Mat phang duong thang mat cau trong khong gian toa do

8/6/2019 Mat phang duong thang mat cau trong khong gian toa do

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(DNG CHO N THI TN C H 2011)

Bm sn.22.03.2011


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CHUYN : VIT PHNG TR NH MT PHNG

A. Kin thc chung 1. Phng tr nh mt phng v cc trng hp c bit - PTTQ(phng tr nh tng qut) mt phng P qua 0 0 0 0( , , ) M x y z v c vtpt(vect php tuyn)

( , , )n A B C l: 0 0 0( ) : ( ) ( ) ( ) 0P A x x B y y C z z Hay ( ) : 0P Ax By Cz D vi 0 0 0( ) D Ax By Cz - PTMP(phng tr nh mt phng) P qua ( ,0,0) ; (0, ,0) ; (0,0, ) A a Ox B b Oy C c Oz c phng tr nh

l: ( ) : 1 x y zPa b c

(Phng tr nh mt phng theo on chn)

- c bit:

+2 2

0( ) / / 0

0

A

P Ox D

B C

+2 2

0( ) / / 0

0

B

P Oy D

A C

+2 2

0( ) / / 0

0

C

P Oz D

A B

- Phng tr nh mt phng(Oxy) l 0 z , (Oyz) l 0 x v (Oxz) l 0 y 2. V tr tng i ca mt thng v mt phng: Cho hai mt phng 1 1 1 1 1( ) : 0 A x B y C z D v 2 2 2 2 2( ) : 0 A x B y C z D TH 1: 1 1 1 11 2

2 2 2 2( ) / /( ) A B C D

A B C D

TH 2: 1 1 1 11 22 2 2 2

( ) ( ) A B C D A B C D

TH 3: 1 2 1 2 1 2 1 2( ) ( ) 0 A A B B C C 3: Phng tr nh chm mt phng: Tp hp cc mt phng( ) cha ng thng ( ) ( ) c gi l chm mt phng xc nh bimt phng( ) v mt phng ( ) Nu 1 1 1 1( ) : 0 A x B y C z D v 2 2 2 2( ) : 0 A x B y C z D th phng tr nh mt phng( ) l:

1 1 1 1 2 2 2 2( ) : ( ) ( ) 0m A x B y C z D n A x B y C z D (*) vi 2 2 0m n phng tr nh (*) c th vit li: ( ) ( ) 0m n 4. Gc v khong cch - Gc ca 2 mt phng:1 1 1 1 1( ) : 0 A x B y C z D v 2 2 2 2 2( ) : 0 A x B y C z D l:

1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2.

A A B B C C cos

A B C A B C

- Gc gia ng thng d v mt phng (P)


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.sin( ,( ))

.u n

d Pu n

- Khong cch t mt im 0 0 0 0; ; M x y z n mt phng : 0P Ax By Cz D

0 0 00 2 2 2,Ax By Cz D

d M P

A B C

B. Mt s dng bi tp

Dng 1:Vit ph ng trnh m t phng (P) i qua im M o(x o ;y o ;z o ) v tho m n iu kin Loi 1 : C m t vect php tuyn Phng php: - Xc nh 0 0 0 0( , , ) M x y z ca mt phng P - Xc nhvtpt ( ; ; )n A B C + Nu / / P QP Q n n

+ Nu P d P d n u

- p dng cng thc: 0 0 0( ) : ( ) ( ) ( ) 0P A x x B y y C z z

Bi tp gii mu:

Bi 1: (SGK 12 BanC BnT89)Trong khng gian vihto Oxyz .Vit phng tr nh mt phng (P):a. i qua im 1; 2; 4 M v nhn vect 2;3;5n lm vect php tuyn b. i qua im 2; 1; 2 M v song song vi mt phng : 2 3 4 0Q x y z Gii:a. Cch 1: Mt phng P i qua im 1; 2; 4 M v c vect php tuyn 2;3;5n c ph ng trnh l :2(x 1) + 3(y + 2) + 5(z 4 ) = 0 hay : 2 3 5 16 0P x y z Cch 2:Mt phng (P) cvtpt 2;3;5n lun c dng 2 3 5 0 x y z D v mt phng (P) i quaim 1; 2;4 2.1 3. 2 5.4 0 16 M D D .Vy mt phng : 2 3 5 16 0P x y z b. Cch 1: Mt phng P i qua im 2; 1; 2 M song song vi mt phng Q nn mt phng P i qua im

2; 1; 2 M v c vtpt 2; 1;3P Qn n nn mt phng P c phng tr nh:2(x 2) 1(y + 1) + 3(z 2) = 0 hay : 2 3 11 0P x y z

Cch 2 :Mt phng(P) c vtpt 2; 1;3Pn lun c dng2 3 0 x y z D v mt phng P i qua im

2; 1; 2 M ' 1 D hay : 2 3 11 0P x y z Hocc thl lun v P song song vi Q nn P lun c dng2 3 0 x y z D v P qua M : 2 3 11 0P x y z


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Bi 2: (SGK Ban C Bn T92)Trong khng gian vi h to Oxyz cho mt phng c phngtrnh: 3x + 5y z 2 = 0 v ng thng d c phng tr nh

12 4: 9 3

1

x t

d y t

z t

a. Tmgiao im M ca ng thng d v mt phng b. Vit phng tr nh mt phng cha im M v vung gc vi ng thng d Gii:a. To im M d l nghim ca phng tr nh

3(12 + 4t) + 5(9 + 3t) (1 + t) 2 = 0 t = 3 .Vy 0;0; 2 M b. Cch 1 : Mt phng i qua im 0;0; 2 M vung gc vi ng thng d nn mt phng i qua im

0;0; 2 M v c vtpt n = d u = (4;3;1) nn mt phng c phng tr nh l:4(x 0) + 3(y 0) + 1(z +2) = 0 hay

: 4 3 2 0 x y z

Cch 2:Mt phng c vtpt n = (4;3;1) lun c dng 4x + 3y + z +D = 0 v mt phng i qua im

0;0; 2 M D = 2 hay : 4 3 2 0 x y z Ch :C th pht biu bi ton di dng nh, cho bit ta 3 im A, B, C. Vit phng tr nh mt phng(P) i qua im A v vung gc vi ng thng BC th khi Pn BC

Nh n xt :- Mt phng c vtpt ; ;n a b c th lun c dng ax + by + cz +D = 0- Nu cho c dng Ax + By + Cz + D = 0 th m song song vi lun c dngAx + By + Cz + D = 0 vi ' 0 D - Hai mt phng song song vi nhau th hai vtpt cng song song (cng phng) vi nhau, mt phngvung gc vi ng thngth vtpt v vtcp cng song song (cng phng) vi nhau . iu ny l giiti sao trong bi 1 cu b li chn Pn = Qn ,thtvy v mt phng P song song vi mt phng (Q)nn hai vtpt cng song song (cng phng) vi nhau hay Pn = k. Qn , v k 0 nn chn k = 1 Pn =

Qn . Tng t nh th trong bi 2b ta chn k = 1 n = d u , t ta c nhn xt+ Hai mt phng song song vi nhau th chng c cng vtpt+ Nu mt phng P cha hai im A v B th AB l mt vtcp ca mt phng P + Nu mt phng P vung gc vi mt phng (Q) th vtpt ca mt phng P l vtcp ca mt

phng (Q) v ngc li + Nu mt phng P vung gc vi vecto AB th vecto AB l mt vtptca mt phng P - Vect php tuyn cng c th cho h nh thcl vung gc vi gi ca vect a no, khi taphi hiu y a l vect ch phng

Bi 3: (SGK Ban C Bn T92)Trong mt phng vi h to Oxyz cho im vect 6; 2; 3a v 1; 2; 3 A . Vit phng tr nh mt phng cha im A v vung gc vi gi ca vect a Hng dn: Lm tngt nh bi 2b ta c : 6 2 3 2 0 x y z


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Bi 4: (SGK Ban C Bn T80)Trong khng gian vi h to Oxyz .Vit phng tr nh mt phng iqua im 2;6; 3 M v ln lt song song vi cc mt phng to Gii:

Nh n xt :- Cc mt phng to y l Oxy; Oyz; Oxz . Thot u ta thy cc mt phng ny khng thy vtpt,nhng thc ra chng cvtpt,cc vtpt ny c xy dng nn t cc vect n v tr n cc trcOx, Oy, Ozln lt l i = (1;0;0) ;j = (0;1;0) ;k = (0;0;1),cc vect ny c coi l cc vtcp- By gi ta s vit phng tr nh mt phng P i qua M v song song vi mt phng 0xy cn cc mtphng khc lm tng tCch 1:Mt phng P i qua 2;6; 3 M v song song vi mt phngOxy mt phng P i qua M vvung gc Oz nn mt phng (P) i qua M nhn vect Pn = k lm vtptc phng tr nh l :

0(x 1) + 0(y 6) + 1(z + 3) = 0 hay : 3 0P z Cch 2:Mt phng P song song vi mt phng0xy mt phng P song song vi hai trcOx v Oy

Pn i v Pn j Pn = [i , j ] = (0;0;1) l vtpt nn : 3 0P z Tng t (P) //Oyz v i qua im M nn : 2 0P x

(P) // Oxz v i qua im M nn : 6 0P y Cch 3:Mt phng P song song vi mt phng Oxy nn mt phng P lun c dng Cx + D = 0 v mtphng P i qua M C. 3 D 0 v C 0 nn chn C = 1 D = 3 .Vy mt phng P c phng tr nh l : 3 0P z Ch : Bi ton c th pht biu l vit phng tr nh (P)i qua M // viOx v Oy P i qua M // vi mtphng 0xy

Loi 2: C mt cp vect ch phng ,a b

(vi , 0a b

c gi song song ho c nm tr n mp ( )P ) - Tm vtpt ,n a b

- P l mp qua 0 0 0 0( , , ) M x y z v c VTPTn - Quay li loi 1

Bi tp gii mu:

Bi 5: (SGK Ban C Bn T80)Trong khng gian vi h to O xyz . Vit phng tr nh mt phng P i qua im 0; 1; 2 A v song song vi gi ca mi vect u = (3;2;1) vv = 3;0;1

Gii:Cch 1:Mt phng P i qua 0; 1; 2 A v song song vi gi ca hai vect u = (3;2;1) ; 3;0;1v

mt phng P i qua A vc Pn u ; Pn v (vi u v v khng cngphng) mt phng P i qua A vc vtpt , 2; 6;6 2 1; 3;3Pn u v

mt phng P c phng tr nh l :


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1(x 0) 3(y + 1) +3(z 2) = 0 hay : 3 3 9 0P x y z Cch 2 :Lm tng t nh bi 1bkhi bit 2; 6;6Pn v 0; 1; 2 A Bi 6: (SBT Ban C Bn T99)Trong khng gian vi h to O xyz . Vit phng tr nh mt phng

i qua im 2; 1; 2 M , song song vi trcOy v vung gc vi mt phng : 2 3 4 0 x y z

Gii:Cch 1:Mt phng i qua im 2; 1; 2 M song song vitrc0y v vung gc vi mt phng

mt phng i qua M v c n j ; n n (vi j v n khng cng phng)mt phng i qua M vc vtpt n = [ j , n ] = (3;0;-2)mt phng c phng tr nh l :

3(x 2) + 0(y + 1) 2(z 2) = 0 hay : 3 2 2 0 x z Cch 2:Lm tng t nh bi 1bkhi bit 3; 0; 2n v 2; 1; 2 M

Cch 3:Gi s mt phng c dng : 2 2 2

0 0 Ax By Cz D A B C mt phng c vtpt ; ;n A B C

- Mt phng i qua im 2; 1; 2 M .2 .( 1) .2 0 1 A B C D - Mt phng song song vi trc Oy . 0 .0 .1 .0 0 2n j A B C - Mt phng vung gc vi mt phng . 0 .2 . 1 .3 0 3n n A B C

Gii h (1), (2) v (3) 3, 0, 2, 2. A B C D Vy mt phng c phng tr nh l :3 2 2 0 x z Bi 7: (SBT Ban C Bn T98)Trong khng gian Oxyz.Vit phng tr nh mt phng i qua im

3; 1; 5 M ng thi vung gc vi hai mt phng : 3 2 2 7 0 x y z v : 5 4 3 1 0 x y z Gii:Cch 1:Mt phng i qua im 3; 1; 5 M ng thi vung gc vi hai mt phng v mtphng i qua im M v c n n ; n n (vi n v n khng cng phng) mt phng

i qua im M v c vtpt n = [ n , n ] = (2;1;-2)mt phng ( ) c phng tr nh l :

2(x 3) + 1(y + 1) 2(z + 5) = 0 hay : 2 2 15 0 x y z

Cch 2:Lm tng t nh bi 1bkhi bit n = 2;1; 2 v 3; 1; 5 M Cch 3:Gi s mt phng c dng : 2 2 20 0 Ax By Cz D A B C mt phng c vtpt ; ;n A B C

- Mt phng i qua im 3; 1; 5 M .3 .( 1) . 5 0 1 A B C D - Mt phng vung gc vi mt phng . 0 .3 . 2 .2 0 2n n A B C

- Mt phng vung gc vi mt phng . 0 .5 . 4 .3 0 3n n A B C


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T (1) v (2) ta c 3 21, 62 2

C B A D B A th vo (3) ta c 2 A B chn

1, 2 2, 15 B A C D Vy phng tr nh mt phng l 2 2 15 0 x y z Bi 8:(H B 2006)Trong khng gian vi h to O xyz, cho im A(0;1;2) v hai ng thng

11 1: , ' : 1 22 1 1

2

x t x y zd d y t

z t

Vit phng tr nh mt phng i qua A ng th i song song vi d v dGii: Cch 1:V 1 20;1; 1 ; 1; 1;2 B d C d v 1 2, , / / B C d d Vecto ch phng ca1 2d v d ln lt l 1 22;1; 1 1; 2;1u v u

vecto php tuyn ca l 1 2, 1; 3; 5n u u

V i qua 0;1;2 : 3 5 13 0 A x y z s: : 3 5 13 0 x y z Cch 2:Gi s mt phng c dng : 2 2 20 0 Ax By Cz D A B C

mt phng c vtpt ; ;n A B C - Mt phng i qua im M .0 .1 .2 0 1 A B C D - Mt phng song song vi ng thngd . 0 .2 .1 . 1 0 2d n u A B C - Mt phng song song vi ng thngd '. 0 .1 . 2 .1 0 3d n u A B C

T (1) v (2) ta c 2 , 4 3C A B D A B th vo (3) ta c 3 A B chn1, 3 5, 13 A B C D Vy phng tr nh mt phng l 3 5 13 0 x y z

Nh n xt: Nu im A d (hoc ' A d ) th bi ton tr thnh vit phng tr nh mt phng cha d (hoc 'd )v song song vi 'd (hocd )

Bi tp t gii:

Bi 1:

a. Trong khng gian vi h to Oxyz cho 3 im 3;4;1 , 2;3;4 , 1;0;2 . M N E Vit phng tr nhmt phng i qua im E v vung gc vi MN. (thi tt nghip BTTHPT ln 2 nm 2007)

b. Vit phng tr nh mt phng i qua 1; 2;1K v vung gc vi ng

thng1

: 1 21 3

x t

d y t

z t

.

