chuyÊn ĐÊ viẾt phƯƠng trÌnh mẶt phẲng - ĐƯỜng thẲng - mẶt cẦu.pdf

Gio vin: Nguyn Thnh Long Gmail: [email protected]

https://www.facebook.com/trithuc.viet.37 1

(DNG CHO N THI TN C H 2011)

Bm sn. 22.03.2011



CHUYN : VIT PHNG TRNH MT PHNG

A. Kin thc chung 1. Phng trnh mt phng v cc trng hp c bit - PTTQ (phng trnh 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 trnh mt phng) P qua ( ,0,0) ; (0, ,0) ; (0,0, )A a Ox B b Oy C c Oz c phng trnh

l: ( ) : 1x y zPa b c (Phng trnh mt phng theo on chn)

- c bit:

+ 2 2

0( ) / / 0

0

AP Ox D

B C

+ 2 2

0( ) / / 0

0

BP Oy D

A C

+ 2 2

0( ) / / 0

0

CP Oz D

A B

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

TH 1: 1 1 1 11 22 2 2 2

( ) / /( ) A B C DA B C D

TH 2: 1 1 1 11 22 2 2 2

( ) ( ) A B C DA B C D

TH 3: 1 2 1 2 1 2 1 2( ) ( ) 0A A B B C C 3: Phng trnh chm mt phng: Tp hp cc mt phng ( ) cha ng thng ( ) ( ) c gi l chm mt phng xc nh bi mt phng ( ) v mt phng ( ) Nu 1 1 1 1( ) : 0A x B y C z D v 2 2 2 2( ) : 0A x B y C z D th phng trnh 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 trnh (*) c th vit li: ( ) ( ) 0m n 4. Gc v khong cch - Gc ca 2 mt phng: 1 1 1 1 1( ) : 0A x B y C z D v 2 2 2 2 2( ) : 0A x B y C z D l:

1 2 1 2 1 2

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

A A B B C Ccos

A B C A B C

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

sin( ,( )).

u nd P

u 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 PA B C

B. Mt s dng bi tp Dng 1: Vit phng trnh mt phng (P) i qua im Mo(xo;yo;zo) v tho mn iu kin Loi 1 : C mt vect php tuyn Phng php: - Xc nh 0 0 0 0( , , )M x y z ca mt phng P - Xc nh vtpt ( ; ; )n A B C

+ Nu / / P QP Q n n

+ Nu P dP 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 Ban C Bn T89) Trong khng gian vi h to Oxyz .Vit phng trnh mt phng (P): a. i qua im 1; 2;4M v nhn vect 2;3;5n lm vect php tuyn b. i qua im 2; 1;2M 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;4M v c vect php tuyn 2;3;5n c phng 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) c vtpt 2;3;5n lun c dng 2 3 5 0x y z D v mt phng (P) i qua im 1; 2;4 2.1 3. 2 5.4 0 16M D D .Vy mt phng : 2 3 5 16 0P x y z b. Cch 1: Mt phng P i qua im 2; 1;2M song song vi mt phng Q nn mt phng P i qua im

2; 1;2M v c vtpt 2; 1;3P Qn n nn mt phng P c phng trnh:

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 dng 2 3 0x y z D v mt phng P i qua im 2; 1;2M ' 1D hay : 2 3 11 0P x y z

Hoc c th l lun v P song song vi Q nn P lun c dng 2 3 0x y z D v P qua M : 2 3 11 0P x y z Bi 2: (SGK Ban C Bn T92) Trong khng gian vi h to Oxyz cho mt phng c phng trnh: 3x + 5y z 2 = 0 v ng thng d c phng trnh



12 4: 9 3

1

x td y t

z t

a. Tm giao im M ca ng thng d v mt phng b. Vit phng trnh mt phng cha im M v vung gc vi ng thng d Gii: a. To im M d l nghim ca phng trnh 3(12 + 4t) + 5(9 + 3t) (1 + t) 2 = 0 t = 3 .Vy 0;0; 2M b. Cch 1 : Mt phng i qua im 0;0; 2M vung gc vi ng thng d nn mt phng i qua im

0;0; 2M v c vtpt n = du

= (4;3;1) nn mt phng c phng trnh l: 4(x 0) + 3(y 0) + 1(z +2) = 0 hay : 4 3 2 0x 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; 2M D = 2 hay : 4 3 2 0x y z

Ch : C th pht biu bi ton di dng nh, cho bit ta 3 im A, B, C. Vit phng trnh mt phng (P) i qua im A v vung gc vi ng thng BC th khi Pn BC

Nhn 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 dng Ax + By + Cz + D = 0 vi ' 0D - Hai mt phng song song vi nhau th hai vtpt cng song song (cng phng) vi nhau, mt phng vung gc vi ng thng th vtpt v vtcp cng song song (cng phng) vi nhau . iu ny l gii ti sao trong bi 1 cu b li chn Pn

= Qn ,tht vy 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

= du , 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 vtpt ca mt phng P

- Vect php tuyn cng c th cho hnh thc l vung gc vi gi ca vect a no , khi ta phi 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; 3A . Vit phng trnh mt phng cha im A v vung gc vi gi ca vect a Hng dn: Lm tng t nh bi 2b ta c : 6 2 3 2 0x y z Bi 4: (SGK Ban C Bn T80) Trong khng gian vi h to Oxyz .Vit phng trnh mt phng i qua im 2;6; 3M v ln lt song song vi cc mt phng to Gii: Nhn 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 c vtpt, cc vtpt ny c xy dng nn t cc vect n v trn cc trc Ox, Oy, Oz ln 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 trnh mt phng P i qua M v song song vi mt phng 0xy cn cc mt phng khc lm tng t Cch 1: Mt phng P i qua 2;6; 3M v song song vi mt phng Oxy mt phng P i qua M v vung gc Oz nn mt phng (P) i qua M nhn vect Pn

