz

ty

tx

zyxyxd

a) (d1) , (d2) c ct nhau hay khngb) Gi B,C ln lt l cc im i xng ca A(1;0;0) qua (d1),(d2) . Tnh din tch tam gic ABC

Bi 9: (HTM-1999): Trong khng gian cho ng thng (d1) v mt phng (P) :

( ) 032:)(P01722

0322:1 =+

=

=zyx

zyx

zyxd

a) Tm im i xng ca im A(3;-1;2) qua ng thng (d)


17/24


- 17 -

b) Vit phng trnh hnh chiu vung gc ca ng thng (d) trn mt phng (P)Bi10: Trong khng gian 0xyz cho bn ng thng (d1), (d2), (d3), (d4) c phng trnh :

( ) 0

:1

=

=

hz

ymxd , ( )

0:2

=

=

hz

ymxd , ( )

0:3

=

=+

hz

ymxd , ( )

0:4

=

=+

hz

ymxd

CMR cc im i xng A1, , A2, , A3, A4 ca A bt k trong khng gian qua (d1), (d2), (d3), (d4) l ngphng. Lp phng trnh mt phng cha chng .Bi ton 7: im v mt phng Bi 1: cho hai im A(1;0;2) ;B(2;-1;3) v mt phng (P): x-2y+z-4=0.Tm im M thuc (P) sao chAM+BM nh nht. Bi 2: cho hai im A(1;1;0) ;B(0;-1;1) v mt phng (P): x-2y+z-4=0.Tm im M thuc (P) sao chAM+BM nh nht.Bi 3: (Hhu /A h cha phn ban 97):Trong khng gian vi h to 0xyz cho mt phng (P): 2x-y+z+1=v hai im A(3;1;0), B(-9;4;9) .Tm to im M trn mt phng (P) sao cho MBMA l ln nht .

Bi 4: (HQG-2000):Cho mt phng(P):x+y+z-1=0 v hai im A(1;-3;0) ,B(5;-1;-2)

a) Chng t rng ng thng i qua A,B ct mt phng (P) ti mt im I, tm to im .

b) Tm to im M trn mt phng (P) sao cho MBMA t gi tr ln nht.

Bi 5: (HMC-97):cho ba im A(1;4;5) B(0;3;1) ,C(2;-1;0) v mt phng (P): 3x-3y-2z-15=0.Gi G l trng tm ABC .CMiu kn cn v M nm trn mt phng (P) c tng cc bnh phng khong cch n cc im A,Bnh nht l im M phi l hnh chiu vung gc ca im G trn mt phng (P) .Xc nh to ca im .Bi 6: Cho mt phng (P) 3x+3y+mz-6-m=0.

a) CMR (P) lun i qua mt im c nh M, Tm to ca M.b) Gi s (P) ct 0x,0y,0z theo th t ti A,B,C .c) Tnh 0A,0B,0C t din 0ABC t gi tr nh nht .d) Tnh 0A,0B,0C 0A+0B+0C l nh nht .

Bi ton 8:im v ng thng

Bi 1: Tm trn ng thng (d) im M(xM,yM,zM) sao cho MMM zyx222

++ nh nht ,bit:

a) ( ) R

tz

ty

tx

d

=

=

+=

t

3

21

2

: b) ( )5

4

3

1

2

3:

=

+=

zyxd c)

( ) 0732

0143:

=+++

=++

zyx

zyxd

Bi 2: Cho ng thng (d) c phng trnh : ( ) 05

03:

=+

=

yx

zyxd .Tm im M thuc (d) sao cho AM +

BM nh nht khi :

a) A(1;2;-1), B(8;1;-2) . b) A(1;2;-1),B(0;1;2).Bi 3: (HBK-98):Cho ng thng (d) v mt phng (P)c phng trnh :