( thi tt nghip THPT ln 2 nm 2007)


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s:a. : 3 5 0 x y z b. : 2 3 8 0 x y z Bi 2:Trong khng gian vi h to Oxyz cho im 1; 1;0 M v mt phng P c phng tr nh:

2 4 0. x y z Vit phng tr nh mt phng i qua M v song song vi P s: : 2 2 0 x y z

( thi tt nghip THPT h phn ban nm 2007)Bi 3:Vit phng tr nh mt phng i qua im 2;3;1 M v vung gc vi hai mt phng

: 2 2 5 0 v : 3 2 3 0P x y z Q x y z (Sch bi tp nng cao h nh hc 12)

s: : 3 4 19 0 x y z Bi 4:Vit phng tr nh mt phng i qua im 2;1; 1 M v qua giao tuyn ca hai mt phng:

4 0 v 3 1 0. x y z x y z (Sch bi tp nng cao h nh hc 12)

s: :15 7 7 16 0 x y z

Dng 2: Vit phng tr nh m t phng (P) i qua im M 1(x1 ;y1 ;z1 ) v M 2(x 2 ;y 2 ;z 2 ) ng thi tho m niu kina. Vung gc vi mt phngb. Song song vi ng thng d (hoc trc Ox, Oy, Oz) c. C khong cch t im M ti l hd. To vi mt gc Q mt gc

Bi tp giimu:

Bi 1: (SGK Ban C Bn T80)Trong khng gian vi h to Oxyz .Vit phng tr nh mt phng

i qua hai im 1;0;1 , 5;2;3 M N v vung gc vi mt phng : 2 7 0 x y z Gii:Cch 1 :Mt phng i qua hai imM(1;0;1); N(5;2;3) v vung gc vi mt phng ( )

mt phng i qua imM v n MN ; n n (vi MN v n khng cng phng) mt phng i qua imM v c vtpt n = [ MN , n ] = 4;0; 8 = 4 1; 0; 2

mt phng c phng tr nh l :1(x 1) + 0(y 0) 2(z 1) = 0 hay : x 2z + 1 = 0

Cch 2:

Gi s mt phng c dng : 2 2 20 0 Ax By Cz D A B C mt phng c vtpt ; ;n A B C

- Mt phng i qua 1;0;1 M .1 .0 .1 0 1 A B C D - Mt phng i qua 5;2;3 N .5 .2 .3 0 2 A B C D - Mt phng vung gc vi mt phng . 0 .2 . 1 .1 0 3n n A B C

T (1) v (2)ta c 2 ,C A B D A B th vo (3) ta c2 0 B chn

1, 0 2, 1 A B C D


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Vy phng tr nh mt phng l 2 1 0 x z Bi 2:Trong khng gian vi h to Oxyz .Vit phng tr nh mt phng (P) i qua hai im

4; 1;1 M ; 3;1; 1 N v cng phng (song song) vi trcOxGii:Cch 1 :

Mt phng (P) i qua im 4; 1;1 M ; 3;1; 1 N v cng phng vi trcOx mt phng (P) i quaim M v Pn MN ; Pn i (vi v i khng cng phng)

mt phng (P) i qua im M v nhnvtpt Pn = [ ,i ] = 0; 2; 2 = 2 0;1;1 mt phng (P) c phng tr nh l :

0(x 4) + 1( y + 1) + 1(z 1) = 0 hay (P): y + z = 0Cch 2:Lm tng t bi 1 (cch2) iu kin y l Pn i Bi 3: (SBT Ban Nng Cao T126)Trong mt phng Oxyz .Vit phng tr nh mt phng (Q) i qua haiim 3;0;0 , 0;0;1 A C v to vi mt phngOxy mt gc= 60o Gii:

Cch 1:Mt phng (Q) i qua A, C v to vi mt phngOxy mt gc bng 60o nn mt phng (Q) ct mt phngOxy ti im B(0;b;0) Oykhc gc to O b 0

mt phng (Q) l mt phng theo on chn c phng trinh l :

113z

b y x hay (Q): bx + 3y + 3bz 3b = 0

mt phng (Q) cvtpt Qn = (b;3;3b)Mt phng 0xy c vtpt k = (0;0;1) .Theo gi thit ,ta c

|cos ( Qn , k )| = cos60o 21

993

2 bb

b

26326999622 bbbbb

Vy c hai mt phng tho mn l :(Q1) : x 26y + 3z 3 = 0(Q2) : x + 26y + 3z 3 = 0

Cch 2:v A Ox v C OzGiAB l giao tuyn ca mt phng (Q) v mt phng 0xy .T O h OIAB .Theo nh l ba ng vung gc ta c ABCI 060OIC

Trong vung OIC ta c OI = OC.tanOIC = 1.tan60o =33

Trong vung OAB ta c 222 111 OBOAOI 232 131

331 OB

OB = 263

B1(0; 26 ;0) Oy hoc B2(0; 26 ;0) Oy .Vy c hai mt phng (Q) tho mn l

113

263

z y x hay (Q) : x 26 y + 3z 3 = 0


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Bi 4:Trong khng gian vi h to Oxyz . Vit phng tr nh mt phng i qua hai im

2;1;3 , 1; 2;1 M N v song song vi ng thng d c phng tr nh l:1

: 23 2

x t

d y t

z t

Gii: Cch 1:Mt phng i qua hai im 2;1;3 , 1; 2;1 M N v song song vi ng thng d

mt phng i qua im M v n MN ; n d u (vi MN v d u khng cng phng) mt phng i qua im M v c vtpt n = [ MN , d u ] = 10; 4;1 mt phng c phng tr nh l :

10(x 2) 4(y 1) + 1(z 3) = 0 hay : 10 4 19 0 x y z Cch 2: Gi s mt phng c dng : 2 2 20 0 Ax By Cz D A B C

mt phng c vtpt ; ;n A B C - Mt phng i qua 2;1;3 M .2 .1 .3 0 1 A B C D - Mt phng i qua 1; 2;1 N .1 . 2 .1 0 2 A B C D - Mt phng song song vi ng thngd . 0 .1 .2 . 2 0 3d n u A B C

T (1) v (2)ta c 1 3 1 7,2 2 2 2

C A B D A B th vo (3) ta c2 5 A B chn

1 195, 2 ,2 2

A B C D

Vy phng tr nh mt phng l 1 195 2 0 10 4 19 02 2 x y z x y z

Bi 5:Trong khng gian vi h trc ta Oxyz, cho cc im A(-1;1;0), B(0;0;-2) v C(1;1;1). Hy vit phng tr nh mt phng (P) qua hai im A v B, ng thi khong cch t C ti mt phng (P) bn

3 .Gii: Gi s mt phng P c dng : 2 2 20 0 Ax By Cz D A B C

mt phng P c vtpt ; ;Pn A B C - Mt phng P i qua 1;1;0 A . 1 .1 .0 0 1 A B C D - Mt phng P i qua 0;0; 2 B .0 .0 . 2 0 2 A B C D

T (1) v (2)ta c 1

,2C A B D A B Nn mt phng P c phng trnh l 1 02

Ax By A B z A B

Theo gi thit

2 22

2 2

172; 3 3 5 2 7 0 151

2

A B A B A B A A

d I P A AB B B B

A B A B


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Vi 1 A B

chn 1, 1 1, 2 : 2 0 A B C D P x y z

Vi 75

A B

chn 7, 5 1, 2 : 7 5 2 0 A B C D P x y z

Nhn xt: Gi Ocban );;( l vct php tuyn ca (P)

V (P) qua A(-1 ;1 ;0) pt : 1 1 0P a x b y cz

M (P) qua B(0;0;-2) 2 0 2a b c b a c

Ta c PT : 2 2 0P ax a c y cz c

2 22 2 22

; 3 3 2 16 14 07( 2 )

a c a cd C P a ac c

a ca a c c

TH 1: ca ta chn 1ca Pt ca : 2 0P x y z

TH 2: ca 7 ta chn a =7; c = 1 Pt ca : 7 5 2 0P x y z Bi 7: Trong khng gian vi h trc to Oxyz cho im A(1;0;1), B(2;1;2) v mt phng

: 2 3 3 0Q x y z . Lp phng tr nh mt phng (P) i qua A, B v vung gc vi (Q). HD: Ta c (1;1;1), (1;2;3), ; (1; 2;1)Q Q AB n AB n

V ; 0Q AB n

nn mt phng (P) nhn ; Q AB n

lm vc t php tuyn Vy (P) c phng tr nh x 2y + z 2 =0Bi 8:Trong khng gian ta Oxyz cho 2 im I( 0;0;1) v K( 3;0;0). Vit phng tr nh mt phng quaI, K v to vi mt phng (xOy) mt gc bng030 Gii:

Gi s mt phng cn c dng : ( ) : 1 ( , , 0) x y z a b ca b c

( ) 1 ( ) 3 ( ) : 13 1 x y z

Do I c v do K ab

000

0

.1 1 3 2( ; ;1) (0;0;1) cos303 2.

x y

x y

x y

n nn v n k b

b n n

( ) : 13 13 22

x y z

Bi 9:(H B 2009 )Trong khng gian vi h to Oxyz, cho t din ABCD c cc nh 1;2;1 , 2;1;3 , 2; 1;1 A B C v 0;3;1 D . Vit phng tr nh mt phng (P) i qua A, B sao cho

khong cch t C n mt phng (P) bng khong cch t D n mt phng (P) Gii: Cch 1:Gi s mt phng P c dng : 2 2 20 0ax by cz d a b c

mt phng P c vtpt ; ;Pn A B C


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- Mt phng P i qua 1;2;1 A .1 .2 .1 0 1a b c d - Mt phng P i qua 2;1;3 B . 2 .1 .3 0 2a b c d

T (1) v (2)ta c 3 1 5,2 2 2c a b d a b

Nn mt phng

P c phng tr nh l

3 1 5 02 2 2

ax by a b z a b

Theo gi thit , ,d C P d D P

2 2

2 2 2 2

3 1 5 5 3 1 5 5.2 . 1 .1 .0 .3 .12 2 2 2 2 2 2 2

3 1 3 12 2 2 22 4

32 0

a b a b a b a b a b a b

a b a b a b a b

a ba b a b

b

Vi 2 4a b chn 1

4, 2 7, 15 : 4 2 7 15 0a b c d P x y z

Vi 2 0b chn 2 23 5 3 50, 1 , : 0 : 2 3 5 02 2 2 2

b a c d P x z P x z

Cch 2:Xt hai trng hp TH1 : (P) // CD. Ta c :AB ( 3; 1;2),CD ( 2;4;0)

(P)c PVT n ( 8; 4; 14) hay n (4;2;7)(P) :4(x 1) 2(y 2) 7(z 1) 0

4x 2y 7z 15 0

TH2 : (P) quaI(1;1;1)l trung im CD Ta c AB ( 3; 1;2), AI (0; 1;0)

(P) c PVT n (2; 0; 3)(P) :2(x 1) 3(z 1) 0 2x 3z 5 0

p s: 1 : 4 2 7 15 0P x y z v 2 : 2 3 5 0P x z Bi tp t gii:

Bi 1:Trong khng gian vi h trc to Oxyz cho im 1;0;1 , 2;1;2 A B v mt phng : 2 3 3 0Q x y z . Lp phng tr nh mt phng (P) i qua A, B v vung gc vi (Q).

s: : 2 2 0P x y z Bi 2:Lp phng tr nh mp(P ) i qua 0;3;0 , 1; 1;1 M N v to vi mt phng : 5 0Q x y z

mt gc vi 1cos3

Bi 3:Lp ph ng trnh mt phng(P) i qua 1; 1;3 , 1;0;4 M N v to vi mt phng : 2 5 0Q x y z mt gc nh nht .

s: : 4 0P y z Bi 4:Vit phng tr nh mt phng i qua hai im 1;2;3 , 2; 2;4 M N v song song vi Oy.

(Ti liu n thi tt nghip nm 2009)


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13

s: : 2 0 x z Bi 5:Trong khng gian vi h to Oxyz cho mt phng : 2 3 7 0P x y z . Vit phng tr nhmt phng ( ) i qua 1;1;0 , 1;2;7 A B v vung gc vi (P).