= k

lm vtpt c phng trnh l : 0(x 1) + 0(y 6) + 1(z + 3) = 0 hay : 3 0P z Cch 2: Mt phng P song song vi mt phng 0xy mt phng P song song vi hai trc Ox 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 mt phng P i qua M C. 3 D 0 v C 0 nn chn C = 1 D = 3 . Vy mt phng P c phng trnh l : 3 0P z Ch : Bi ton c th pht biu l vit phng trnh (P) i qua M // vi Ox v Oy P i qua M // vi mt phng 0xy

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

(vi , 0a b

c gi song song hoc nm trn mp ( )P )

- Tm vtpt ,n a b

- P l mp qua 0 0 0 0( , , )M x y z v c VTPT n

- Quay li loi 1 Bi tp gii mu: Bi 5: (SGK Ban C Bn T80) Trong khng gian vi h to Oxyz . Vit phng trnh mt phng P i qua im 0; 1;2A v song song vi gi ca mi vect u = (3;2;1) v v = 3;0;1 Gii: Cch 1: Mt phng P i qua 0; 1;2A v song song vi gi ca hai vect u = (3;2;1) ; 3;0;1v mt phng P i qua A v c Pn

u ; Pn

v (vi u v v khng cng phng)

mt phng P i qua A v c vtpt , 2; 6;6 2 1; 3;3Pn u v

mt phng P c phng trnh l : 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 1b khi bit 2; 6;6Pn

v 0; 1;2A Bi 6: (SBT Ban C Bn T99) Trong khng gian vi h to Oxyz . Vit phng trnh mt phng i qua im 2; 1;2M , song song vi trc Oy v vung gc vi mt phng : 2 3 4 0x y z



Gii: Cch 1: Mt phng i qua im 2; 1;2M song song vi trc 0y 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 v c vtpt n

= [ j

, n ] = (3;0;-2)

mt phng c phng trnh l : 3(x 2) + 0(y + 1) 2(z 2) = 0 hay : 3 2 2 0x z Cch 2: Lm tng t nh bi 1b khi bit 3;0; 2n

v 2; 1;2M Cch 3: Gi s mt phng c dng : 2 2 20 0Ax By Cz D A B C mt phng c vtpt ; ;n A B C

- Mt phng i qua im 2; 1;2M .2 .( 1) .2 0 1A 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 trnh l : 3 2 2 0x z Bi 7: (SBT Ban C Bn T98) Trong khng gian Oxyz.Vit phng trnh mt phng i qua im

3; 1; 5M ng thi vung gc vi hai mt phng : 3 2 2 7 0x y z v : 5 4 3 1 0x y z Gii: Cch 1: Mt phng i qua im 3; 1; 5M ng thi vung gc vi hai mt phng v mt phng 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 trnh l : 2(x 3) + 1(y + 1) 2(z + 5) = 0 hay : 2 2 15 0x y z Cch 2: Lm tng t nh bi 1b khi bit n

= 2;1; 2 v 3; 1; 5M Cch 3: Gi s mt phng c dng : 2 2 20 0Ax By Cz D A B C mt phng c vtpt ; ;n A B C

- Mt phng i qua im 3; 1; 5M .3 .( 1) . 5 0 1A 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

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

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

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

11 1: , ' : 1 2

2 1 12

x tx y zd d y t

z t



Vit phng trnh mt phng i qua A ng thi song song vi d v d Gii: Cch 1: V 1 20;1; 1 ; 1; 1;2B d C d v 1 2, , / /B C d d Vecto ch phng ca 1 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 0A x y z s: : 3 5 13 0x y z Cch 2: Gi s mt phng c dng : 2 2 20 0Ax By Cz D A B C mt phng c vtpt ; ;n A B C

- Mt phng i qua im M .0 .1 .2 0 1A B C D - Mt phng song song vi ng thng d . 0 .2 .1 . 1 0 2dn u A B C

- Mt phng song song vi ng thng d '. 0 .1 . 2 .1 0 3dn u A B C

T (1) v (2) ta c 2 , 4 3C A B D A B th vo (3) ta c 3A B chn

1, 3 5, 13A B C D Vy phng trnh mt phng l 3 5 13 0x y z Nhn xt: Nu im A d (hoc 'A d ) th bi ton tr thnh vit phng trnh mt phng cha d (hoc 'd ) v song song vi 'd (hoc d ) 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 trnh mt phng i qua im E v vung gc vi MN.

( thi tt nghip BTTHPT ln 2 nm 2007) b. Vit phng trnh mt phng i qua 1; 2;1K v vung gc vi ng

thng 1

: 1 21 3

x td y t

z t

.

( thi tt nghip THPT ln 2 nm 2007) s: a. : 3 5 0x y z b. : 2 3 8 0x y z Bi 2: Trong khng gian vi h to Oxyz cho im 1; 1;0M v mt phng P c phng trnh:

2 4 0.x y z Vit phng trnh mt phng i qua M v song song vi P s: : 2 2 0x y z ( thi tt nghip THPT h phn ban nm 2007) Bi 3: Vit phng trnh mt phng i qua im 2;3;1M 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 hnh hc 12) s: : 3 4 19 0x y z



Bi 4: Vit phng trnh mt phng i qua im 2;1; 1M 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 hnh hc 12) s: :15 7 7 16 0x y z Dng 2 : Vit phng trnh mt phng (P) i qua im M1(x1;y1;z1) v M2(x2;y2;z2) ng thi tho mn iu kin a. Vung gc vi mt phng b. Song song vi ng thng d (hoc trc Ox, Oy, Oz) c. C khong cch t im M ti l h d. To vi mt gc Q mt gc Bi tp gii mu: Bi 1: (SGK Ban C Bn T80) Trong khng gian vi h to Oxyz .Vit phng trnh mt phng i qua hai im 1;0;1 , 5; 2;3M N v vung gc vi mt phng : 2 7 0x y z Gii: Cch 1 : Mt phng i qua hai im M(1;0;1); N(5;2;3) v vung gc vi mt phng ( ) mt phng i qua im M v n