( ) R

tz

ty

tx

d

=

=

+=

t

3

2

21

: , ( ) : 2 - - 2 1 0P x y z + =

a) Tm to cc im thuc ng thng(d) sao cho khong cch tmi im n mt phng (P) bng 1.b) Gi K l im i xng ca im I(2;-1;3) qua ng thng (d) .Xc nh to K.Bi 4: (HHng c -2000): Cho ng thng (d) v mt phng (P) c phng trnh :


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- 18 -

( ) R

tz

ty

tx

d

=

+=

+=

t

2

1

1

: v (P): x+2y+z-1=0.

a) Tm to cc im thuc ng thng(d) sao cho khong cch tmi im n mt phng (P) bng

6 .b) Gi K l im i xng ca im I(2;0;-1) qua ng thng (d) .Xc nh to K.Bi 5: (H nng -2000): Cho im A(-4;4;0),B(2;0;4),C(1;2;-1),D(7;-2;3).

a) CMR A,B,C,D ng phng . b) Tnh khong cch t Cnng thng (AB)Bi ton 9: Gc trong khng gianBi 1: Xc nh s o gc gia 2 ng thng (d1),(d2) c phng trnh :

a) ( ) 015z-x

019-y4x:)(d&

46

32

23

: 21

=+

=+

+=

+=

+=

tz

ty

tx

d b) ( )

33

2

12

:1

+=

+=

+=

tz

ty

tx

d ,

( ) 31

23

2

:2

+=

+=

+=

uzuy

ux

d

c) ( ) 01

012:1

=+

=++

zyx

yxd ( )

012

033:2

=+

=++

yx

zyxd

Bi 2: (HHH-2000): Cho ba ng thng (d1),(d2), (d3) c phng trnh :

( ) R

tz

ty

tx

d

+=

+=

+=

t

32

42

1

:1 , ( ) 012

034:2

=+

=+

zyx

zyxd ( )

1

5

1

1

3:3

=

=

zyxd

a) Xc nh cosin gc gia (d1),(d2).

b) Lp phng trnh ng thng (d) song song vi (d3) ng thi ct c (d1),(d2).Bi 3: Xc nh s o gc gia ng thng (d) v mt phng (P) c phng trnh cho bi :

( ) 015

0194:

=+

=+

zx

yxd v (P):x+y-7z-58=0.

Bi 4: (CSP TP.HCM-99): Cho ng thng (d) v mt phng (P) c phng trnh :

( )1

3

2

4

1

3:

+=

=

zyxd v (P):2x+y+z-1=0

a) Xc nh s o gc gia ng thng (d) v mt phng (P) .b) Tm to giao im A ca ng thng (d) v mt phng (P).c) Lp phng trnh tng qut ca ng thng (d1) i qua A vung gc vi (d) v nm trong mt phng (P).Bi 5: (HAN-CS-98): Cho ng thng (d) v mt phng (P) c phng trnh :

( )2

1

2

3

1

1:

+=

=

zyxd v (P): x+z+2=0

a) Xc nh s o gc gia ng thng (d) v mt phng (P) .b) Lp phng trnh ng thng (d1) l hnh chiu vung gc ca (d) ln mt phng (P).

Bi ton 10: Tam gic trong khng gianBi 1: Cho ABC bt A(1;2;5), B(1;4;3), C(5;2;1) v mt phng (P):x-y-z-3=0.a) Lp phng trnh ng trung tuyn ,ng cao v ng phn gic trong k t nh A.