(Ti liu n thi tt nghip nm 2009) s: :11 8 2 19 0 x y z

Dng 3:Vit phng tr nh m t phng (P) i qua ba im M o(x o ;y o ;z o ) M 1(x1 ;y 1 ;z1 ) v M 3(x 3 ;y 3 ;z 3 ) khng th ng h ng cho trcPhng php: Cch 1:- Tm hai vecto 0 1 0 2, M M M M

- Tm vtpt 0 1 0 2,n M M M M

- P l mt phng qua 0 M v c VTPTn Cch 2:

- Gi s phng tr nh mt phng P l 0 1 Ax By Cz D 2 2 2

( 0) A B C - V P i qua ba im 0 1, M M v 2 M thay ta vo phng tr nh (1)c h 3 n, 3 phng tr nhtheo , A B v C . Gii h ny ta c, A B v C - Thay vo phng tr nh (1) tac phng tr nh mt phng P

Bi tp gii mu:

Bi 1: (SGK Ban C Bn T80)Trong khng gian vi hto Oxyz .Vit phng tr nh mt phng i qua ba im M 3;0;0 ; 0; 2;0 N v 0;0; 1P

Gii:Cch 1:Mt phng i qua ba im 3;0;0 M ; 0; 2;0 N v 0;0; 1P mt phng i qua im Mv n MN ; n MP (vi MN v MP khng cng phng)

mt phng i qua im M v nhnvtpt n = [ MN , MP ] = (2;3;6)mt phng c phng tr nh l :

2(x + 3) + 3(y 0 ) + 6(z 0) = 0 hay : 2x + 3y + 6z + 6 = 0Cch 2:Gi s mt phng c dng 2 2 20 ( 0) Ax By Cz D A B C - Mt phng i qua M 3;0;0 . 1 .0 .0 0 1 A B C D - Mt phng i qua 0; 2;0 N .0 . 2 .0 0 2 A B C D - Mt phng i qua 0;0; 1P .0 .0 . 1 0 3 A B C D Gii h (1), (2) v (3) ta c A = 2, B = 3, C = 6 v D = 6 .Vy mt phng c phng tr nh l 2 3 6 6 0 x y z Cch 3:Nhn thy M 3;0;0 Ox ; N 0; 2;0 Oy vP 0;0; 1 Oz nn phng tr nh mt phng lmt phng theo on chn c dng :


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14

1123

z y x hay : 2 3 6 6 0 x y z

Dng 4: Vi t phng tr nh m t phng trung trc ca on MN , bit M v N c to cho trc Phng php: - Tnh ta trung im I ca MN v tnh MN - Mt phng trung trc ca on MN l mt phng i qua I v c vtpt Pn MN - Bit mt im v mt vtpt ta c phng tr nh mt phng cn t m

Bi tp gii mu:

Bi 1: (SGK Ban C Bn T80)Trong khng gian vi h to Oxyz .Vit phng tr nh trung trcca on thng AB vi A(2;3;7)v B(4;1;3)Gii:Cch 1:Gi I l trung im ca on thng ABI(3;2;5) .Mt phng trung trc(P) ca on AB i trung im

I ca A,B v vung gc vi on thng AB mt phng trung trc (P) ca on AB i qua I v nhn vect AB = 2; 2; 4 = 2 1; 1; 2 lm vtptmt phng trung trc(P)c phng tr nh l:

1(x 3) 1(y 2) 2(z 5 ) = 0 hay : 2 9 0P x y z Cch 2:(Phng php qu tch ) Mi im M(x;y;z) thuc mt phng trung trc(P) ca on AB MA = MB

2 2 2 2 2 22 2 2 3 7 4 1 3 MA MB x y z x y z 2 9 0 x y z

Cch 3:Mt phng trung trc(P) nhn AB lm vtpt lun c dng x y 2z + D = 0 ,v I mt phng trung

trc 3 2 2.5 + D D = 9 mt phng trung trc (P)c phng tr nh l : x y 2z + 9 = 0Cch 4:Mt phng trung trc (P) nhn AB lm vtpt lun c dng x y + 2z + D = 0 v mt phng (P) cch u

, , , A B d A P d B P

411'7.232 D

=411

'3.214 D 9'13' D D D = 9

Vy mt phng tr ung trc (P) c phng tr nh l: 2 9 0 x y z Nh n xt :- Bi ton ny thc cht l bi ton vit phng tr nh mt phng i qua mt im v vung gc gi camt vect (thuc dng 1)

- Vect i qua hai im cho trc c coi l mt vtcpca ng thngBi tp t gii:

Bi 1:Trong khng gian vi h to O xyz cho 2 im 1; 4;5 , 3;2;7 E F . Vit phng tr nh mt

phng ( ) l trung trc ca on thng EF. ( thi tt nghip THPT h phn ban ln 2 nm 2007)

s: : 3 5 0 x y z


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Dng 5: Vit phng tr nh m t phng (P) song song v cch u hai hai ng thng ( 1 ) v ( 2 )cho trcPhng php: - Mt phng (P) song song vi hai ng thng1 2v nn c vtpt 1 2;Pn u u

- mt phng (P) cch u hai ng thng1 2v nn (P) i qua trung im ca I ca MN vi1 2 M v N ...Quay v dng 4

Bi tp gii mu:

Bi 1: (HSP HN2 98)Trong khng gian Oxyz cho hai ng thng c phngtrnh l

d :t z

t y

t x

212

v d :03

022 y

z x

a. Chng minh rng d v d cho nhaub. Vit phng tr nh mt phng (P) song song v ng thi cch u d v dGii:a. Chn im M(2;1;0) d v d c vtcp 1; 1; 2d u ,chn im N(0;3;1) d v d c vtcp

' 2;0;1d u .Tnh n = [ d u , 'd u ] = 1; 5; 2 (vi d u v 'd u khng cng phng) v 2;2;1 MN . Xt . 1. 2 5.2 2.1 10 0n MN

d v d cho nhau

b. Gi I

21;2;1 l trung im ca M v N .Mt phng (P) song song v cch u d v d

mt phng (P) i qua I v c vtpt Pn = n mt phng (P) c phng tr nh l :

1(x 1) 5(y 2) 2

21

z = 0 hay (P) : x + 5y + 2x 12 = 0

Nh n xt : - Mt phng (P) song song v ng thi cch u d v d thc cht l mt phng trung trc ca on M vN nn c th p dng cc cch bi (dng4 )Bi 2:Vit phng tr nh mt phng cch u hai ng thng d1 v d2 bit:

1

2: 2

3

x t

d y t

z t

v 21 2 1:

2 1 5 x y z

d

Gii:

ng thng d2 c phng tr nh tham sl:1 2 '2 '1 5 '

x t

y t

z t

vect ch phng ca d1 v d2 l: 1 2(1;1; 1), (2;1;5)u u

VTPT ca mp( ) l

1 2. (6; 7; 1)d d n u u

pt mp( ) c dng 6x 7y z + D = 0

ng thng d1 v d2 ln lt i qua 2 M(2; 2; 3) v N(1; 2; 1)( , ( )) ( , ( )) |12 14 3 | | 6 14 1 |

| 5 | | 9 | 7d M d N D D

D D D


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Vy PT mp( ) l: 3x y 4z +7 0

Dng 6 : Vit phng tr nh m t phng (P) song song v cch u hai mt phng (Q1 ) v Q 2 (vi Q1 vQ 2 song song v i nhau) Ch :- S dng cng thc khong cch - Khong cch gia hai mt phng song song chnh l khong cch t mt im thuc mt phng ny timt phng kia

Bi tp gii mu:

Bi 1:Trong khng gian Oxyz cho hai mt phng (P) v (Q) c phng tr nh l(P) : 3x y + 4z + 2 = 0 v (Q) : 3x y + 4z + 8 = 0

Vit phng tr nh mt phng ( ) song song v cch u(P), (Q)Gii:V Pn = Qn = (3;-1;4) v 2 8 nn (P) // (Q), chn im M(0;2;0)(P) v im N(0;8;0)(Q)Mt phng song song vi (P) v (Q) lun c dng 3x y + 4z + D = 0, v cchu (P) v (Q)nn , ,d M d N

1619'0.420.3 D

=1619

'0.480.3 D 8'2' D D D = 4

Vy mt phng ( ) c phng tr nh l :3x y + 4z + 4 = 0

Dng 7: Vi t ph ng trnh m t phng (P) tip xc vi mt mt cu (S)v th a m n m t iu kin cho trc Phng php:- Bc 1: Xc nh tm I v bn knh R ca mt cu (S)v vtpthocvtcp- Bc 2:T iu kin cho trc xc nh vtptPn , gis ; ;Pn a b c khi mt phng P c dng

'

ax 0by cz D vi'

0 D (1)- Bc 3: Mt phng (P) tip xc vi mt cu (S) ,d I P R , t y c phng tr nh theo D,gii phng tr nh (ti tuyt i) c D thay vo (1) ta c phng tr nh mt phng P cn t m- Bc 4: Kt lun (thng c hai mt phng tha mn)Ch : iu kin cho trc l- Song song vi mt phng Q cho trc P Qn n

- Vung gc vi ng thng d cho trc P d n u

- Song song vi hai ng thng d1 v d2 cho trc 1 2,Pn u u

- Vung gc vi hai mt phng Q v R cho trc 1 2,Pn n n

- Song song vi ng thng d v vung gc vi mt phng Q cho trc ,P d Qn u n

Ch :Nu mt phng P tip xc vi mt cu(S) ti M S th mt phng P i qua im M v c VTPTl MI

Bi tp gii mu:


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17

Bi 1: (SGK Ban C Bn T93)Trong khng gian vi h to Oxyz. Vit phng tr nh mt phng

tip xcvi mt cu (S) c phng tr nh 2 2 2: 10 2 26 170 0S x y z x y z v song song vi hai ng thng

5 2

: 1 313 2

x t

d y t z t

7 3 '

: 1 2 '8

x t

d y t z

Gii:Ta c 2; 3; 2d u v ' 3; 2; 0d u .Mt cu (S) (x 5)2 + (y + 1)2 + (z + 13)2 = 25

mt cu (S) c tm 5; 1; 13 I v bn knh R = 5Mt phng song song vi d v d

mt phng c n d u ; n 'd u (vi d u v 'd u khng cng phng ) mt phng c vtpt n = [ d u , 'd u ] = (4;6;5)

mt phng lun c dng 4x + 6y + 5z + D = 0Mt phng ( ) tip xc vi mt cu (S) d(I,( )) = R

5253616

')13.(5)1.(65.4 D 77552' D D = 52 5 77

Vy c hai mt phng ( ) tha mn bi l : 1 : 4x + 6y + 5z + 51 + 577 = 0 2 : 4x + 6y + 5z + 51 577 = 0

Bi 2: (SBT Ban Nng Cao T138)Trong khng gian Oxyz. Vit phng tr nh mt phng (P) tip xcvi mt cu (S) v vung gc vi ng thng d c phng tr nh ln lt l :

(S): x2+ y

2+ z

2 10x + 2y + 26z 113 = 0 v

5 1 13: 2 3 2 x y z

d

Gii:ng thng d cvtcp 2; 3;2d u .Mt cu (S) (x 5)2 + (y + 1)2 + (z + 13)2 = 308

mt cu (S) c tm 5; 1; 13 I v bn knh 308 R Mt phng (P) vung gc vi ng thng d nn c vtpt 2; 3;2P d n u

mt phng (P) lun c dng 2x 3y + 2z + D = 0

Mt phng (P) tip xc vi mt cu (S) ,d I P R 2.( 5) 3.( 1) 2.( 13) '

308 ' 13 5236 ' 13 52364 9 4 D

D D

Vy c hai mt phng (P) tha mnu bi l :(P1): 2x 3y + 2z + 13 + 5236 = 0(P2): 2x 3y + 2z + 13 5236 = 0

Bi 3: (SGK Ban Nng Cao T90 HGTVT 1998 )Trong khng gian Oxyz. Vit phng tr nhmt phng (P) song song vi mt phng (Q) v tip xc vi mt cu (S) c phng tr nh ln lt l :

: 4 3 12 1 0Q x y z v 2 2 2: 2 4 6 2 0S x y z x y z Gii:


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Mt phng (Q) cvtpt 4;3; 12Qn .Mt cu (S) (x 1)2 + (y 2)2 + (z 3)2 = 16

mt cu (S) c tm I(1;2;3) v c bn kinh R = 4Mt phng (P) song song vi mt phng (Q)

mt phng (P) lun c dng 4x + 3y 12z + D = 0Mtphng (P) tip xc vi mt cu (S)

,d I P R

4.1 3.2 12.3 ' ' 264 ' 26 52

' 7816 9 144 D D

D D

Vy c hai mt phng tha mnu bi l :(P1): 4x + 3y 12z + 78 = 0(P2): 4x + 3y 12z 26 = 0

Bi 4: (Ti Liu n Thi Tt Nghip 2009)Trong khng gian vi h to Oxyz . Vit phng tr nhmt phng ( ) song song vi trcOz, vung gc vi mt phng(P): x + y + z = 0 v tip xc vi mtcu (S): x2 + y2 + z2 2x + 2y 4z 3 = 0Gii:Mt phng (P) c vtpt Pn = (1;1;1) .

Mt cu (S) (x 1)2 + (y + 1)2 + (z 2)2 = 9mt cu (S) c tm 1; 1; 2 I v c bn knh R = 3

Mt phng ( ) song song vi trcOz v vung gc vi mt phng (P)mt phng ( ) c n k ; n Pn (vi k v Pn khng cng phng ) mt phng ( ) c vtpt n = [k , Pn ] = 1; 1;0 mt phng ( ) lun c dng x y + D = 0

Mt phng ( ) tip xc vi mtcu (S) ,d I P R 1.1 1.( 1) '

3 ' 2 3 2 2 3 21 1

D D D

Vy c hai mt phng tho mnu bi l: 1 : x y 2 + 3 2 = 0 2 : x y 2 3 2 = 0

Bi 5: (SBT Ban Nng Cao T126)Trong khng gian vi h to O xyz cho mt cu(S): x2 + y2 + z2 6x 2y + 4z+ 5 = 0 v im M(4;3;0) .Vit phng tr nh mt phng (P) tip xc vimt cu (S) v i qua im M Gii:V M(4;3;0) (S) nn mt phng (P) i qua M v tip xc vi mt cu (S) l mt phng i qua M vnhn 1;2;2 IM lm vtpt vi 3;1; 2 I l tm ca mt cu (S)

mt phng (P) c phng tr nh l:

1(x 4) + 2(y 3) + 2(z 0 ) = 0 hay (P): x + 2y + 2z 10 = 0Bi 6:Trong khng gian vi h to Oxyz cho mt cu 2 2 2( ) : 2 6 4 2 0S x y z x y z . Vit phng tr nh mt phng (P) song song vi gi ca vc t (1;6;2)v , vung gc vi mtphng ( ) : 4 11 0 x y z v tip xc vi (S). Gii: Ta c mt cu (S) c tm 1; 3; 2 I v bn knh 4 R Vc t php tuyn ca( ) l (1;4;1)n V ( ) ( )P v song song vi gi cav nn nhn vc t (2; 1;2) pn n v

lm vtpt.