MN ; n

n (vi MN v n

khng cng phng)

mt phng i qua im M v c vtpt n = [ MN , n

] = 4;0; 8 = 4 1;0; 2

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

- Mt phng i qua 1;0;1M .1 .0 .1 0 1A B C D - Mt phng i qua 5; 2;3N .5 .2 .3 0 2A 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 c 2 0B chn

1, 0 2, 1A B C D Vy phng trnh mt phng l 2 1 0x z Bi 2: Trong khng gian vi h to Oxyz .Vit phng trnh mt phng (P) i qua hai im

4; 1;1M ; 3;1; 1N v cng phng (song song) vi trc Ox Gii: Cch 1 : Mt phng (P) i qua im 4; 1;1M ; 3;1; 1N v cng phng vi trc Ox mt phng (P) i qua im M v Pn MN

; Pn

i

(vi v i

khng cng phng)

mt phng (P) i qua im M v nhn vtpt Pn = [ , i

] = 0; 2; 2 = 2 0;1;1

mt phng (P) c phng trnh l : 0(x 4) + 1( y + 1) + 1(z 1) = 0 hay (P): y + z = 0 Cch 2: Lm tng t bi 1 (cch 2) iu kin y l Pn

i



Bi 3: (SBT Ban Nng Cao T126) Trong mt phng Oxyz .Vit phng trnh mt phng (Q) i qua hai im 3;0;0 , 0;0;1A C v to vi mt phng Oxy mt gc = 60o Gii: Cch 1: Mt phng (Q) i qua A, C v to vi mt phng Oxy mt gc bng 60o nn mt phng (Q) ct mt phng Oxy ti im B(0;b;0) Oy khc gc to O b 0 mt phng (Q) l mt phng theo on chn c phng trinh l :

113

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

mt phng (Q) c vtpt Qn = (b;3;3b)

Mt phng 0xy c vtpt k

= (0;0;1) .Theo gi thit ,ta c

|cos ( Qn , k

)| = cos60o

21

99

32

bb

b

263

269996 22 bbbbb

Vy c hai mt phng tho mn l : (Q1) : x 26 y + 3z 3 = 0 (Q2) : x + 26 y + 3z 3 = 0 Cch 2: v A Ox v C Oz Gi AB l giao tuyn ca mt phng (Q) v mt phng 0xy .T O h OI AB . Theo nh l ba ng vung gc ta c AB CI 060OIC

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

Trong vung OAB ta c 222111

OBOAOI 232

131

33

1OB

OB =

263

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

113

263

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

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

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

: 23 2

x td y t

z t

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

; n

du (vi MN

v du

khng cng phng) mt phng i qua im M v c vtpt n

= [ MN

, du ] = 10; 4;1

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



- Mt phng i qua 2;1;3M .2 .1 .3 0 1A B C D - Mt phng i qua 1; 2;1N .1 . 2 .1 0 2A B C D - Mt phng song song vi ng thng d . 0 .1 .2 . 2 0 3dn 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 c 2 5A B chn

1 195, 2 ,2 2

A B C D

Vy phng trnh 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 trnh mt phng (P) qua hai im A v B, ng thi khong cch t C ti mt phng (P) bng

3 . Gii: 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

- Mt phng P i qua 1;1;0A . 1 .1 .0 0 1A B C D - Mt phng P i qua 0;0; 2B .0 .0 . 2 0 2A B C D

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

C 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 2

22 2

172; 3 3 5 2 7 0 151

2

A B A B A BA Ad I P A AB BB B

A B A B

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

Vi 75

AB chn 7, 5 1, 2 : 7 5 2 0A 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 2

2; 3 3 2 16 14 0

7( 2 )

a c a cd C P a ac c

a ca a c c

TH 1: ca ta chn 1 ca 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 trnh 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 QAB n AB n

V ; 0QAB n

nn mt phng (P) nhn ; QAB n

lm vc t php tuyn

Vy (P) c phng trnh x 2y + z 2 =0 Bi 8: Trong khng gian ta Oxyz cho 2 im I( 0;0;1) v K( 3;0;0). Vit phng trnh mt phng qua I, K v to vi mt phng (xOy) mt gc bng 030 Gii:

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

( ) 1 ( ) 3 ( ) : 13 1x y zDo I c v do K a

b

000

0

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

x yx y

x y

n nn v n k b

b n n

( ) : 13 13 2

2

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;1A B C v 0;3;1D . Vit phng trnh 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

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

T (1) v (2) ta c 3 1 5,2 2 2

c a b d a b

Nn mt phng P c phng trnh l 3 1 5 02 2 2

ax by a b z a b

Theo gi thit , ,d C P d D P

2 22 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 14, 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) qua I (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;2A B v mt phng : 2 3 3 0Q x y z . Lp phng trnh mt phng (P) i qua A, B v vung gc vi (Q). s: : 2 2 0P x y z Bi 2: Lp phng trnh mp(P) i qua 0;3;0 , 1; 1;1M N v to vi mt phng : 5 0Q x y z

mt gc vi 1cos3

Bi 3: Lp phng trnh mt phng (P) i qua 1; 1;3 , 1;0;4M 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 trnh mt phng i qua hai im 1;2;3 , 2; 2;4M N v song song vi Oy.

(Ti liu n thi tt nghip nm 2009) s: : 2 0x z Bi 5: Trong khng gian vi h to Oxyz cho mt phng : 2 3 7 0P x y z . Vit phng trnh mt phng ( ) i qua 1;1;0 , 1;2;7A B v vung gc vi (P).