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- 19 -

b) Gi G l trng tm ABC .CMR iu kn cn v im M nm trn mt phng (P) c tng cc bnhphng khong cch n cc im A,B,C nh nht l im M phi l hnh chu vung gc ca im G trnmt phng (P) .Xc nh to ca im M .Bi 2: Cho mt cu ( ) 0642: 222 =++ zyxzyxS .a) Gi A,B,C ln lt l giao im (khc gc to ) ca mt cu (S) vi 0x,0y,0z .Cc nh to ca A,B,Cv lp phng trnh mt phng (ABC).b) Lp phng trnh cc ng trung tuyn , ng cao v ng phn gic trong k t nh A ca ABC.c) Xc nh to tm v tnh bn knh ng trn ngoi tip ABC.Bi 3 Cho cc im A(3;1;0), B(2;2;4) ,C(-1;21).a) Lp phng trnh mt phng (ABC).b) Lp phng trnh cc ng trung tuyn ,ng cao v ng phn gic trong k t nh A ca ABC.c) Xc nh to tm v tnh bn knh ng trn ngoi tip ABC.VI. Mt cu Bi ton 1. Phng trnh mt cuBi 1: Trong cc phng trnh sau y ,phng trnh no l phng trnh ca mt cu ,khi ch r to tv bn knh ca n ,bit:

a) ( ) 02642: 222 =++++ zyxzyxS b) ( ) 09242: 222 =++++ zyxzyxS

c) ( ) 03936333:222

=++++ zyxzyxS d) ( ) 7524:222

=++ zyxzyxSe) ( ) 022: 222 =+++ yxzyxS

Bi 2: Cho h mt cong (Sm) c phng trnh: ( ) 04624:2222

=++++ mmzmymxzyxSm

a) Tm iu kin ca m (Sm) l mt h mt cu .b) CMR tm ca (Sm) lun nm trn mt ng thng c nh.Bi 3: Cho h mt cong (Sm) c phng trnh: ( ) 05824:

22222=+++ mymmxzyxSm

a) Tm iu kin ca m (Sm) l mt h mt cu .b) Tm qu tch tm ca h (Sm) khi m thay i.c) Tm im c nh M m (Sm) lun i qua.

Bi 4: Cho h mt cong (Sm) c phng trnh: ( ) 03cos2sin2:222

=++ mymxzyxSm

a) Tm iu kin ca m (Sm) l mt h mt cu .b) CMR tm ca (Sm) lun chy trn mt ng trn (C) c nh trong mt phng 0xy khi m thay ic) Trong mt phng 0xy, (C) ct 0y ti A v B. ng thng y=m(-1


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- 20 -

c) Lp phng trnh mt cu (S) c ng knh l on vung gc chung ca (d1) v (d2).d) Vit phng trnh tng qut ca mt phng cch u (d1) v (d2).

Bi ton 2: Mt cu tip xc vi mt phngBi 1: Vit phng trnh mt cu (S) bit :

a) Tm I(1;2;-2) v tip xc vi mt phng (P):6x-3y+2z-11=0.b) (CGTVT-2000): Tm I(1;4;-7) v tip xc vi mt phng (P) :6x+6y-7z+42=0.c) Bn knh R = 9 v tip xc vi (P): x+2y+2z+3=0 ti im M(1;1;-3).

Bi 2: Vit phng trnh mt cu c tm I trn ng thng (d) v tip xc vi hai mt phng ( )1P v ( )2P ,bit

a) (HL-95): ( )2

1

2

1

3

2:

=

=

zyxd b) ( )1P :x+2y-2z-2=0. v ( )2P :x+2y-

2z+4=0.

c) ( ) 01445

0724:

=++

=++

zyx

zyxd , d) ( )1P :2x+2y-z-12=0. v

( )2P :-2x+2y-z+8=0.

e) ( ) Rtz

ty

tx

d

=

+=

+=

t

2

3

21

: , f) ( )1

P :3x4y+2z-10=0 ( )2

P :2x

3y+4z-10=0Bi 3: (HLN-97): Cho ng thng (d) v hai mt phng ( )1P , ( )2P ,bit :

( )2

1

3

1

2:

+=

=

zyxd , ( )1P :x+y-2z+5=0. v ( )2P :2x-y+z+2=0

a) Gi A l giao im ca (d) vi ( )1P v ( )2P .Tnh di on AB.