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Do : 2 2 0P x y z m

V (P) tip xc vi (S) nn21

( ,( )) 4 ( ,( )) 43

md I P d I P

m

Vy c hai mt phng: 1 : 2 2 21 0P x y z v 2 : 2 2 3 0P x y z

Bi tp t gii:

Bi 1:Vit phng tr nh mt phng tip xc vi mt cu

2 2 2: 2 1 1 9S x y z v vung gc vi ng thng1

: 1 21 3

x t

d y t

z t

s: 2 3 7 3 14 0 x y z v 2 3 7 3 14 0 x y z Bi 2:Trong khng gian vi h to Oxyz cho mt cu 2 2 2: 4 2 4 7 0S x y z x y z

v hai ng thng 4 0

:3 1 0

x y zd

x y z

v 1 2 :

1 2 2 x y z

d . Vit phng tr nh mt phng

l tip din ca (S) ng thi song song vi d v d.s: 4 7 12 2 0 x y z v 4 7 12 2 0 x y z Bi 3:Vit phng tr nh mt phng / / : 2 2 4 0P x y z v tip xc vi mt cu (S) c phng tr nh: 2 2 2 2 2 4 3 0 x y z x y z s: 2 2 17 0 x y z v 2 2 1 0 x y z Bi 4:Vit phng tr nh mt phng / / : 2 2 1 0P x y z v tip xc vi mt cu (S)c phng tr nh: 2 2 22 1 2 4. x y z s: 2 2 6 0 x y z v 2 2 6 0 x y z Bi 5:Vit phng tr nh mt phng tip xc vi mt cu

2 2 2: 2 2 4 3 0S x y z x y z v vung gc vi ng thng 1 2: 1 2 2 x y z

d

s: 2 2 6 0 x y z v 2 2 12 0 x y z

Bi 6:Vit phng tr nh mt phng song song vi 2 1: 1 3 1 x y z

d , vung gc vi

: 2 1 0P x y z v tip xc vi mt cu 2 2 2: 2 1 9S x y z s: 4 3 5 11 15 2 0 x y z v 4 3 5 11 15 2 0 x y z Bi 7:Trong khng gian vi h to Oxyz cho mt cu 2 2 2: 2 2 4 3 0S x y z x y z

v hai ng thng 2 2 0:2 0

x yd x z

v

' 1:1 1 1

x y zd

. Vit phng tr nh mt phng l tip

din ca (S) ng thi song song vi d v d.s: 3 3 2 0 y z v 3 3 2 0 y z Bi 8:Trong khng gian ta Oxyz, lp phng tr nh mt phng i qua hai im 0; 1;2 , A

1;0;3 B v tip xc vi mt cu S c phng tr nh: 2 2 2( 1) ( 2) ( 1) 2 x y z


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Dng 8: Vi t phng tr nh m t phng cha mt ng thng cho trc v tho m n iu kinLoi 1:Vit phng tr nh mt phng cha ng thng v vung gc vi mt phng Phng php:

1. Tm VTPT ca l n v VTCP ca l u

2. VTPT ca mt phng l: n n u

3. Lymt im M tr n 4. p dng cch vit phng tr nh mt phng i qua 1 im v c 1 VTPT

Ch :Thc cht y l bi ton vit phng tr nh mt phng i qua hai im v vung gc vi mt mtphngLoi 2:Vit phng tr nh mt phng cha ng thng v song song vi ( , cho nhau)Phng php:

1. Tm VTCP ca v l u v 'u 2. VTPT ca mt phng l: 'n u u

3. Ly mt im M tr n

4. p dng cch vit phng tr nh mt phng i qua 1 im v c 1 VTPTCh :Thc cht y l bi ton vit phng tr nh mt phng i qua hai im v song song vi mtng thngLoi 3: Vit phng tr nh mt phng cha ng thng v 1 im M Phng php:

1. Tm VTCP ca l u , ly 1 i m N tr n . Tnh ta MN 2. VTPT ca mt phng l: n u MN

3. p dng cch vit phng tr nh mt phng i qua 1 im v c 1 VTPTCh :Thc cht y l bi ton vit phng tr nh mt phng i qua ba im phn bit cho trc Loi 4:Vit phng tr nh mt phng cha ng thng v to vi mt phng (hoc ng

thngd ) mt gc Loi 5:Vit phng tr nh mt phng cha ng thng v cch mt im M khng thuc mt khong h

Bi tp gii mu:

Bi 1: (SBT Ban Nng Cao T125)Trong khng gian vi h to Oxyz .Vit phng tr nh mt phng P

a. i qua im 2;1; 1o M v qua giao tuyn ca hai mt phng Q v R c phng tr nh ln lt l: 4 0 x y z v 3 1 0 x y z

b. Qua giao tuyn ca hai mt phng : 3 2 0 x y z v : 4 5 0 x y ng thi vung gcvi mt phng : 2 7 0 x z Gii: a. Cch 1: Gi l giao tuyn ca (Q) v (R) c phng tr nh

:01304

z y x

z y x


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21

chn hai im 3 11; ;02 2

M

v 3 11;0;

2 2 N

Mt phng (P) i qua giao tuyn ca (Q) v (P) mt phng (P) cha giao tuyn mt phng (P) i qua ba im Mo; M v N

(P) i qua im Mo v c vtpt Pn = [ 0 M M , 0 M N ] = 7;7;154

11

4

77;

4

77;

4

165

(vi

0 M M v 0 M N khng cng phng ) mt phng (P) c phng tr nh l :

15(x 2) 7(y 1) + 7(z + 1) = 0 hay : 15 7 7 16 0P x y z Cch 2:Gi l giao tuyn ca Q v R c phng tr nh

4 0:

3 1 0 x y z

x y z

Chn hai im 3 11; ;02 2

M

v 3 11;0;2 2

N

Gi s mt phng P c dng : 2 2 20 0 Ax By Cz D A B C mt phng P c vtpt ; ;Pn A B C

- Mt phng P i qua 3 11; ;02 2 M

3 11. . .0 0 12 2

A B C D

- Mt phng P i qua 3 11;0;2 2 N

3 11. .0 . 0 22 2

A B C D

- Mt phng P i qua 2;1; 1o M .2 .1 . 1 0 3 A B C D Gii h (1), (2) v (3) ta c 15, 7, 7, 16 : 15 7 7 16 0 A B C D P x y z Cch 3:S dng phngphp chm (tham kho phn sau) Nhn xt: Thc cht bi ton ny chnh l bi ton ny chnh l bi ton vit phng trnh mt phng iqua ba im (trong hai im cn li thuc giao tuyn ca hai mt phng)b. Cch 1:Gi l giao tuyn ca v ( ) c phng tr nh

:054

023 y x

z y x

Chn hai im M 5;0; 13 v N(1;1;0) Mt phng (P)i qua giao tuyn ca hai mt phng( ), v vung gc vi mt phng

mt phng (P) cha giao tuyn v vung gc vi mt phng

mt phng (P) i qua im M v c vtpt Pn = [ MN , n ] = 1;22; 2 mt phng (P) c phng tr nh l :

1(x 5) + 22(y 0 ) 2(z + 13) = 0 hay (P) : x 22y + 2z + 21 = 0Hocc th tnh ,Pn u n

Nh n xt : Thccht bi ton ny chnh l bi ton ny chnh l bi ton vit phng tr nh mt phng i qua haiim v vung gc vi mt mt phng (trong hai im cn li thuc giao tuyn ca hai mt phng)Cch 2:. Gi l giao tuyn ca v c phng tr nh


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22

:054

023 y x

z y x

Chn hai im 5; 0; 13 M v 1;1;0 N Gi s mt phng P c dng : 2 2 20 0 Ax By Cz D A B C

mt phng

P c vtpt ; ;

Pn A B C

- Mt phng P i qua 5; 0; 13 M .5 .0 . 13 0 1 A B C D - Mt phng P i qua 1;1;0 N .1 .1 .0 0 2 A B C D

T (1) v (2)ta c 413

A BC v D A B

Nn mt phng P c vtpt 4; ; 13P A B

n A B

Mt phng c vtpt 2;0; 1n , mt phng P vung gc vi

4. .2 .0 . 1 0 2213P A B

n n A B A B

chn 1, 22 2, 21 A B C D

Vy mt phng P c phng tr nh l 22 2 21 0 x y z Cch 3:S dng phng php chm (tham kho phn sau) Bi 2: (H A 2002)Trong khng gian vi h to vung gc Oxyz cho hai ng thng

12 4 0

:2 2 4 0

x y z

x y z

21

: 21 2

x t

y t

z t

Vit phng tr nh mt phng P cha ng thng1 v song song vi ng thng2 Gii:Cch 1:Chn M 0; 2;0 1 v 1 c vtcp 1u = (2;3;4), 2 c vtcp 2u = (1;1;2)Mt phng (P) cha ng thng1 v song song vi ng thng 2

mt phng (P) i qua im M v c vtpt Pn = [ 1u , 2u ] = (2;0;-1)mt phng (P) c phng tr nh l :

2(x 0 ) + 0(y + 2) 1(z 0 ) = 0 hay2 0 x z HocC th tnh vtpt l 2,Pn MN u

vi 1, M N Cch 2:

Chn hai im 4 8;0;3 3

M

v 0; 2;0 N 1

Gis mt phng P c dng : 2 2 20 0 Ax By Cz D A B C mt phng P c vtpt ; ;Pn A B C

- Mt phng P i qua 4 8;0;3 3 M

4 8. .0 . 0 13 3 A B C D

- Mt phng P i qua 0; 2;0 N .0 . 2 .0 0 2 A B C D

T (1) v (2)ta c 1 32 4

C A B v 2 D B

Nn mt phng P c vtpt 1 3; ; 2 4Pn A B A B


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23

ng thng 2 c vtcp 2 1;1;2u , mt phng P song song vi ng thng2

21 3. .1 .1 .2 0 5 02 4P

n u A B A B B

chn 11, 0 , 02

A B C D

Vy mt phng P c phng tr nh l 1 0 2 02 x z x z

Cch 3:S dng phng php chm (tham kho phn sau) Bi 4:Trong khng gian vi h to Oxyz. Vit phng tr nh mt phng P i qua giao tuyn ca haimt phng : 3 0 x y z v : 3 5 1 0 x y z ng thi song song vi mtphng : 2 3 0 x y z Gii: Cch 1:Gi l giao tuyn ca ( ) v ( ) c phng tr nh

:0153

03 z y x

z y x

Chn M 34;34;3 Mt phng (P) i qua giao tuyn ca ( ) v ( ) ng thi song song vi mt phng ( )

mt phng (P) cha giao tuynv song song vi mt phng ( )mt phng (P) i qua im M v lun c dng: x + y + 2z + D = 0

P i qua im M nn 3 +

34 + 2

34 + D = 0 D = 1

Vy mt phng (P) c phng tr nh x + y + 2z + 1 = 0Hoc:Mt phng (P) i qua im M v c vtpt ,Pn u n

Hoc:Mt phng (P) i qua im M v c vtpt ,Pn MN n

vi , M N Cch 2:Gi l giao tuyn ca v c phng tr nh

3 0:

3 5 1 0 x y z

x y z

Gi s mt phng P c dng : 2 2 20 0 Ax By Cz D A B C mt phng P c vtpt ; ;Pn A B C

Chn hai im 1 7;0; 4 M v 2 1; 2;0 M - Mt phng P i qua 1 7;0; 4 M .7 .0 . 4 0 1 A B C D

- Mt phng P i qua 2 1; 2;0 M .1 . 2 .0 0 2 A B C D T (1) v (2) ta c 3

2 B A

C v 2 D B A

Nn mt phng P c vtpt 3; ; 2P B A

n A B

Mt phng c vtpt 1;1;2n , mt phng P song song vi

Pn v n cng phng 2.23

11 A B B A chn 1, 1 2, 1 A B C D


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Vy mt phng P c phng tr nh l 2 1 0 x y z Cch 3:S dng phng php chm (tham kho phn sau)

Nh n xt : Thc cht bi ton ny chnh l bi ton ny chnh l bi ton vit phng tr nh mt phng i qua mtim v song song mt mt phng (trong mt im cn li thuc giao tuyn ca hai mt phng)Bi 5:Trong khng gian Oxyz cho hai im 1;3; 2 , 3;7; 18 A B v mt phng

: 2 1 0P x y z . Vit phng tr nh mt phng cha AB v vung gc vi mp (P). Gii: Gi (Q) l mt phng cn t mTa c ( 2, 4, 16) AB cng ph ng vi ( 1,2, 8)a Mt phng(P) c vtpt 1 (2; 1;1)n Mt phng (Q) cha ng thng AB v vung gc vi mt phng (P) nn cvtpt , 6;15;3 3 2;5;1Qn n a

Chnvtpt ca mt phng (Q) l 2 (2,5,1)n

Mp(Q) cha AB v vung gc vi (P) i qua A nhn 2 (2,5,1)n l vtpt c phng tr nh l:2(x + 1) + 5(y 3) + 1(z + 2) = 0 2x + 5y + z 11 = 0

Bi 6:Trong khng gian vi h ta Oxyz cho hai ng thng d v d ln lt c phng tr nh l:

d : z y x12 v d :

153

22 z

y x .

Vit phng tr nh mt phng )( i qua d v to vid mt gc 030 Gii: - ng thng d i qua im )0;2;0( M v c vtcp (1; 1;1)u - ng thng d i qua im )5;3;2(' M v c vtcp '(2;1; 1)u

Gi s mt phng )( c vtpt ( ; ; )n A B C Mt phng )( phi i qua im M v c vtptn vung gc vi v u

ng thi to vi ng thngd mt gc 030 tc l2160cos)';cos( 0un

Ta c h

21

62

0

222 C B A

C B A

C B A

02)(632 22222 C AC A

C A B

C C A A A

C A B

Gii phng tr nh 2 22 0 ( )(2 ) 02 A C

A AC C A C A C A C

.

- Nu C A , chn 1 A C , khi 2 B , tc l (1;2;1)n v mt phng( ) c phng tr nh l0)2(2 z y x hay 2 4 0 x y z

- Nu C A2 , chn 2,1 C A , khi 1 B , tc l (1; 1; 2)n v mt phng( ) c phng tr nhl 02)2( z y x hay 2 2 0 x y z

Bi 7:Trong khng gian vi h ta Oxyz cho hai ng thng 11 1:

2 1 1 x y z

d v

22 1:

1 1 1 x y z

d . Vit phng tr nh mt phng cha 1d v hp vi 2d mt gc 300


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Gii: Gi s mt phng P c dng : 2 2 20 0 Ax By Cz D A B C

mt phng P c vtpt ; ;Pn A B C Trn ng thng1d ly 2 im 1;0; 1 , 1;1;0 M N

Do P qua , M N nn:0 20

A C D C A B

A B D D A B

Nn ( ) : (2 ) 0P Ax By A B z A B .

Theo gi thit ta c 02 2 2 2 2 2

1. 1. 1.(2 )1 sin302 1 ( 1) 1 . (2 )

A B A B

A B A B

2 2 2 22 3 2 3(5 4 2 ) 21 36 10 0 A B A AB B A AB B

D thy 0 B nn chn 1 B , suy ra: 18 11421

A

Vy c 2 mt phng tha mn: 18 114 15 2 114 3 114 021 21 21

x y z .