(Ti liu n thi tt nghip nm 2009) s: :11 8 2 19 0x y z Dng 3: Vit phng trnh mt phng (P) i qua ba im Mo(xo;yo;zo) M1(x1;y1;z1) v M3(x3;y3;z3) khng thng hng cho trc Phng 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 0M v c VTPT n

Cch 2: - Gi s phng trnh mt phng P l 0 1Ax By Cz D 2 2 2( 0)A B C - V P i qua ba im 0 1,M M v 2M thay ta vo phng trnh (1) c h 3 n, 3 phng trnh theo ,A B v C . Gii h ny ta c ,A B v C - Thay vo phng trnh (1) ta c phng trnh mt phng P Bi tp gii mu:



Bi 1: (SGK Ban C Bn T80) Trong khng gian vi h to Oxyz .Vit phng trnh mt phng i qua ba im M 3;0;0 ; 0; 2;0N v 0;0; 1P Gii: Cch 1: Mt phng i qua ba im 3;0;0M ; 0; 2;0N v 0;0; 1P mt phng i qua im M v n

MN

; n

MP

(vi MN

v MP

khng cng phng)

mt phng i qua im M v nhn vtpt n = [ MN

, MP

] = (2;3;6)

mt phng c phng trnh l : 2(x + 3) + 3(y 0 ) + 6(z 0) = 0 hay : 2x + 3y + 6z + 6 = 0 Cch 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 1A B C D - Mt phng i qua 0; 2;0N .0 . 2 .0 0 2A B C D - Mt phng i qua 0;0; 1P .0 .0 . 1 0 3A 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 trnh l 2 3 6 6 0x y z Cch 3: Nhn thy M 3;0;0 Ox ; N 0; 2;0 Oy v P 0;0; 1 Oz nn phng trnh mt phng l mt phng theo on chn c dng :

1123

zyx hay : 2 3 6 6 0x y z

Dng 4: Vit phng trnh mt 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 trnh mt phng cn tm Bi tp gii mu: Bi 1: (SGK Ban C Bn T80) Trong khng gian vi h to Oxyz .Vit phng trnh trung trc ca on thng AB vi A(2;3;7) v B(4;1;3) Gii: Cch 1: Gi I l trung im ca on thng AB I(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 vtpt

mt phng trung trc (P) c phng trnh l: 1(x 3) 1(y 2) 2(z 5 ) = 0 hay : 2 9 0 P 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 3MA 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 trnh l : x y 2z + 9 = 0 Cch 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' DD D = 9

Vy mt phng trung trc (P) c phng trnh l: 2 9 0 x y z Nhn xt : - Bi ton ny thc cht l bi ton vit phng trnh mt phng i qua mt im v vung gc gi ca mt vect (thuc dng 1) - Vect i qua hai im cho trc c coi l mt vtcp ca ng thng Bi tp t gii: Bi 1: Trong khng gian vi h to Oxyz cho 2 im 1; 4;5 , 3; 2;7E F . Vit phng trnh mt

phng ( ) l trung trc ca on thng EF.

( thi tt nghip THPT h phn ban ln 2 nm 2007) s: : 3 5 0x y z Dng 5: Vit phng trnh mt phng (P) song song v cch u hai hai ng thng ( 1 ) v ( 2 ) cho trc Phng php: - Mt phng (P) song song vi hai ng thng 1 2v nn c vtpt 1 2;Pn u u

- mt phng (P) cch u hai ng thng 1 2v nn (P) i qua trung im ca I ca MN vi

1 2M v N ...Quay v dng 4

Bi tp gii mu: Bi 1: (HSP HN 2 98) Trong khng gian Oxyz cho hai ng thng c phng trnh l

d :

tztytx

212

v d :

03022

yzx

a. Chng minh rng d v d cho nhau b. Vit phng trnh mt phng (P) song song v ng thi cch u d v d Gii: a. Chn im M(2;1;0) d v d c vtcp 1; 1; 2du

,chn im N(0;3;1) d v d c vtcp ' 2;0;1du

.Tnh n = [ du , 'du

] = 1; 5; 2 (vi du v 'du

khng cng phng) v 2;2;1MN

. 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 trnh l :



1(x 1) 5(y 2) 2

21z = 0 hay (P) : x + 5y + 2x 12 = 0

Bi 2: Vit phng trnh mt phng cch u hai ng thng d1 v d2 bit:

1

2: 2

3

x td y t

z t

v 21 2 1:

2 1 5x y zd

Gii:

ng thng d2 c phng trnh tham s l: 1 2 '2 '1 5 '

x ty tz 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 dn 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

Vy PT mp( ) l: 3x y 4z + 7 0 Dng 6: Vit phng trnh mt phng (P) song song v cch u hai mt phng (Q1) v Q2 (vi Q1 v Q2 song song vi 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 ti mt phng kia Bi tp gii mu: Bi 1: Trong khng gian Oxyz cho hai mt phng (P) v (Q) c phng trnh l (P) : 3x y + 4z + 2 = 0 v (Q) : 3x y + 4z + 8 = 0 Vit phng trnh 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 cch u (P) v (Q) nn , ,d M d N

1619

'0.420.3

D =

1619'0.480.3

D 8'2' DD D = 4

Vy mt phng ( ) c phng trnh l :3x y + 4z + 4 = 0 Dng 7: Vit phng trnh mt phng (P) tip xc vi mt mt cu (S) v tha mn mt iu kin cho trc Phng php: - Bc 1: Xc nh tm I v bn knh R ca mt cu (S) v vtpt hoc vtcp - Bc 2: T iu kin cho trc xc nh vtpt Pn

, gi s ; ;Pn a b c khi mt phng P c dng

'ax 0by cz D vi ' 0D (1) - Bc 3: Mt phng (P) tip xc vi mt cu (S) ,d I P R , t y c phng trnh theo D, gii phng trnh (ti tuyt i) c D thay vo (1) ta c phng trnh mt phng P cn tm



- 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 dn 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 VTPT l MI

Bi tp gii mu: Bi 1: (SGK Ban C Bn T93) Trong khng gian vi h to Oxyz. Vit phng trnh mt phng tip xc vi mt cu (S) c phng trnh 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 td y t

z t

7 3 '

: 1 2 '8

x td y t

z

Gii : Ta c 2; 3; 2du

v ' 3; 2;0du .