b) Vit phng trnh mt cu cod tm I trn ng thng (d) v tip xc vi hai mt phng ( )1P v

( )2P . Bi ton 3: Mt cu ct mt phng

Bi 1: Lp phng trnh mt cu c tm to giao im I ca mt phng (P) v ng thng (d) sao cho mtphng (Q) ct khi cu theo tht din l hnh trn c din tch 12 ,bit :

a) ( ) Rtz

ty

tx

d

+=

=

+=

t

2

3

1

: ,(P):x-y-z+3=0 b) ( ) 01

03:

=

=++

y

zyxd , (P):x+y-2=0.

Bi 2: Lp phng trnh mt cu c tm thuc ng thng (d) v ct mt phng (P) theo thit din l ng

trn ln c bn knh bng 18.bit: ( ) Rtz

ty

tx

d

+=

+=

+=

t

1

39

412

: v (P):y+4z+17=0.

Bi 3: Trong khng gian 0xyz , cho hai im A(0;0;-3),B(2;0;-1) ,v mt phng (P):3x-8y+7z-1=0 .a) (HVNH-2000): Tm to im C nm trn mt phng (P) sao cho tam gic u .b) Lp phng trnh mt cu (S) i qua 3 im A,B,C v c tm thuc mt phng (P):x-y-z-2=0.

Bi ton 4:Mt cu tip xc vi ng thngBi 1: Vit phng trnh mt cu (S) bit :

a) Tm I(1;2;-1) v tip xc vi ng thng (d) c phng trnh : ( ) Rz

ty

tx

d

=

=

=

t

1

1

:


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- 21 -

b) Tm I(3;-1;2) v tip xc vi ng thng (d) c phng trnh : ( ) 017322

0322:

=

=

zyx

zyxd

Bi 2: Trong khng gian 0xyz, cho hai ng thng (d1),(d2) ,bit :

( ) R

tz

ty

tx

d

+=

=

+=

t

32

1

21

:1 , ( ) 012

043:2

=+

=

zyx

yxd

Lp phng trnh mt cu (S) tip xc vi (d 1) ti im H(3;1;3) v c tm thuc ng thng (d2).Bi 3: Trong khng gian 0xyz, cho hai ng thng (d1),(d2) ,bit :

( ) 01

012:1

=+

=++

zyx

yxd , ( )

012

033:2

=+

=++

yx

zyxd

a) CMR hai ng thng ct nhau .Xc nh ta giao im I ca chng .b) Vit phng trnh tng qut ca mt phng (P) i qua hai ng thng (d1) v (d2).c) Lp phng trnh mt cu tip xc vi (d1),(d2) v c tm thuc ng thng (d) c phng trnh :

( ) R

tz

ty

tx

d

+=

+=

+=

t

33

2

21

:


( ) R)(t

46

32

23

:1

+=

+=

+=

tz

ty

tx

d , ( ) 015

0194:2

=+

=+

zx

yxd

a) CMR hai ng thng ct nhau .Xc nh ta giao im I ca chng .b) Vit phng trnh tng qut ca mt phng (P) i qua hai ng thng (d1) v (d2).c) Lp phng trnh mt cu tip xc vi (d1),(d2) v c tm thuc ng thng (d) c phng trnh :

( )4

9

1

5

3

7:

=

=

+ zyxd


( )4

1

32

2:1

+=

=

zyxd , ( )

129

2

6

7:2

zyxd =

=

a) CMR hai ng thng song song vi nhau.b) Vit phng trnh tng qut ca mt phng (P) i qua hai ng thng (d1) v (d2).c) Lp phng trnh mt cu tip xc vi (d1),(d2) v c tm thuc ng thng (d) c phng trnh :

( ) R

z

ty

tx

d

=

=

=

t

1

1

:


( )4

9

1

5

3

7:1

=

=

+ zyxd , ( )