Bi 8:Trong khng gian ta Oxyz cho 2 ng thng c phng tr nh:

1 22 3 5 0 2 2 3 17 0

: :2 0 2 2 3 0

x y z x y zd v d

x y z x y z

Lp phng tr nh mt phng i qua1d v song song vi 2d .Gii: Gi (Q) l mt phng cn t mng thng1d v 2d c vtcp ln lt l 1 2(1; 1; 1); (1; 2;2)u u

Mt phng (Q) i qua1d v song song vi 2d nn c vtpt l 1 2, ( 4; 3; 1) 1(4;3;1)Qn u u

Chn (4;3;1)Q

n v1

(2; 1;0) I d Mt khc: 2(0; 25;11) J d ta thy (0; 25;11) J Q Vy mt phng (Q) c phng tr nh l( ) : 4( 2) 3( 1) 0 ( ) : 4 3 5 0Q x y z hay Q x y z

Bi 9:Cho ng thng 2 0:1 0

yd

z

. Vi

t phng tr nh mt phng P i qua d v to vi mt phng

Oxy mt gc 045s:Mt phng : 2 1 0P y z

Bi 10:Vit phng tr nh mt phng i qua ng thng 2 0:1 0

x y zd

x y

v cch im 0;0;2 M

mt khong 12

h

s:C hai mt phng tha mn l 1 0, 5 4 3 1 0 x y x y z

Bi tp t gii:

Bi 1:Trong khng gian Oxyz cho hai im 1;3; 2 , 3;7; 18 A B v mtphng : 2 1 0P x y z . Vit phng tr nh mt phng cha AB v vung gc vi mp (P).


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s:2 5 11 0 x y z

Bi 2:Vit phng tr nh mt phng (P) cha ng thng8 11 8 30 0:2 0

x y zd

x y z

v c khong cch

n im 1; 3; 2 A bng 29 Bi 3:Lp phng tr nh mt phng(P) cha trcOz v to vi mt phng :2 5 0Q x y z mt gc

060 s: 1 2: 3 0 ; :3 0P x y P x y

Bi 4:Vit phng tr nh mt phng () cha 1 1 2:2 1 5

x y zd

sao cho khong cch t 5;1;6 A

n () ln nht. s: : 2 1 0 x y z Bi 5:Vit phng tr nh mt phng (P ) i qua A( 1; 2; 2) v B(1; 6 ; 4) bit khong ccht M (3 ;2; 1) n mt phng (P) bng 11

Bi 6:Trong khng gian vi h to O xyz cho ng thng d c phng tr nh3

12

11

2 z y x v

mt phng : 3 2 0P x y z . Vit phng tr nh mt phng cha d v vung gc vi (P).

( thi tt nghip THPT nm 2007) s: : 3 5 0 x z

Dng 9: Vi t phng tr nh m t phng cha hai ng thng 1 v 2 ct nhau hoc song song vi nhau Loi 1:Vit phng tr nh mt phng cha 2 ng thng ct nhauv Phng php:

1. Tm VTCP ca v l u v 'u 2. VTPT ca mt phng l: 'n u u 3. Ly mt im M tr n 4. p dng cch vit phng trnh mt phng i qua 1 im v c 1 VTPT

Loi 2:Vit phng tr nh mt phng cha 2 song song v Phng php gii

1. Tm VTCP ca v l u v 'u , ly , ' M N 2. VTPT ca mt phng l: n u MN

3.p dng cch vit phng tr nh mt phng i qua 1 imv c 1 VTPT

Bi tp gii mu: Bi 1: (H D 2005)Trong khng gian vi h to Oxyz cho hai ng thng

11 2 1:

3 1 2 x y z

d v 2

2 0:

3 12 0 x y z

d x y

. Chng minh d1 v d2 song song vi nhau.Vit

phng tr nh mt phng (P) cha c hai ng thng d1 v d2 Gii:- Chng minh d1 v d2 song song v i nhau ,ta cd1 i qua im M(1;-2;-1) v c vtcp 1u = (3;-1;2)


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27

d2 c vtcp 2u = (3;-1;2) = 1u v M1 d2 vy d1 // d2 - Vit phng tr nh mt phng (P) cha c d1 v d2 Cch 1:chn hai im N(-3;5;0) v Q(12;0;10) d2 . Mt phng (P) cha d1 // d2 mt phng (P) i qua ba im M,N v Q mt phng (P) i qua im M v c vtpt ,Pn MN MQ

75;55; 85 5 15;11; 17 (vi MN v MQ khng cng phng ) mt phng (P) c phng tr nh l :

15(x 1) + 11(y + 2) 17(z + 1) = 0 hay (P) : 15x + 11y 17z 10 = 0Cch 2:chn N(-3;5;0) d2 .Mt phng (P) cha d1 // d2 mt phng (P) i qua im M v c vtpt Pn =[ 1u , MN ] = (15;11;-17) (vi 1u v MN khng cng phng)

mt phng (P) c phng tr nh l :15(x 1) + 11(y + 2) 17(z + 1) = 0 hay (P) : 15x + 11y 17z 10 = 0

Cch 3:S dng phng php chm (tham kho) Mt phng (P) cha d1 // d2 mt phng (P) i qua im v cha d2 Mt phng (P) chm mt phng xc nh bi d2 c dng(x + y z 2) +(x + 3y 12) = 0 (2 + 2 0)

( + )x +( + 3)y z 2 12 = 0 v M (P)( + ).1 + ( + 3)(-2) (-1) 2 12 = 0 - 2 - 17 = 0 chn = 17 v = -2

Vy mtphng (P) c phng tr nh l : 15x + 11y 17z 10 = 0Bi 2: (SBT Ban C Bn T115)Trong khng gian Oxyz cho hai ng thng

d1 : 45

32

21 z y x v d2 :

t z

t y

t x

212237

a. Chng minh rng d1 v d2 cng nmtrong mt mt phng ()b. Vit phng tr nh mt phng ( )

Gii:a. Chn M1(1;-2;5) d1 v d1 c vtcpn = (2;-3;4) ,chn M2(7;2;1) d2 v d2 c vtcp 2u = (3;2;-2). Tnhn = [ 1u , 2u ] = (-2;16;13) v 1 2 M M = (6;4;-4)Xt n . 1 2 M M = (-2).6 + 16.4 +13.(-4) = 0

d1 v d2 cng nm tr n mt phng ( ) ,mt khc ta c 1u 2u d1 v d2 ct nhaub. Cch 1: Mt phng ( ) cha d1 v d2 mt phng( ) i qua M1 v c vtptn = n

mt phng ( ) c phng tr nh l :- 2(x 1) + 16(y + 2) + 13(z 5) = 0 hay ( ) : 2x 16y 13z + 31 = 0

Cch 2:

C th ly vtptn = [ 1u , 1 2 M M ] = (-4;32:26) = -2(2;-16;-13)Mt phng ( ) i qua M v c vtptn mt phng ( ) c phng tr nh l :2x 16y 13z + 31 = 0

Cch 3:

Chuyn d1 v dng tng qut l d1 : 07340123

z y

y x chn I

37;0;

31 d1 .Mt phng ( ) cha d1 v

d2 l mt phng i qua ba im M1,M2 v Ihng dn :Lm tng t nh dng

Cch 4:S dng phng php chm (tham kho) Mt phng ( ) cha d1 v d2 mt phng ( ) cha d1 v i qua im M2 d2


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hng dn :Lm tng t nh bi 1(cch 3) dng Nh n xt :Bi 3:Trong khng gian ta Oxyz cho 2 ng thng c phng tr nh:

1 2

5 27 0

: 1 :2 3 16 0

5

x t x y z

d y t v d x y z

z t

Vit phngtrnh mt phng cha1 2d v d Gii: Gi s mt phng cn lp l (Q) ta c:

1

1 2

( ) ( )

(5;1;5) ; (5;2;0) (0;1; 5)

. (0;1; 5) ( ) : 3( 5) 5( 1) 5 0

( ) : 3 5 25 0Q d

M d N d MN

v n u MN Q x y z

hay Q x y z

Dng 10: Vit phng tr nh mt phng P song song vi mt phng Q v cch mt phng Q mt khong h

Ch :S dng cng thc khong cch t mt im n mt mt phng

Bi tp gii mu:

Bi 1:Vit phng tr nh mt phng P song song vi mt phng : 2 0Q x y z v cch n mtkhong 3h Gii: Mt phng P song song vi mt phng Q c dng 0 x y z C Ly im 2;0;0 M Q .

Ta c 2 1, , 3 53m m

d P Q d M Pm

Vy phng trnh mt phng Q cn t m l 1 0; 5 0 x y z x y z

Bi tp t gii:

Bi 1:Vit phng tr nh mt phng / / P Q v cch P mt khong h trong cc trng hp sau: a. Mt phng P c phng tr nh 2 2 0 x y z v 3h b. Mt phng P c phng tr nh 4 0 x v 2h

s: a. 2 2 9 0;2 2 9 0 x y z x y z b. 2 0; 6 0 x x

Dng 11: Vit phng tr nh mt phng i qua im v to vi hai ng thng1 2d v d (hoc haimt phng 1 2P v P ) cc gc v Ch :S dng cng thc gc gia hai mt phng


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29

Bi tp gii mu:

Bi 1: Vit phng tr nh mt phng P i qua im 1;2;3 M v to vi mt phngOx, Oy cc gctng ng l 0 045 , 30 Gii:

Gi ; ;n A B C l vtpt ca mt phng P . Cc vtcp ca trc Ox v Oy l 1;0;0i v 0;1;0 j .Theo gi thit ta c h

02 2 2

2 2 20

2 2 2

1sin45222

1sin302

A

A B A B A B C B C B A B C

A B C

Chn 1 B ta c 2, 1 A C Vy phng tr nh mt phng P i qua im 1;2;3 M l

2 1 2 3 0; 2 1 2 3 0 x y z x y z Bi 2:Cho mt phng P c phng tr nh 2 0 x y z v im 2; 3;1 M . Vit phng tr nh mtphng Q i qua M vung gc vi mt phng v to vi mt phng mt gc045Gii: Gi ; ;n A B C l vtpt ca mt phng Q . Theo gi thit ta c h phng tr nh

2 2 2

2 012

A B C

A

A B C

. Gii h trn ta c 1;1;0 , 5; 3;4n n

Vy phng tr nh mt phng Q i qua im 2; 3;1 M l

1 0 x y hoc 5 2 3 3 4 1 0 x y z

Bi tp t gii:

Bi 1:Vit phng tr nh mt phng mt Q trong cc trng hp sau: a. Vung gc vi cc mt phng 1 2,P P c phng tr nh ln lt l 2 0; 1 0 x z x y z v iqua gc ta b. To vi ccmt phng 1 2,P P c phng tr nh ln lt l 2 0; 1 0 x z x y z cc gc

tng ng l 045 v gc vi 1cos3

ng thi i qua gc ta

s: a. 0 x z b. 0; 0 x z

Bi tpgii mutng hp:


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Bi 1: Trong khng gian vi h ta Oxyz cho im 10; 2; 1 A v ng thng d c phng

trnh1 2

1 3

x t

y t

z t

. Lp phng tr nh mt phng (P) i qua A, song song vi d v khong cch t d ti (P) l

ln nht. Gii: Gi H l hnh chiu ca A tr n d, mt phng (P) i qua A v / / P d , khi khong cch gia d v (P) lkhong cch t H n (P). Gi s im I l hnh chiu ca H ln (P), ta c HI AH HI ln nht khi I A Vy (P) cn t m l mt phng i qua A v nhn AH lm vc t php tuyn.

)31;;21( t t t H d H v H l hnh chiu ca A tr n d nn )3;1;2((0. uu AH d AH l vc t ch phng ca d) )5;1;7()4;1;3( AH H Vy (P): : 7 10 2 5 1 0P x y z

7 5 77 0 x y z

Bi 2:Cho ng thng D c phng tr nh:2

: 22 2

x t D y t

z t

.Gi l ng thng qua im 4;0; 1 A

song song vid v 2;0;2 I l hnh chiu vung gc ca A tr n D . Trong cc mt phng qua, hy vit phng tr nh camt phng c khong cch nl ln nht. Gii: Gi (P) l mt phng i qua ng thng, th ( ) //( )P D hoc( )P D . Gi H l hnh chiu vung gc ca I tr n (P). Ta lun c IH IA v IH AH .

Mt khc , ,d D P d I P IH

H P

Trong mt phng P , IH IA;do max IH IA H A . Lc ny (P) v tr (P0) vunggc vi IA ti A. Vect php tuyn ca (P0) l 6;0; 3n IA

, cng phng vi 2;0; 1v .

Phng tr nh ca mt phng (P0) l: 2 4 1. 1 2 9 0 x z x z .

Bi 3:Trong khng gian vi h trc ta Oxyz, cho ng thng 1 3:1 1 4

x y z v im

0; 2;0 . M Vit phng tr nh mt phng (P) i qua im M song song vi ng thngng thikhong cch gia ng thngv mt phng (P) bng 4. Gii: Gi s ( ; ; )n a b c l mt vect php tuyn ca mt phng (P). Phng tr nh mt phng : 2 0P ax by cz b .ng thng i qua im A(1; 3; 0) v c mt vect ch phng (1;1;4)u


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T gi thit ta c2 2 2

. 4 0 / /( ) (1)| 5 | 4( ;( )) 4 (2)

n u a b c

Pa b

d A Pa b c

Th 4b a c vo (2) ta c2 2 2 2 2

4

( 5 ) (2 17 8 ) 2 8 0 2

ac

a c a c ac a ac c ac

Vi 4ac

chn 4, 1 8a c b . Phng tr nh mt phng : 4 8 16 0.P x y z

Vi 2ac

chn 2, 1 2a c b . Phng tr nh mt phng : 2 2 4 0.P x y z

Bi 4:Trong khng gian vi h ta Oxyz, cho mt phng (P) v ng thngd ln lt c phng

trnh: : 2 2 2 0P x y z v 1 2: 1 2 1 x y z

d

1. Vit phng tr nh mt cu c tm thuc ng thngd , cch mt phng (P) mt khong bng 2 v vt

mt phng (P) theo giao tuyn l ng tr n c bn knh bng 3. 2. Vit phng tr nh mt phng (Q) cha ng thngd v to vi mt phng (P) mt gc nh nht.