Mt cu (S) (x 5)2 + (y + 1)2 + (z + 13)2 = 25 mt cu (S) c tm 5; 1; 13I v bn knh R = 5 Mt phng song song vi d v d mt phng c n

du ; n

'du (vi du

v 'du khng cng phng )

mt phng c vtpt n = [ du

, 'du ] = (4;6;5)

mt phng lun c dng 4x + 6y + 5z + D = 0 Mt 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 + 5 77 = 0 2 : 4x + 6y + 5z + 51 5 77 = 0 Bi 2: (SBT Ban Nng Cao T138) Trong khng gian Oxyz. Vit phng trnh mt phng (P) tip xc vi mt cu (S) v vung gc vi ng thng d c phng trnh ln lt l :

(S): x2 + y2 + z2 10x + 2y + 26z 113 = 0 v 5 1 13:2 3 2

x y zd

Gii: ng thng d c vtcp 2; 3;2du

. Mt cu (S) (x 5)2 + (y + 1)2 + (z + 13)2 = 308



mt cu (S) c tm 5; 1; 13I v bn knh 308R Mt phng (P) vung gc vi ng thng d nn c vtpt 2; 3;2P dn 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 5236

4 9 4D

D D

Vy c hai mt phng (P) tha mn u 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 trnh mt phng (P) song song vi mt phng (Q) v tip xc vi mt cu (S) c phng trnh 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: Mt phng (Q) c vtpt 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 = 4 Mt phng (P) song song vi mt phng (Q) mt phng (P) lun c dng 4x + 3y 12z + D = 0 Mt phng (P) tip xc vi mt cu (S) ,d I P R

4.1 3.2 12.3 ' ' 264 ' 26 52

' 7816 9 144D D

DD

Vy c hai mt phng tha mn u 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 trnh mt phng ( ) song song vi trc Oz, vung gc vi mt phng (P): x + y + z = 0 v tip xc vi mt cu (S): x2 + y2 + z2 2x + 2y 4z 3 = 0 Gii: Mt phng (P) c vtpt Pn

= (1;1;1) . Mt cu (S) (x 1)2 + (y + 1)2 + (z 2)2 = 9 mt cu (S) c tm 1; 1; 2I v c bn knh R = 3 Mt phng ( ) song song vi trc Oz 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 mt cu (S) ,d I P R

1.1 1.( 1) '3 ' 2 3 2 2 3 2

1 1D

D D

Vy c hai mt phng tho mn u 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 trnh mt phng (P) tip xc vi mt 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 v nhn 1;2;2IM

lm vtpt vi 3;1; 2I l tm ca mt cu (S)

mt phng (P) c phng trnh l: 1(x 4) + 2(y 3) + 2(z 0 ) = 0 hay (P): x + 2y + 2z 10 = 0 Bi 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 trnh mt phng (P) song song vi gi ca vc t (1;6; 2)v

, vung gc vi mt phng

( ) : 4 11 0x y z v tip xc vi (S). Gii: Ta c mt cu (S) c tm 1; 3;2I v bn knh 4R Vc t php tuyn ca ( ) l (1;4;1)n

V ( ) ( )P v song song vi gi ca v

nn nhn vc t (2; 1;2)pn n v

lm vtpt.

Do : 2 2 0P x y z m

V (P) tip xc vi (S) nn 21

( , ( )) 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 trnh mt phng tip xc vi mt cu

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

: 1 21 3

x td y t

z t

s: 2 3 7 3 14 0x y z v 2 3 7 3 14 0x 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 0x y z

dx y z

v 1 2 :1 2 2

x y zd

. Vit phng trnh mt phng

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

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

Bi 6: Vit phng trnh mt phng song song vi 2 1:1 3 1

x y zd

, vung gc vi

: 2 1 0 P 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 0x y z v 4 3 5 11 15 2 0x 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 trnh mt phng l tip

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

Dng 8: Vit phng trnh mt phng cha mt ng thng cho trc v tho mn iu kin Loi 1: Vit phng trnh 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. Ly mt im M trn 4. p dng cch vit phng trnh mt phng i qua 1 im v c 1 VTPT

Ch : Thc cht y l bi ton vit phng trnh mt phng i qua hai im v vung gc vi mt mt phng Loi 2: Vit phng trnh 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


Ch : Thc cht y l bi ton vit phng trnh mt phng i qua hai im v song song vi mt ng thng Loi 3: Vit phng trnh mt phng cha ng thng v 1 im M Phng php:

1. Tm VTCP ca l u

, ly 1 i m N trn . Tnh ta MN

2. VTPT ca mt phng l: n u MN

3. p dng cch vit phng trnh mt phng i qua 1 im v c 1 VTPT

Ch : Thc cht y l bi ton vit phng trnh mt phng i qua ba im phn bit cho trc Loi 4: Vit phng trnh mt phng cha ng thng v to vi mt phng (hoc ng thng d ) mt gc Loi 5: Vit phng trnh 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 trnh mt phng P a. i qua im 2;1; 1oM v qua giao tuyn ca hai mt phng Q v R c phng trnh ln lt l:

4 0x y z v 3 1 0x y z



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

:

01304

zyxzyx

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 = [ 0M M

, 0M N

] = 7;7;15411

477;

477;

4165

(vi

0M M

v 0M N

khng cng phng ) mt phng (P) c phng trnh 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 trnh

4 0

:3 1 0x y zx 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 0Ax 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; 1oM .2 .1 . 1 0 3A B C D Gii h (1), (2) v (3) ta c 15, 7, 7, 16 : 15 7 7 16 0A B C D P x y z Cch 3: S dng phng php 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 i qua 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 trnh

:

054023

yxzyx

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 trnh l : 1(x 5) + 22(y 0 ) 2(z + 13) = 0 hay (P) : x 22y + 2z + 21 = 0



Hoc c th tnh ,Pn u n

Nhn xt: Thc cht bi ton ny chnh l bi ton ny chnh l bi ton vit phng trnh mt phng i qua hai im 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 trnh