4

18

1

4

3:2

+=

+=

zyxd

a) CMR hai ng thng song song vi nhau.b) Vit phng trnh tng qut ca mt phng (P) i qua hai ng thng (d1) v (d2).c) Lp phng trnh mt cu tip xc vi (d1),(d2) v c tm thuc ng thng (d) c phng trnh :

( ) R

tz

ty

tx

d

=

=

+=

t

1

3

23

:


22/24


- 22 -


( ) R)(t

33

2

21

:1

+=

+=

+=

tz

ty

tx

d , ( )

31

23

2

:2

+=

+=

+=

uz

uy

ux

d

a) CMR hai ng thng cho nhau.b) Vit phng trnh ng vung gc chung ca(d1) v (d2).

c) Tnh khong cch gia (d1) v (d2).d) Lp phng trnh mt cu tip xc vi (d1),(d2) v c tm thuc mt phng (P) : xy+z-2=0


( ) 01

03:1

=+

=++

zx

zyxd , ( )

01

0922:2

=+

=+

zy

zyxd

a) CMR hai ng thng cho nhau.b) Vit phng trnh ng vung gc chung ca(d1) v (d2).c)Lp phng trnh mt cu tip xc vi (d1),(d2) v c tm thuc mt phng (P):2x-y+3z-6=0.

Bi ton 5: Mt cu ct ng thng

Bi 1: (HQG-96): Cho im I(2;3;-1) v ng thng (d) c phng trnh : ( ) 0843

020345:

=+

=++

zyx

zyxd

a) Xc nh VTCP a ca (d) suy ra phng trnh mt phng (P) qua I v vung gc vi (d):b) Tnh khong cch t I n (d) t suy ra phng trnh mt cu (S) c tm sao cho (S) ct (d) ti h

im phn bit A,B tho mn AB = 40.

Bi 2: Cho ng thng (d) v mt phng (P) c phng trnh : ( ) R

tz

ty

tx

d

=

=

+=

t

3

2

21

: , (P):2x-y-2z+1=0.

a) (HBK-98):Tm to cc im thuc ng thng (d) sao cho khong cch t mi im nmt phng (P) bng 1.

b) (HBK-98):Gi K l im i xng ca im I(2;-1;3) qua ng thng (d) .Xc nh to K.

c) Lp phng trnh mt cu tm I ct ng thng (d) ti hai im phn bit A,B sao cho AB=12.d) Lp phng trnh mt cu tm I tip xc vi mt phng (P).e) Lp phng trnh mt cu tm I ct mt phng (P) theo giao tuyn l mt ng trn c din tch

bng 16 Bi ton 6: Mt cu ngoi tip khi a dinBi 1: (H Hu-96): Trong khng gian vi h to 0xyz ,cho bn im A(1;0;1), B(2;1;2),C(1;-1;1),D(4;5;-5)

a) Vit phng trnh tham s ca ng thng i qua D v vung gc vi mt phng (ABC).b) Vit phng trnh mt cu ngoi tip t din ABCD.

Bi 2: Cho bn im O(0;0;0),A(6;3;0), B(-2;9;1), S(0;5;8)a) (HKT-99): CMR SB vung gc SA.b) (HKT-99): CMR hnh chiu ca cnh SB ln mt phng (0AB) vung gc vi cnh 0A. Gi K l

giao im ca hnh chiu vi 0A. Hy xc nh to d ca K.c) Vit phng trnh mt cu ngoi tip t din ABCD.d) (HKT-99): Gi P,Q ln lt l im gia ca cc cnh S0,AB . Tm to ca im M trn SB sa

cho PQ v KM ct nhau.Bi 3: Trong khng gian vi h to 0xyz ,cho bn im A(4;4;4), B(3;3;1), C(1;5;5), D(1;1;1).

a) (HVKTQS-98): Tm hnh chiu vung gc ca D ln (ABC) v tnh th tch t din ABCD.b) (HVKTQS-98): Vit phng trnh tham s ng thng vung gc chung ca AC v BD.c) Vit phng trnh mt cu ngoi tip t din ABCD.d) Tnh th tch t din ABCD.