Gii:

1. ng thng c phng tr nh tham s l: : 1 2 ;2

x t

y t t R

z t

Gi tm mt cu l I . Gi s ( ; 1 2 ;2 ) I t t t V tm mt cu cch mt phng (P) mt khong bng 3 nn:

2| 2 1 2 4 2 2 | | 6 5 | 3

( ; ) 3 73 33

t t t t t

d I t

C hai tm mt cu: 2 1 8 7 17 1; ; ; ; ;3 3 3 3 3 7

I I

V mt phng (P) ct mt cu theo ng tr n c bn knh bng 4 nn mt cu c bn knh l R = 5.Vy phng tr nh mt cu cn t m l:

2 2 2 2 2 22 1 8 7 17 125 ; 253 3 3 3 3 3

x y z x y z

2. ng thng c VTCP ( 1;2;1)u c phng tr nh tng qut l:2 1 0

:2 0

x y

x z

Mt phng (P) c VTPT (2; 1; 2)n

Gc gia ng thng () v mt phng (P) l: | 2 2 2 | 6sin33. 6

Gc gia mt phng (Q) v mt phng (Q) cn t m l 6 3cos 19 3

Gi s (Q) i qua c dng:m(2 x + y + 1) +n( x + z 2) = 0 (m2+ n2 > 0)(2m + n) x + my + nz + m 2n = 0


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Vy gc gia (P) v (Q) l:2 2

| 3 | 3cos33. 5 2 4

m

m n mn

m2 + 2mn + n2 = 0 (m + n)2 = 0 m = n.Chnm = 1,n = 1, ta c: mt phng (Q) l: x + y z + 3 = 0Bi 5 :Trong khng gian vi h ta Oxyz, cho 1;2;3 . M Lp phng tr nh mt phng i qua Mct ba tia Ox ti A, Oy ti B, Oz ti C sao cho th tch t din OABC nh nht. Gii:

Mt phng ct 3 tia Ox,Oy, Oz ti A(a;0;0),B(0;b;0), C(0;0;c) c dng : 1, , , 0 x y z a b ca b c

Do M nn:cos

31 2 3 61 3. 162

y

abca b c abc

Th tch: min3

1 27 27 66

9

a

V abc V b

c

Mt phng cn t m

: 6 3 2 18 0 x y z Bi 7:Vit phng tr nh mt phng (P) qua O, vung gc vi mt phng : 0Q x y z vcch im 1; 2; 1 M mt khong bng2 Gii: Phng tr nh mt phng (P) qua O nn c dng : Ax + By + Cz = 0 vi 2 2 2A B C 0 V (P) (Q) nn1. 1. 1. 0 0 A B C A B C C A B (1)Theo :

2 2 2 22 2 22

; 2 2 ( 2 ) 2( ) A B C d M P A B C A B C A B C

(2)

Thay (1) vo (2) , ta c : 2 08AB 5 0 8B =5

B B A

(1)0 . B C A Chn 1, 1 A C th : 0P x z 8B =5 A . Chn (1)A 5 , B 1 3C th : 5 8 3 0P x y z

Bi 8:Trong khng gian Oxyz, cho mt cu (S) v mt phng (P) c ph ng trnh(S): 2 2 2 4 4 2 16 0 x y z x y z ; ( ) : 2 2 1 0.P x y z Vit ph ng trnh mt phng (Q) songsong vi (P) v khong cch t tm mt cu (S) n mt phng(Q) bng 3. Gii:

(S):2 2 2

4 4 2 16 0 x y z x y z (S) c tm I(2;2;-1)ph ng trnh mt phng (Q) cdng:2 2 0 x y z D iu kin 1(*) D

( , ( )) 3d I P2 2 2

| 2.2 1.2 2( 1) | 32 1 ( 2)

D

1| 8 | 9

17 D

D D

Kt hp vi iu kin (*) ta c D =-17


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Vy ph ng trnh ca : 2 2 17 0Q x y z Bi 9: (H D 2010)Trong khng gianto Oxyz, cho hai mt phng (P): x + y + z 3 = 0 v (Q): x

y + z 1 = 0. Vit phngtrnh mt phng (R) vung gc v i (P) v (Q) sao chokhong cch tO n(R) bng 2. Gii:

PVT (1;1;1)Pn ; PVT (1; 1;1)Qm ; PVT (2;0; 2) 2(1;0; 1) Rk n m

Phngtrnh (R)c dng : x z + D = 0. Ta c : d (0;(R)) = 2 2 2 2

2 D

D

Phngtrnh (R) : 2 2 0 2 2 0 x z hay x z Bi 10:Trong khng gian Oxyz, cho mt cu (S) c phngtrnh 2 2 21 2 3 14 x y z vim 1; 3; 2 M . Lp phngtrnh mt phng (P)i qua sao cho(P) ct (S) theo mt giao tuyn l ng trn c bn knhnhnht.Gii: Ta thy M thuc min trongca (S) v (S) c tm 1; 2; 3 , 14 I R . Do ,

(P) qua M ct (S) theo mt giao tuyn l ng trn c bn knhnhnht2 2 R IH nhnht (H l hnh chiu vung gcca I trn mt phng (P)) IH l n nht

0;1; 1 M H IM l VTPTca (P). Vy (P) c phngtrnh l 1 0. y z Bi 11:Trong khng gian Oxyz, vit phng tr nh mt phng (P) cha ng

thng 2 0:2 6 0 x y

d x z

sao cho giao tuynca mt phng (P) v mt

cu 2 2 2: 2 2 2 1 0S x y z x y z l ng tr n c bn knh r = 1.Gii: Mt phng (P) cha d c dng: m(x y 2) + n(2x z 6) = 0

( ) : ( 2 ) 2 6 0P m n x my nz m n - Mt cu (S) c tm I(-1; 1; -1), bn knh R = 2.- (P) ct (S) theo mt ng tr n giao tip (C) c bn knh r = 1

2 2( ; ) 3d I P R r 2 2

2 2 2

2 2 63 4 7 3. 2 5 4 .

( 2 )m n m n m n

m n m n m nm n m n

2 25 22 . 17 0m m n n

Cho 2 171 5 22 17 0 15

n m m m hay m

Vy, c 2 mt phng (P): 12

( ) : 4 0

( ) : 7 17 5 4 0

P x y z

P x y z

Bi 12:Trong khng gian Oxyz cho mt phng : 2 5 0d x y z . Vit phng tr nh mt phng (P)qua giao tuyn cad v mt phng Oxyv (P) to vi 3 mt phng ta mt t din c th t

bng12536

.

Gii: Phng tr nh mt phng Oxy: z = 0Phng tr nh mtphng (P) thuc chm xc nh bid v Oxy c dng:

2 5 0m x y z nz : 2 5 0P mx my m n z m


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34

- Giao im A, B, C ca (P) v 3 trc Ox, Oy, Oz ln lt c ta :5 5; 0; 0 , (0; 5; 0), 0; 0;2

m A B C

m n

- Th tch t din OABC bng125 1 1 5 5 125. . . . .5.36 6 6 2 36

mV OA OB OC

m n

3 1, 23 3 1, 4m n m m nm n mm n m m n

Vy, c 2 phng tr nh mt phng (P):1

2

( ) : 2 3 5 0 ( 1; 2)( ) : 2 3 5 0 ( 1; 4)P x y z m n

P x y z m n

Bi 13:Trong khng gian ta Oxyz cho im G(1;1;1) a. Vit phng tr nh mt phng (P) qua G v vung gc vi OG b. Mt phng (P) cu (1) ct cc trc Ox, Oy, Oz ln lt ti A,B,C. Chng minh rng : ABC l tam gic u.Gii:

a. Do ( )( ) nn (1;1;1;)POG P n OG

( ) :1( 1) 1( 1) 1( 1) 0 hay ( ) : 3 0P x y z P x y z

b. V trc 0Ox : (3;0;0)0

y A

z

Tng t: (0;3;0) (0;3;0) B v C Ta c: 3 2 AB BC CA ABC l tam gic u Bi 14: (SGK Ban Nng Cao T89)Trong khng gian Oxyz . Vit phng tr nh mt phng (P)i qua im G(1;2;3) v ct cc trc to Ox,Oy, Oz ln lt ti A,B, C sao cho G l trng tm catam gic ABCi qua im H(2;1;1) v ct cc trc to Ox,Oy,Oz ln lt ti A,B, C sao cho H l trc tm ca tam

gic ABCGii: a. Gi giao im ca mt phng (P) v cc trc O x,Oy,Oz ln lt l A(a;0;0),B(0;b;0),C(0;0;c) vi a,b,c> 0Trng tm G ca tam gic ABC c to l:

G

3;

3;

3cba G(1;2;3) a = 3,b = 6,c = 9

Vy mt phng (P) c phng tr nh l :1

c z

b y

a x 1

963z y x hay (P) : 18x + 3y + 2z 18 = 0

b. Ta c AB CH (v H l trc tm ca ABC) v AB OC (v OC (Oxy)) AB) AB OH

(1)tng t BC OH (2) .T (1) v (2) OH (ABC) .Vy mt phng (ABC)(P) i qua H v nhnOH = (2;1;1) lm vtpt c phng tr nh l :2(x 2) + 1(y 1) + 1(z 1) = 0 hay (P) : 2x + y + z 6 = 0Bi 15:Trong khng gian vi h ta Oxyz, vit phng tr nh mt phng (P) i qua im D(1; 1; 1)v ct ba trc ta ti cc im M, N, P khc gc O sao cho D l trc tm ca tam gic MNP. Gii: Theo gi thit ta c M(m; 0; 0)Ox , N(0; n; 0) Oy , P(0; 0; p) Oz.


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Ta c :

1; 1; 1 ; ; ;0 .1; 1; 1 ; ;0; .

DP p NM m n DP NM m n

DN n PM m p DN PM m p

.

Phng tr nh mt phng 1 x y zPm n p

. V D (P) nn: 1 1 1 1m n p

.

D l trc tm ca MNP . 0

0 3. 0 03

( ) ( ) 1 1 1 1

DP NM DP NM m n m

DN PM DN PM m pn p

D P D Pm n p

Kt lun, phng tr nh ca mtphng : 13 3 3 x y z

P .

Bi 16:Trong khng gian vi h ta Oxyz, cho 2 ng thng d: 12 1 1 x y z v

d: 1 21 2 1

x y z . Vit phng tr nh mt phng (P) vung gc vi d, ct trc Oz v d theo mt on

thng c di nh nht. Gii: Do mt phng(P) vung gc vi d, nn c pt 2x y + z + c = 0Mt phng(P) ct Oz ti A(0; 0;-c), ct d ti B(1- c; -2c; -2-c)

Theo gi thit2

2 2

1 124 1245 2 5 55 25 25

AB c c c

suy ra AB nh nht khi 15

c tha mn.

Vy phng tr nh mt phng(P): 2x y + z + 1/5 = 0

Bi 17: Trong khng gian vi h ta Oxyz, cho ng thng: 11 1 4 x y z v im 0;3; 2 M .

Vit phng tr nh mt phng (P) qua M, song song v khong cch gia ng thng v mt phng (P) bng 3.

HD:Gi s phng tr nh mt phng(P) c dng Ax + By + Cz + D = 0

Tgi thitta c h

2 2 2

3 2 04 0

3

B C D

A B C

C D

A B C

28

B C

B C

TH 1: 2 B C chn 1, 2 2, 8C B A D TH 2: 8 B C chn 1, 8 4, 26C B A D ( ( ; ) ( , )d P d M P , vi M(0; 0; 1) )Vy c 2 mp (P)tha mn l: 2 2 8 0; 4 8 26 0. x y z x y z


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Bi 18: Trong khng gian vi h to Oxyz , cho mt cu 2 2 2: 2 2 4 3 0S x y z x y z v hai

ng thng 12

: 1 x t

y t t R

z t

, 21:

1 1 1 x y z . Vit phng tr nh tip din ca mt cu S ,

bit tip din song song vi c hai ng thng 1

v 2

.Gii: Mt cu S c tm 1; 1; 2 , 3 I R ng thng 1 2, ln lt c cc vct ch phng 2; 1;1 , 1; 1;1u v

mp P c vct php tuyn , 0; 1; 1Pn u v

: 0P y z m m

3 2 33

, 32 3 3 2

mmd I P R

m

Vy 1 2( ) : 3 3 2 0; : 3 3 2 0P y z P y z

Bi 19:Trong khng gian vi h to Oxyz , cho cc im 0;3;0 , 4;0; 3 B M . Vit phng tr nhmt phng( )P cha , B M v ct cc trc ,Ox Oz ln lt ti cc im A v C sao cho th tch khi tdinOABC bng3 (O l gc to ). Gii: Gi ,a c ln lt l honh , cao ca cc im, A C .

V 0;3;0 B Oynn : 13 x y z

Pa c

.

4 34;0; 3 1 4 3 M P c a aca c

(1)

1 1 1. .3. 3 63 3 2 2OABC OAC

acV OB S ac ac (2)

T (1) v (2) ta c h46 6 234 3 6 4 3 6 32

aac ac a

c a c a cc

Vy 1 22: 1; : 1

4 3 3 2 3 3 x y z x y z

P P

Bi 20:Cho im M (1; 2; 3). Lp phng tr nh mt phng (P) bit rng (P) ct ba trcOx, Oy, Oz lnlt tiA, B, C sao cho M l trng tm ca tam gic ABC .Gii: Do A, B, C ln lt thucOx, Oy, Oz nn ta gs A( x A ; 0 ; 0), B(0 ; y B ; 0),C (0 ; 0 ; zC ).

V M l trng tm ca tam gic ABC nn ta c M C B A

M C B A

M C B A

z z z z

y y y y

x x x x

333

963

C

B

A

z

y

x

.

Mt phng (P) i qua A(3 ; 0 ; 0), B(0 ; 6 ; 0),C (0 ; 0 ; 9) Nn (P) c phng tr nh l 1

963z y x .018236:)( z y xP

(phng tr nh m t phng theo on chn)


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Bi 21:Trong khng gian vi h trc ta Oxyz cho hai im(0; 1;2) M v ( 1;1;3) N . Vit phngtrnh mt phng (P) i qua M, N sao cho khong cch t 0;0;2K n (P) t gi tr ln nht Gii: Gi , ,n A B C 2 2 2 0 A B C l mt vect php tuyn ca mt phng (P).Phng tr nh mt phng (P) c dng;

1 2 0 2 0 Ax B y C z Ax By Cz B C 1;1;3 3 2 0 2 N P A B C B C A B C : 2 2 0P B C x By Cz B C

Khong cch t K n mp(P) l: , 2 24 2 4 B

K P B C BC

d

- Nu B = 0 th d(K,(P)) = 0 (loi)

- Nu 0 B th 2 2 21 1,

24 2 42 1 2

Bd K P

B C BC C B

Du = xy ra khi B = C. Chn C = 1v B = 1Vy phng tr nh mt phng(P): x + y z + 3 = 0Bi 22:Trong khng gian vi h to Oxyz cho im A(1;2;0), B(0;4;0),C(0;0;3) . Vit phng tr nhmt phng (P) cha OA, sao cho khong cch t B n (P) bngkhong cch t C n (P)Gii: Mt phng (P) c PT dng: Ax + By + Cz + D = 0 (A2 + B2 + C2 0)V (P) cha OAsuy ra(P) i qua 2 im O(0;0;0) v A(1; 2; 0).