:

054023

yxzyx

Chn hai im 5;0; 13M v 1;1;0N 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

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

T (1) v (2) ta c 413A BC v D A B

Nn mt phng P c vtpt 4; ;13P

A Bn A B

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

4. .2 .0 . 1 0 2213P

A Bn n A B A B

chn 1, 22 2, 21A B C D

Vy mt phng P c phng trnh l 22 2 21 0x 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

1

2 4 0:

2 2 4 0x y zx y z

2

1: 2

1 2

x ty tz t

Vit phng trnh mt phng P cha ng thng 1 v song song vi ng thng 2 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 thng 1 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 trnh l : 2(x 0 ) + 0(y + 2) 1(z 0 ) = 0 hay 2 0x z Hoc C 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;0N 1


- 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;0N .0 . 2 .0 0 2A B C D

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

C A B v 2D B

Nn mt phng P c vtpt 1 3; ;2 4P

n A B A B

ng thng 2 c vtcp 2 1;1;2u

, mt phng P song song vi ng thng 2

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 trnh 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 trnh mt phng P i qua giao tuyn ca hai mt phng : 3 0x y z v : 3 5 1 0x y z ng thi song song vi mt phng : 2 3 0x y z Gii: Cch 1: Gi l giao tuyn ca ( ) v ( ) c phng trnh

:

015303

zyxzyx

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 tuyn v 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 trnh x + y + 2z + 1 = 0 Hoc: 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 trnh

3 0

:3 5 1 0x y zx y z


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

T (1) v (2) ta c 32

B AC v 2 D B A

Nn mt phng P c vtpt 3; ;2P

B An A B



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

Pn v n

cng phng

2.23

11ABBA

chn 1, 1 2, 1A B C D

Vy mt phng P c phng trnh l 2 1 0x y z Cch 3: S dng phng php 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 i qua mt im 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; 18A B v mt phng : 2 1 0P x y z . Vit phng trnh mt phng cha AB v vung gc vi mp (P). Gii: Gi (Q) l mt phng cn tm Ta c ( 2, 4, 16)AB

cng phng 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 c vtpt , 6;15;3 3 2;5;1Qn n a

Chn vtpt 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 trnh 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 trnh l:

d : zyx

12 v d :

153

22

zyx .

Vit phng trnh mt phng )( i qua d v to vi d 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 vtpt n

vung gc vi v u

ng thi to vi ng thng d mt gc 030 tc l 2160cos)';cos( 0 un

Ta c h

21

6

20

222 CBACBA

CBA

02)(632 22222 CACACAB

CCAAA

CAB

Gii phng trnh 2 22 0 ( )(2 ) 02A C

A AC C A C A CA C

.

- Nu CA , chn 1A C , khi 2B , tc l (1;2;1)n

v mt phng ( ) c phng trnh l 0)2(2 zyx hay 2 4 0x y z

- Nu CA 2 , chn 2,1 CA , khi 1B , tc l (1; 1; 2)n

v mt phng ( ) c phng trnh l 02)2( zyx hay 2 2 0x y z

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

2 1 1x y zd

v



22 1:

1 1 1x y zd

. Vit phng trnh mt phng cha 1d v hp vi 2d mt gc 30

0

Gii: 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

Trn ng thng 1d ly 2 im 1;0; 1 , 1;1;0M N

Do P qua ,M N nn: 0 20

A C D C A BA 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 sin 302 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 0A B A AB B A AB B

D thy 0B nn chn 1B , 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 trnh:

1 2

2 3 5 0 2 2 3 17 0: :

2 0 2 2 3 0x y z x y z

d v dx y z x y z

Lp phng trnh mt phng i qua 1d v song song vi 2d . Gii: Gi (Q) l mt phng cn tm ng thng 1d v 2d c vtcp ln lt l 1 2(1; 1; 1); (1; 2; 2)u u

Mt phng (Q) i qua 1d 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)Qn

v 1(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 trnh 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

. Vit phng trnh mt phng P i qua d v to vi mt phng

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

Bi 10: Vit phng trnh mt phng i qua ng thng 2 0

:1 0

x y zd

x y

v cch im 0;0;2M

mt khong 12

h

s: C hai mt phng tha mn l 1 0, 5 4 3 1 0x y x y z Bi tp t gii: Bi 1: Trong khng gian Oxyz cho hai im 1;3; 2 , 3;7; 18A B v mt phng : 2 1 0P x y z . Vit phng trnh mt phng cha AB v vung gc vi mp (P).



s: 2 5 11 0x y z

Bi 2: Vit phng trnh mt phng (P) cha ng thng 8 11 8 30 0

:2 0

x y zd

x y z

v c khong cch

n im 1;3; 2A bng 29 Bi 3: Lp phng trnh mt phng (P) cha trc Oz 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 trnh mt phng () cha 1 1 2:2 1 5

x y zd sao cho khong cch t 5;1;6A

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

Bi 6: Trong khng gian vi h to Oxyz cho ng thng d c phng trnh 3

12

11

2

zyx v

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

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

Dng 9: Vit phng trnh mt phng cha hai ng thng 1 v 2 ct nhau hoc song song vi nhau Loi 1: Vit phng trnh mt phng cha 2 ng thng ct nhau v Phng php:

1. Tm VTCP ca v l u

v 'u

2. VTPT ca mt phng l: 'n u u


Loi 2: Vit phng trnh 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 trnh mt phng i qua 1 im v 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 2x y zd

v 2

2 0:

3 12 0x y z

dx y

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

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

= (3;-1;2) d2 c vtcp 2u

= (3;-1;2) = 1u v M1 d2 vy d1 // d2

- Vit phng trnh 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 trnh l : 15(x 1) + 11(y + 2) 17(z + 1) = 0 hay (P) : 15x + 11y 17z 10 = 0 Cch 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 trnh 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 mt phng (P) c phng trnh l : 15x + 11y 17z 10 = 0 Bi 2: (SBT Ban C Bn T115) Trong khng gian Oxyz cho hai ng thng

d1 : 4

532

21

zyx v d2 :

tztytx

212237

a. Chng minh rng d1 v d2 cng nm trong mt mt phng ( ) b. Vit phng trnh mt phng ( ) Gii: a. Chn M1(1;-2;5) d1 v d1 c vtcp n