Bi 4: Cho bn im A(-1;3;2), B(4;0;-3), C(5;-1;4), D(0;6;1).


23/24


- 23 -

a) (HVNHTPHCM-99):Vit phng trnh tham s ca ng thng BC .H AH vung gc BC .Tm t ca im H.

b) (HVNHTPHCM-99):Vit phng trnh tng qut ca (BCD) .Tm khong cch t A n mt phng(BCD).

c) Vit phng trnh mt cu ngoi tip t din ABCD.Bi 5: Trong khng gian 0xyz, cho hnh chp .bit to bn nh S(5;5;6), A(1;3;0), B(-1;1;4), C(1;-1;4),D(3;1;0).

a) Lp phng trnh cc mt ca hnh chp. b) Lp phng trnh mt cu (S) ngoi tip hnhchp .c) Tnh th tch hnh chp SABCD

Bi 6: (HVKTMM-97) Cho bn im A(1;2;2), B(-1;2;-1), C(1;6;-1), D(-1;6;2).a) CMR t din ABCD c cp cnh i din bng nhau . b) Xc nh to trng tm G ca t

din.c) Vit phng trnh mt cu ngoi tip ,ni tip t din ABCD.

Bi ton 7:Mt cu ni tip khi a dinBi 1: Lp phng trnh mt cu ni tip hnh chp SABCD ,bit:

a)4

( ;0;0)3

S ,A(0;-4;0), B(0;-4;0),C(3;0;0). b) S 0,A(a;0;0),B(0;b;0), C(0;0;c), vi a,b,c>0.

Bi 2: Cho hnh chp SABCD .nh )4,29,

21(S y ABCD l hnh vung c A(-4,5,0) ,ngf cho BD c

phng trnh : ( ) 0

087:

=

=+

z

yxd

a) Tm to cc nh ca hnh chp . b) Lp phng trnh nt cu ngotip hnh chp.

c) Lp phng trnh mt cu ni tp hnh chp.Bi 3: Cho ba im A(2;0;0), B(0;2;0), C(0;0;3).

a) Vit phng trnh tng qut cc mt phng (0AB), (0BC), (0CA), (ABC).b) Xc nh tm I ca mt cu ni tip t din 0ABC .c) Tm to im J i xng vi I qua mt phng (ABC).

Bi 4: (HVKTMM-99):Cho bn im A(1;2;2), B(-1;2;-1), C(1;6;-1), D(-1;6;2).a) CMR t din ABCD c cc cp cnh i din bng nhau.b) Xc nh to trng tm G ca t din .c) Vit phng trnh mt cu ngoi tip t din ABCD.d) Vit phng trnh mt cu ni tip t din ABCD.

Bi ton 8: V tr tng i ca im v mt cuBi 1: Cho mt cu ( ) 034: 222 =++ zyxzyxS .xt v tr tpng i ca im A i vi mt cu (S)trong cc trng hp sau:

a) im A(1;3;2). b) im A(3;1;-4). c) im A(-3;5;1).Bi 2: Tm to im M thuc mt cu ( ) 03242: 222 =+++ zyxzyxS .Sao cho khong cch MAt gi tr ln nht ,nh nht,bit:

a) im A(1;-2;0). b) im A(1;1;-2).Bi ton 9: V tr tng i ca ng thng v mt cu

Bi 1: Cho mt cu ( ) 06222: 222 =++ zyxzyxS .Tm to im M thuc (S) sao cho khongcch t M n (d) t gi tr ln nht, nh nht,bit:

a) ( ) R

tz

ty

tx

d

=

+=

=

t

1

1

2

: b) ( ) 012

032:

=+

=++

zy

zyxd

Bi ton 10: V tr tng i ca mt phng v mt cu


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