0 02 0 2

D D

A B A B

Vy mp(P) c phng tr nh l: 2 0 Bx By Cz

Theo gi thit th : 2 2 2 24 3, , 5 5 B C d B P d C P

B C B C

34 3 4 34

B B C B C

C

Chn C = 4 suy raB = 3Vy c 2 mp tho mn: 1 2: 6 3 4 0 ; : 6 3 4 0.P x y z P x y z Bi 23:Trong h trc Oxyz cho M(2;4;1). Vit phng tr nh mt phng (P) qua M v ct cc trc Ox,Oy, Oz ti A, B, C tng ng vi honh , tung v cao dng sao cho 4OA =2OB = OC.Gii:

A,B,Cl im nm tr n Ox, Oy,Oz tng ng c honh , tung v cao dng v4OA = 2OB = OC suy ra A(a;0;0) , B(0;2a;0) v C(0;0;4a) vi 0a

Phng tr nh (ABC) 1 4 2 4 02 4

x y z x y z a

a a a

(ABC) qua M(2;4;1) suy ra 4.2 +2.4 +1 4a = 0 174

a

Vy phng tr nh (ABC) l 4 2 17 0. x y z


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Bi 24:Trong khng gian vi h ta Oxyz. Cho mt cu (S) : 921 222 z y x .

Lp phng tr nh mt phng (P) vung gc vi ng thng 1:1 2 2 x y z

d v ct mt cu (S) theo

ng tr n c bn knh bng 2 . Gii: (S) c tm )2,0,1( J bn knh R = 3

+ ng thngd c vtcp (1,2, 2)u , (P) vung gc vi t a nn (P) nhn u lm vtptPhng tr nh mt phng(P) c dng : 2 2 0 x y z D + (P) ct (S) theo ng tr n c bn knh r = 2 nn 2 2, 5d J P R r

nn ta c :1 2.0 2.( 2) 5 3 5

53 5 3 5

D D

D

KL : C 2 mt phng : (P1) : 053522 z y x v (P2) : 053522 z y x Bi 25:Trong khng gian vi h to Oxyz, cho mt cu (S) c phng

trnh 011642222

zyxzyx v mt phng ( ) c phng tr nh 2 x + 2 y z + 17 = 0. Vit phng tr nh mt phng ( ) song song vi ( ) v ct (S) theo giao tuyn l ng tr n c chu vi bng 6 . Gii: Do ( ) // ( ) nn ( ) c phng tr nh 2x + 2y z + D = 0 (D17)Mt cu (S) c tm I(1;2; 3), bn knh R = 5ng tr n c chu vi 6 nn c bn knh r = 3.Khong cch t I ti () l 2 2 2 25 3 4h R r

Do D DD

D (loai)2 2 2

2.1 2( 2) 3 74 5 12172 2 ( 1)

Vy ( ) c phng tr nh 2x + 2y z 7 = 0Bi 26:Trong khng gian ta Oxyz cho mt cu 2 2 2: 2 4 4 5 0S x y z x y z , mt phng (Q): 2x + y 6z + 5 = 0. Vit phng tr nh mt phng (P). Bitrngmt phng (P) i qua A(1;1;2),vung gc vi mt phng (Q) v tip xc vi mt cu (S). Gii: Mt phng (P) qua A(1;1;2) c phng trnh : 2 2 21 1 2 0 ( 0)a x b y c z a b c Mt cu (S) c tm I(1;-2;2) bn knh R = 2Mt phng (Q) c VTPT (2;1; 6)Qn

Ta c (P) vung gc vi (Q) v tip xc (S) nn

2 2 2

2 6 03

2

a b c

b

a b c

2 2 2 2 2 2

22 6 22 6 2 6

(I)2 59 4 4 4 3 10 0 5 112

a ca c b b c

a c b a c bb c b cb a b c b bc cb c

a c

Chn c = 0 th a = b = 0 (loi) Nn 0c . T (I) Phng tr nh mt phng : 2 1 2 1 2 0P c x c y c z


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2 2 6 0 x y z

Hoc 11 1 5 1 2 0 11 10 2 5 02 c x c y c z x y z

Bi tp tng hpt gii:

Bi 1:Cho im 2;5;3 A v ng thng1 2

: .2 1 2 x y z

d

Vit phng tr nh mt phng cha d sao cho khong cch t A n ln nht.

s: 4 3 0 x y z Bi 2:Trong khng gian vi h ta Oxyzcho hai ng thngd vd ln lt c phng tr nh :

2:1

yd x z v 2 5 : 3

2 1 x z

d y . Vit phng tr nh mt phng )( i quad v to vid

mt gc 030 s: 2 4 0 ; 2 2 0 x y z x y z Bi 3:Trong khng gian vi h trc ta Oxyz, cho cc im A(-1;1;0), B(0;0;-2) v C(1;1;1). Hy vit phng tr nh mt phng (P) qua hai im A v B, ng thi khong cch t C ti mt phng (P) bn

3 .s: 2 0 ;7 5 2 0 x y z x y z Bi 4:Trong khng gian vi h ta Oxyz, cho M(1;2;3).Lp phng tr nh mt phng i qua M ct batia Ox ti A, Oy ti B, Oz ti C sao cho th tch t din OABC nh nht. s: 6 3 2 18 0 x y z

Bi 5:Cho ng thngd c phng tr nh:2

: 22 2

x t

d y t

z t

.Gi l ng thng qua im A(4;0;-1)

song song vid v I(-2;0;2) l hnh chiu vung gc ca A tr n d. Trong cc mt phng qua , hy vit phng tr nh ca mt phng c khong cch nd. l ln nht.

s: 2x - z - 9 = 0.Bi 6:Trong kgian vi h ta Oxyz cho im A(10; 2;-1) v ng thng d c phng

trnh1 2

1 3

x t

y t

z t

. Lp phng tr nh mtphng (P) i qua A, song song vi d v khong cch t d ti (P) l

ln nht. s:7 5 77 0 x y z

Bi 7:Cho im 2;5;3 A v ng thng 1 2: .2 1 2 x y z

d Vit phng tr nh mt phng

cha d sao cho khong cch t A n ln nht.

s:2 2 15 0 x y z Bi 8:Trong khng gian vi h trc ta Oxyz, cho ng thng 1 3:

1 1 4 x y z v

im 0; 2;0 . M Vit phng tr nh mt phng (P) i qua im M song song vi ng thngngthi khong cch gia ng thngv mt phng (P) bng 4.s:4 8 16 0 x y z hay 2 2 4 0. x y z Bi 9:Vit phngtrnh mt phng cch u hai ng thng d1 v d2 bit:


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1

2: 2

3

x t

d y t

z t

v 21 2 1:

2 1 5 x y z

d

s: 3 4 7 0 x y z

Bi 10:Trong khng gian vi h ta Oxyz, cho ng thng2 0

: 2 6 0 x y

d x y

v

mtcu 2 2 2: 2 2 2 1 0S x y z x y z . Vit phng tr nh mt phng (P) cha dsao cho giaotuyn ca mt phng (P) v mt cu (S) l ng tr n c bn knh r = 1.Bi 11:Trong khng gian vi h ta Oxyz cho mt cu 2 2 2: 2 2 4 3 0S x y z x y z v hai

ng thng 12 2 0

:2 0

x y

x z

, 21:

1 1 1 x y z

Vit phng tr nh tip din vi mt cu (S), bit n song song vi1 v 2.

Bi 12: Lp phng tr nh mt phng cha ng thng8 11 8 30 02 0

x y z

x y z

v ti

p xc vi mt

cu 2 2 2: 2 6 4 15 0S x y z x y z .Bi 13:Cho mt cu (S): 2 2 2: 10 2 26 170 0S x y z x y z ;

2

5 2: 1 3

13 2

x t

y t

z t

v

1

1 1

7: 1 2

8

x t

y t

z

Vit phng tr nh )( tip xc mt cu (S) v song song vi 1 v 2 Bi 14:Lp phng tr nh mt phng (P) i qua hai im 1;2;3 A v 2;3;4 B v ct mt cu

2 2 2: 2 6 4 15 0S x y z x y z theo giao tuyn l mt ng tr n c chu vi 8

Bi 15:Lp phng tr nh mt phng (P) song song vi hai ng thng15 1 13

: 2 3 2 x y z

d

27 1 8:

3 2 0 x y z

d ng thi tip xc vi mt cu 2 2 2: 10 2 26 113 0S x y z x y z

Bi 16:Lp phng tr nh mt phng (P) i qua im 2,3,1 A v vung gc vi mt phng : 1 0Q x y z ng thi tip xc vi mt cu 2 2 2: 2 2 4 1 0S x y z x y z

Bi 17: Lp phng tr nh mt phng (P) i qua im 2,1, 1 A ng thi song song vi hai ng

thng 1 3 5:1 2 1

x y zd

; 2 1 0:3 2 3 1 0 x y z

x y z

Bi 18: Lp phng tr nh mt phng (P) i qua ng thng 1 2:1 2 1

x y zd v to vi mt phng

: 2 2 2 0Q x y z mt gc nh nht Bi 19:Lp phng tr nh mt phng i qua hai im 0;0;1 ; 3;0;0 A B ng thia. To vi mt phngOxy mt gc 60o b. Vung gc vi mt phng : 2 3 1 0P x y z Bi 20:Lp phng tr nh mt phng i qua giao tuyn ca hai mt phng : 1 0P x y z v mt phng : 2 3 2 0Q x y z ng thi


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a. i qua A (1,3,-2)b. Vung gc vi mt phng: : 2 4 1 0 x y z

c. Song song vi ng thng 1 3 5:1 2 1

x y zd

Bi 21:Trong khng gian vi h trc ta vung gcOxyz, cho 2;5;3 A v ng thng

1 2: 2 1 2 x y zd

1. Vit phng tr nh mt phng Q cha d sao cho khong cch tA n Q ln nht. 2. Vit phng tr nh mt cu S c tm nm trn ng thngd ng thi tip xc vi hai mtphng : 3 4 3 0, : 2 2 39 0 x y x y z .Bi 22:Trong cc mt phng i qua cc im 1;2; 1 , 1;1;2 A B ,vit phng tr nh mt phng to vi mt xOy mt gc nh nht. S: : 6 3 5 7 0 x y z Bi 23:Trong khng gian vi h ta Oxyz, vit phng tr nh mt phng i qua im 1;2;4 A v ct chiu dng ca cc trc ta Ox, Oy, Oz ln lt M, N, P khc gc ta sao cho t diOMNP c din tch nh nht

s: : 13 6 12 x y z

Bi 24:Trong khng gian vi h ta Oxyz, vit phng tr nh mt phng i qua im

1;2;3 M v ct cc trc ta Ox, Oy, Oz ti A, B, C sao cho2 2 21 1 1

OA OB OC nh nht

s: : 2 3 14 0 x y z Bi 25:Trong khng gian vi h ta Oxyz, vit phng tr nh mt phng i qua im 2;5;3 M v ct chiu dng ca cc trc ta Ox, Oy, Oz ln lt A, B, C sao choOA OB OC nh nht

s: : 12 6 10 5 10 15 3 6 15

x y z

Bi 26:Trong cc mt phng i qua im 2; 1;0 A v song song vi ng thng1 2 1:

1 1 1 x y z

d . Vit phng tr nh mt phng to vi mt phng xOy mt gc nh nht

s: : 2 1 0 x y z Bi 27:Trong cc mt phng i qua 1;1; 1 A v vung gc vi mt phng : 2 2 0 x y z . Vit phng tr nh mt phng to vi Oy mt gc ln nht.

s :

: 0

5 1: 3 02 2

y z

x y z

CHUYN : VIT PHNG TR NHNG THNG TRONGKHNG GIAN

1. Vi t phng tr nh tham s hoc chnh tc


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42

- vit phng tham s hoc chnh tc ta phi bit c mt im 0 0 0 0; ; M x y z v mt vtcp

; ;u a b c phng tr nh tham s0

0

0

x x at

y y bt

z z ct

vi t R l tham s

phng tr nh chnh tc0 0 0 x x y y z z

a b c

(khi , , 0a b c )

2. Vi t phng tr nh t ng qut ca ca ng thng Cch 1:ng thng d chnh l giao tuyn ca hai mt phng phn bit gi s

1 1 1 1: 0a x b y c x d v 2 2 2 2: 0a x b y c x d . Khi d hay1 1 1 1

2 2 2 2

0:

0a x b y c z d

d a x b y c z d

- Tm vecto ch phngu ca ng thng dCch 1.1:Vtcp ;u n n

vi n v n ln lt l vtpt ca v Cch 1.2:Chn hai im M v N phn bit thuc ng thng d, khi vtcpu cng phng vi vecto

MN hay vecto MN chnh l vtcp ca d Cch 2:T phng tr nh chnh tc hoc tham s chuyn v phng tr nh tng qut

Cc dng bi tp

Dng 1: Vit phng tr nh ng thng i qua im 0 0 0 0; ; M x y z v th a m n iu kin chotrc Phng php: Loi 1: C mt vtcp cho trc, khi iu kin l

- C mt vecto ; ;u a b c cho trc - Song song vi mt ng thng d cho trc d u u

- Vung gc vi mt mt phng (P) cho trc Pu n

Bi tp gii mu:

Bi 1:Trong khng gian vi h trc to Oxyz.Cho tam gic ABC c 1; 2;3 , 2;1;0 , A B 0; 1; 2 .C Vit ph ng trnh tham s ng cao t ngng vi nh A ca tam gic ABC.

Gii: Gi d l ng cao tng ng vi nh A ca ABC

d l giao tuyn ca ABC vi qua A v vung gc vi BC.

Ta c: 1; 3; 3 AB , 1;1; 5 AC , 2; 2; 2 BC v , 18;8;2 AB AC

mp ABC c vtpt

1 , 3;2;14

n AB AC

mp c vtpt '1 1;1;12

n BC

- ng thng d c vtcp ', 1;4; 5u n n


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Vy phng tr nhng thng1

: 2 43 5

x t

d y t

z t

Bi 2:Trong khng gian vi h trc ta cc vung gc Oxyz, cho 3 imA(0,0,1); B(-1,-2,0) ;C(2,1,-1).