= (2;-3;4) ,chn M2(7;2;1) d2 v d2 c vtcp 2u = (3;2;-2). Tnh

n = [ 1u , 2u ] = (-2;16;13) v 1 2M M

= (6;4;-4)

Xt n . 1 2M M

= (-2).6 + 16.4 +13.(-4) = 0 d1 v d2 cng nm trn mt phng ( ) ,mt khc ta c 1u

2u d1 v d2 ct nhau

b. Cch 1: Mt phng ( ) cha d1 v d2 mt phng ( ) i qua M1 v c vtpt n

= n

mt phng ( ) c phng trnh l : - 2(x 1) + 16(y + 2) + 13(z 5) = 0 hay ( ) : 2x 16y 13z + 31 = 0 Cch 2: C th ly vtpt n = [ 1u

, 1 2M M

] = (-4;32:26) = -2(2;-16;-13) Mt phng ( ) i qua M v c vtpt n mt phng ( ) c phng trnh l : 2x 16y 13z + 31 = 0 Cch 3:

Chuyn d1 v dng tng qut l d1 :

07340123

zyyx

chn I

37;0;

31 d1 .Mt phng ( ) cha d1 v

d2 l mt phng i qua ba im M1,M2 v I hng 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 hng dn : Lm tng t nh bi 1 (cch 3) dng Nhn xt : Bi 3: Trong khng gian ta Oxyz cho 2 ng thng c phng trnh:



1 2

5 27 0

: 1 :2 3 16 0

5

x tx y z

d y t v dx y z

z t

Vit phng trnh mt phng cha 1 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 0

Q d

M d N d MN

v n u MN Q x y z

hay Q x y z

Dng 10: Vit phng trnh 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 trnh mt phng P song song vi mt phng : 2 0Q x y z v cch n mt khong 3h Gii: Mt phng P song song vi mt phng Q c dng 0x y z C Ly im 2;0;0M Q .

Ta c 2 1

, , 353

m md P Q d M P

m

Vy phng trnh mt phng Q cn tm l 1 0; 5 0x y z x y z Bi tp t gii: Bi 1: Vit phng trnh mt phng / /P Q v cch P mt khong h trong cc trng hp sau: a. Mt phng P c phng trnh 2 2 0x y z v 3h b. Mt phng P c phng trnh 4 0x v 2h s: a. 2 2 9 0;2 2 9 0x y z x y z b. 2 0; 6 0x x Dng 11: Vit phng trnh mt phng i qua im v to vi hai ng thng 1 2 d v d (hoc hai mt phng 1 2 P v P ) cc gc v Ch : S dng cng thc gc gia hai mt phng Bi tp gii mu: Bi 1: Vit phng trnh mt phng P i qua im 1;2;3M v to vi mt phng Ox, Oy cc gc tng 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;0j

. Theo gi thit ta c h

0

2 2 2

2 2 20

2 2 2

1sin 45222

1sin 302

AA BA BA B C

B C BA B CA B C

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

2 1 2 3 0; 2 1 2 3 0x y z x y z Bi 2: Cho mt phng P c phng trnh 2 0x y z v im 2; 3;1M . Vit phng trnh mt phng Q i qua M vung gc vi mt phng v to vi mt phng mt gc 045 Gii: Gi ; ;n A B C

l vtpt ca mt phng Q . Theo gi thit ta c h phng trnh

2 2 2

2 0

12

A B CA

A B C

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

Vy phng trnh mt phng Q i qua im 2; 3;1M l 1 0x y hoc 5 2 3 3 4 1 0x y z

Bi tp t gii: Bi 1: Vit phng trnh mt phng mt Q trong cc trng hp sau: a. Vung gc vi cc mt phng 1 2,P P c phng trnh ln lt l 2 0; 1 0x z x y z v i qua gc ta b. To vi cc mt phng 1 2,P P c phng trnh ln lt l 2 0; 1 0x z x y z cc gc

tng ng l 045 v gc vi 1cos3

ng thi i qua gc ta

s: a. 0x z b. 0; 0x z Bi tp gii mu tng hp: Bi 1: Trong khng gian vi h ta Oxyz cho im 10;2; 1A v ng thng d c phng

trnh1 2

1 3

x ty tz t

. Lp phng trnh 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 trn d, mt phng (P) i qua A v / /P d , khi khong cch gia d v (P) l khong cch t H n (P). Gi s im I l hnh chiu ca H ln (P), ta c HIAH HI ln nht khi IA



Vy (P) cn tm l mt phng i qua A v nhn AH lm vc t php tuyn. )31;;21( tttHdH v H l hnh chiu ca A trn d nn )3;1;2((0. uuAHdAH l vc t

ch phng ca d) )5;1;7()4;1;3( AHH Vy (P): : 7 10 2 5 1 0P x y z

7 5 77 0x y z

Bi 2: Cho ng thng D c phng trnh:2

: 22 2

x tD y t

z t

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

song song vi d v 2;0; 2I l hnh chiu vung gc ca A trn D. Trong cc mt phng qua , hy vit phng trnh ca mt phng c khong cch n l 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 trn (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) vung gc vi IA ti A. Vect php tuyn ca (P0) l 6;0; 3n IA

, cng phng vi 2;0; 1v

.

Phng trnh ca mt phng (P0) l: 2 4 1. 1 2 9 0x 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 trnh mt phng (P) i qua im M song song vi ng thng ng thi khong cch gia ng thng v mt phng (P) bng 4. Gii: Gi s ( ; ; )n a b c

l mt vect php tuyn ca mt phng (P).