1. Vit phng tr nh ca mt phng (P) i qua 3 im A,B ,C 2. Vit phng tr nh tham s ca ng thng i qua trng tm ca tam gic ABC v vung gc vi mtphng (P ).3. Xc nh chn ng cao h t A xung ng thng BC Gii: 1. Phng tr nh mt phng (P) i qua A,B,C. Ta c VTP (P) l: , (5, 4,3)Pn AB AC

Phng tr nh mt phng (P): 5x 4y + 3z 3 = 0

To trng tm tam gic ABC l 1 1; ;03 3

G

ng thng d i qua G v d (P): (5, 4,3)Pd a n

Phng tr nh tham s ca d l:

1 53

1 43

3

x t

y t

z t

2. Chn ng cao H h t A xung ng thng BC. Ta c: (3,3, 1) BC

Phng tr nh tham s ca BC l: 1 32 3 x t y t

z t

Ly 1 3 ; 2 3 ; H t t t BC

H l hnh chiu ca A 19. 0 3(1 3 ) 3(2 3 ) 1(1 ) 19 88

HA BC t t t t t

Vy 5 14 8; ;19 19 19

H

Bi 3: (H B 2009)Trong khng gian vi h to Oxyz, cho mt phng (P): x 2y + 2z 5 = 0 vhai im A(-3;0;1), B(1;-1;3). Trong cc ng thng i qua A v song song vi (P), hy vit phngtrnhng thng m khong cch t B n ng thng l nh nht Gii: Ta c (4; 1;2); (1; 2;2)P AB n

Pt mt phng (Q) qua A v // (P) : 1(x + 3) 2(y 0) + 2(z 1) = 0x 2y + 2z + 1 = 0. Gi l ng thng bt k qua A

Gi H l hnh chiu ca B xung mt phng (Q).Ta c :d(B, ) BH; d (B, ) t min qua A v H.


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Pt tham s1

: 1 23 2

x t

BH y t

z t

Ta H = BH (Q) tha h phng tr nh :1 , 1 2 , 3 2 10 1 11 7

; ;2 2 1 0 9 9 9 9 x t y t z t

t H x y z

qua A (-3; 0;1) v c 1 VTCP 1 26;11; 29a AH

Phng tr nh 3 0 1:26 11 2

x y z

p s:ng thng c phng tr nh l 3 0 1:26 11 2

x y z

Bi tp t gii:

Bi 1:Vit phng tr nh chnh tc ca ng thngi qua im 1;1;2 M v song song vi ng

thng 3 2 7 0:3 2 3 0

x y zd

x y z

p s: 1 1 2:2 4 5

x y z

Bi 2:Vit phng tr nh tham s ca ng thngi qua im 1;1;1 M v vung gc vi mt phng : 2 3 12 0P x y z

p s:1

: 1 21 3

x t

y t

z t

Bi 3:Trong khng gian Oxyz cho ba im 1;3;2 ; 1;2;1 A B v 1;1;3C . Vit phng tr nhngthng d i qua trng tm ca tam gic ABC v vung gc vi mt phng cha tam gic

p s:1 3

: 22

x t

d y

z

Bi 4:Vit phng tr nhng thngd i qua im 1;2; 1 A v song song vi ng thng giao tuynca hai mt phng : 3 0 x y z v : 2 5 4 0 x y z

p s: 1 4: 2 71 3

x t d y t

z t

Loi 2: C mt cp vecto khng cng phng cho trc, khi iu kin l- C mt cp vecto ch phnga v b cho trc - Vung gc vi hai ng thng d1 v d2 cho trc 1 2;u u u

- Song song vi hai mt phng (P1) v (P2) cho trc 1 2;u n n


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- Vung gc vi mt ng thng d v song song vi mt mt phng (P) ;d Pu u n

- Nu i qua hai im phn bitv A B u AB

Ch : - Nu gi thit l vung gc vi mt vectoc bt k th hiu c l vtcp,- Nu gi thit l song song vi mt vectod th hiud l vtpt

Bi tp gii mu:

Bi 1:Vit phng tr nh tham s ca ng thngi qua im 1;1; 2 A song song vi mt phng

: 1 0P x y z v vung gc vi ng thng 1 1 2: 2 1 3 x y z

d

Gii: - Mt phng (P) c vtpt 1; 1; 1Pn v ng thng d c vtcp 2;1;3d u - ng thng song song vi mt phng (P) v vung gc vi ng thng d nn c vtcp

, 2;5; 3P d u n u

- ng thng i qua 1;1; 2 A v c vtcpu c phng tr nh l1 2

: 1 52 3

x t

y t

z t

Ch : Nu bi yu cu vit phng tr nhng thng dng tng qutCch 1:ng thng chnh l giao tuyn ca hai mt phng

- Mt phng i qua im A v song song vi mt phng (P) - Mt phng i qua im A v vung gc vi ng thng d

Cch 2:T phng tr nh tham s chuyn v phng tr nh tng qut Bi 2:Vit phng tr nh tham s ca ng thngi qua im 1;1;1 M ng thi vung gc vi hai

ng thng11 2:

8 1 1 x y z

d v 2

2 3 0:

1 0 x y z

d x y z

Gii: - ng thng1d c vtcp 1 8;1;1u v 2d c vtcp 2 2; 3;1u - ng thng vung gc vi hai ng thng1d v 2d nn c vtcp 1 2, 4;6; 26u u u

- ng thng i qua 1;1;1 M v c vtcpu c phng tr nh l1 4

: 1 61 26

x t

y t

z t


- Mt phng i qua im M v vung gc vi ng thng d1 - Mt phng i qua im M v vung gc vi ng thng d2

Cch 2:T phng tr nh tham s chuyn v phng tr nh tng qut

Bi tp t gii:


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Bi 1:Vit phng tr nh chnh tc ng thng i qua 1;2;5 M ng thi song song vi hai mtphng : 3 5 8 0P x y z v : 2 1 0Q x y z

p s: 1 2 5:4 7 5

x y z


- Mt phng i qua im M v song song vi mt phng (P) - Mt phng i qua im M v song song vi mt phng (Q)

Cch 2:T phng tr nh tham s chuyn v phng tr nh tng qut Bi 2:Vit phng tr nhng thngd i qua im 2; 1;1 A v vung gc vi hai ng thng lnlc c vtcp l 1 1;1; 2u v 2 1; 2; 0u

p s: 2 4

: 1 2

1

x t

d y t

z t

Bi 3: (H TCKT 1999)Vit phng tr nh chnh tc ca ng thngi qua im 1;1; 2 A songsong vi mt phng (P) v vung gc vi ng thng d bit

: 1 0P x y z v 1 1 2: 2 1 3 x y z

d

p s: dng chnh tc l 1 1 22 5 3

x y z

Dng 2: Vit phn g trnh ng thng i qua mt im 0 0 0 0; ; M x y z ct ng thng d v th a mn iu kin cho trc iu kin cho trc l- Vung gc vi ng thng1cho trc

- Song song v i mt mt phng (P) cho trc Ch :- Nu gi thit l vung gc vi mt vectoc bt k th hiu c l vtcp,- Nu gi thit l song song vi mt vectod th hiud l vtptPhng php chung:Trng hp 1: Nu b i yu c u vit phng tr nh ng thng dng tng qut Cch 1:Xc nh cc vtcp v vtptng thng chnh l giao tuyn ca hai mt phng

- Mt phng i qua im 0 M v cha ng thngd - Mt phng i qua im 0 M v tha mniu kin cho trc

Kt lun :- Nu th bi ton c v s nghim- Nu th / / d th bi ton v nghim, ctd th l ng thng cn dng

Trng hp 2: Nu bi yu c u vit phng tr nh ng thng dng tham s hoc chnh tc Cch 2:Xc nh cc vtcp v vtpt

- Mt phng i qua imM 0 v tha mniu kin cho trc(hin nhin mt phng chang thng )


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- Gi M d , ta M l nghim ca h+ Nu khng tn ti giao im th bi ton v nghim+ Nu cv s nghim (tc l d ) th bi ton c v s nghim+ Nu c nghim duy nht th tnh vecto 0 M M

Kt lun:ng thng i qua im 0 0 0 0; ; M x y z v tha mniu kin cho trc chnh l ng

thng i qua im 0 M v c vtcp 0 M M Cch 3:Xc nh cc vtcp v vtpt

- Do ng thng i qua im 0 M v ct ng thng d nn chn im M d nn ng thng chnh l ng thng0 M M

- Tnh vecto 0 M M - T iu kin cho trc ta dn n mt phng tr nh bc nht theo tham s t, t m t 0 M M

Kt lun: ng thng chnh l ng thng i qua im0 M v c vtcp 0 M M Ch :

- Vi cch ny thng thng d phi dng tham s (nu d dng tng qut hay chnhtc th chuynv dng tham s)

- Ta im M phi theo tham s t - t m ta im M ta c th lm nh sau ng thngct ng thng d nn M d , tnh0 M M , t iu kin cho trc t m rac im M, suy ra ng thng

Bi tp gii mu:

Bi 1:Cho im 1;2;3 A v hai ng thng12 2 3:

2 1 1 x y z

d v

2 1 1 1: 1 2 1 x y zd . Vit phng tr nh ng thngd i qua A, vung gc vid 1 v ctd 2

Gii:

ng thng2d c phng tr nh tham s 21

: 1 21

x t

d y t

z t

ng thng d1 c vtcp l: )1;1;2(u , Gi B l giaoim cad vid2 th 2 1 ;1 2 ; 1 B d B t t t )4;12;( t t t B A

Theo gi thit 1 1. 0 1d d AB u t

Vyd qua 1;2;3 A c vtcp (1; 3; 5) AB nn ph ng trnh l: 53321 1z y x

Bi 2:Vit phng tr nhng thng d i qua M(1;1;1)ct ng thng12 1:

3 1 2 x y z

d

v vung gc vi ng thng22 2

: 52

x t

d y t

z t

( Rt ).

Gii: - Vtcp ca d2 l 2 2; 5;1u v cng l vtpt ca mp(P) i qua M v vung gc vi d2.


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Phng tr nh mp (P) l: 0252 z y x - GiN l giao im ca d1 v mp(P) nn 12 3 ; ;1 2 N t t t P d Thay vo phng tr nh mp(P) th 1 5; 1;3t N - ng thng d cn lp pt c VTCP 3;1; 1u do 6; 2; 2 MN

Vy phng tr nhng thng d l: 11

11

31 z y x

(vd d 2)

Bi 3:Trong khng gian vi h ta Oxyz cho, ng thng 1 2:1 2 1 x y z

d v mt phng (P): x

+ 3 y + 2 z + 2 = 0. Lp phng tr nhng thngd i qua im M (2; 2; 4), song song vi mt phng (P)v ct ng thngd .HD:Chn ( ;1 2 ;2 ) ( 2;2 1; 2) N d N t t t MN t t t .

1 3 3 / / ( ) . 0 ( ) 1 (1;3;3) ' :1 1 1P x y z

MN P MN n do M P t N d

.

Bi 4:Trong khng gian vi h trc ta Oxyz, cho cc im 1;1;1 A v ng thng1 1:

1 2 1 x y z

d

. Vit phng tr nhng thng qua A v ctd, sao cho khongcch t gc to

O n nh nht. Gii: ngthng thucmp (P) qua A v chad, nn : 3 2 4 0P x y z . GiH l hnh chiuvung gc caO ln (P ). To H tha mn h:

32 6 4 2; ;

7 7 73 2 0

x t

y t H

z t

x y z

GiK l hnh chiuvung gc caA ln : ( ; )d A AK AH khongcch nhnhtth K H H . Suy ra qua A,H nn c pt :

1 2 1:1 3 9

x y z , d thy ct d nn l ng thngcn t m.

Bi 5:Trong khng gian vi h trc Oxyz, cho hai mt phng( ) : 5 0 x y , ( ) : 3 0 y z , imM(1; 1; 0). Vit phng tr nhng thng d qua M vung gc vi giao tuyn ca( ) v ( ) , ng thid ct( ) v ( ) ln lt ti A, B sao cho M l trung im ca AB. Gii: ngthngd thucmtphng(P) qua M v vung gc vigiao tuynca ( ) v ( ) .Ph ng trnh (P): 0 x y z .Ly ixng ( ) qua Mc( ') : 1 0 x y . Suy ra B l giaoimcaba mtphng ( )P , ( ) v ( ') . To B l nghim ca h

01 4 53 0 ; ;3 3 3

1 0

x y z

y z B

x y

Vy ngthngqua B, M lngthngd cntm: 1 12 7 5

x y z .


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Bi tp t gii:

Bi 1:Vit phng tr nh tham s ng thng i qua 1;2;3 A ng thi vung gc vi d1 v ct

2d bit 16 2

: 1 4

4

x t

d y t

z t

v 21 2 3:

2 1 1 x y z

d

p s:1 6

: 3 235 11

x t

y t

z t

Bi 2: (CGT 2005)Trong khng gian vi h ta Oxyz. Cho im 1;2 1 M , ng thng3 3:

1 3 2 x y z

d v mt phng : 3 0P x y z . Vit phng tr nhng thng i qua im

M song song vi mt phng (P) v ct ng thng d Bi 3:Vit phng tr nhng thng i qua im 1; 2; 3 A vung gc vi vecto 6; 2; 3a v

ct ng thng2 1 1 3: 3 2 5 x y zd

p s: dng chnh tc l 1 2 32 3 6

x y z , dng tng qut l 6 2 3 1 03 28 13 4 0 x y z

x y z

Dng 3:Vit phng tr nh ng thng i qua mt im 0 0 0 0; ; M x y z vung gc v c t ng thng d cho trc (chnh l trng hp c bit ca dng 2)Phng php:Trng hp 1: Nu b i yu c u vit phng tr nh ng thng dng tng qut Xc nh vtcp ca ng thngd ng thng chnh l giao tuyn ca hai mt phng

- Mt phng i qua im 0 M v cha ng thngd - Mt phng i qua im 0 M v vung gc vi ng thngd

Trng hp 2: Nu b i yu c u vit phng tr n h ng thng dng tham s hoc chnh tc - ng thng i qua 0 M ct ng thngd ti im M d tc l 0 M M (M chnh l hnh

chiu ca 0 M trn ng thngd )- Tm ta im M theo tham s ca ng thng d, tnh vecto0 M M - T iu kin 0 d d M M u ta c phng tr nh bc nht

Mat phang duong thang mat cau trong khong gian toa do

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