Phng trnh 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

T gi thit ta c 2 2 2

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

n u a b cP

a bd A Pa b c

Th 4b a c vo (2) ta c 2 2 2 2 24

( 5 ) (2 17 8 ) 2 8 02

aca c a c ac a ac cac

Vi 4ac chn 4, 1 8a c b . Phng trnh mt phng : 4 8 16 0.P x y z

Vi 2ac chn 2, 1 2a c b . Phng trnh 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 thng d ln lt c phng

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

x y zd



1. Vit phng trnh mt cu c tm thuc ng thng d, cch mt phng (P) mt khong bng 2 v vt mt phng (P) theo giao tuyn l ng trn c bn knh bng 3. 2. Vit phng trnh mt phng (Q) cha ng thng d v to vi mt phng (P) mt gc nh nht. Gii:

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

x ty t t Rz 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

tt t t td It

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 trn c bn knh bng 4 nn mt cu c bn knh l R = 5. Vy phng trnh mt cu cn tm 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 trnh tng qut l: 2 1 0

:2 0

x yx 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 tm l 6 3cos 19 3

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

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

| 3 | 3cos33. 5 2 4

mm n mn

m2 + 2mn + n2 = 0 (m + n)2 = 0 m = n. Chn m = 1, n = 1, ta c: mt phng (Q) l: x + y z + 3 = 0 Bi 5 : Trong khng gian vi h ta Oxyz, cho 1;2;3 .M Lp phng trnh mt phng i qua M ct 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, , , 0x y z a b ca b c

Do M nn: cos

31 2 3 61 3. 162

yabc

a b c abc

Th tch: min

31 27 27 66

9

aV abc V b

c

Mt phng cn tm : 6 3 2 18 0x y z



Bi 7: Vit phng trnh mt phng (P) qua O, vung gc vi mt phng : 0Q x y z v cch im 1; 2; 1M mt khong bng 2 Gii: Phng trnh mt phng (P) qua O nn c dng : Ax + By + Cz = 0 vi 2 2 2A B C 0 V (P) (Q) nn 1. 1. 1. 0 0A B C A B C C A B (1) Theo :

2 2 2 22 2 2

2; 2 2 ( 2 ) 2( )

A B Cd M P A B C A B C

A B C

(2)

Thay (1) vo (2) , ta c : 20

8AB 5 0 8B =5

BB A

(1)0 .B C A Chn 1, 1A C th : 0P x z

8B =5A

. 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 phng trnh (S): 2 2 2 4 4 2 16 0x y z x y z ; ( ) : 2 2 1 0.P x y z Vit phng trnh mt phng (Q) song song 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 0x y z x y z (S) c tm I(2;2;-1) phng trnh mt phng (Q) c dng: 2 2 0x y z D iu kin 1(*)D

( , ( )) 3d I P 2 2 2

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

D

1| 8 | 9

17D

DD

Kt hp vi iu kin (*) ta c D = -17 Vy phng trnh ca : 2 2 17 0Q x y z Bi 9: (H D 2010) Trong khng gian to Oxyz, cho hai mt phng (P): x + y + z 3 = 0 v (Q): x y + z 1 = 0. Vit phng trnh mt phng (R) vung gc vi (P) v (Q) sao cho khong cch t O 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

Phng trnh (R) c dng : x z + D = 0. Ta c : d (0;(R)) = 2 2 2 22

DD

Phng trnh (R) : 2 2 0 2 2 0x z hay x z

Bi 10: Trong khng gian Oxyz, cho mt cu (S) c phng trnh 2 2 21 2 3 14x y z v im 1; 3; 2M . Lp phng trnh mt phng (P) i qua sao cho (P) ct (S) theo mt giao tuyn l ng trn c bn knh nh nht. Gii: Ta thy M thuc min trong ca (S) v (S) c tm 1; 2; 3 , 14I R . Do , (P) qua M ct (S) theo mt giao tuyn l ng trn c bn knh nh nht

2 2R IH nh nht (H l hnh chiu vung gc ca I trn mt phng (P)) IH ln nht 0;1; 1M H IM

l VTPT ca (P). Vy (P) c phng trnh l 1 0.y z



Bi 11: Trong khng gian Oxyz, vit phng trnh mt phng (P) cha ng

thng 2 0

:2 6 0x y

dx z

sao cho giao tuyn ca mt phng (P) v mt

cu 2 2 2: 2 2 2 1 0S x y z x y z l ng trn 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 trn 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 nm n m n m n

m 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 0P x y zP x y z

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

bng 12536

.

Gii: Phng trnh mt phng Oxy: z = 0 Phng trnh mt phng (P) thuc chm xc nh bi d v Oxy c dng: 2 5 0 m x y z nz : 2 5 0P mx my m n z m

- 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

mA B Cm n

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

mV OA OB OCm n

3 1, 23

3 1, 4m n m m n

m n mm n m m n

Vy, c 2 phng trnh mt phng (P): 1

2

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

Bi 13: Trong khng gian ta Oxyz cho im G(1;1;1) a. Vit phng trnh 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 0

Ox : (3;0;0)0

yA

z

Tng t: (0;3;0) (0;3;0)B v C



Ta c: 3 2AB BC CA ABC l tam gic u Bi 14: (SGK Ban Nng Cao T89) Trong khng gian Oxyz . Vit phng trnh 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 ca tam gic ABC i 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 ABC Gii: 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 > 0 Trng tm G ca tam gic ABC c to l:

G

3

;3

;3

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

Vy mt phng (P) c phng trnh l :

1cz

by

ax 1

963

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

b. Ta c AB CH (v H l trc tm caABC) 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 nhn OH = (2;1;1) lm vtpt c phng trnh l : 2(x 2) + 1(y 1) + 1(z 1) = 0 hay (P) : 2x + y + z 6 = 0 Bi 15: Trong khng gian vi h ta Oxyz, vit phng trnh 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.

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 trnh mt phng 1x y zPm n p . V D (P) nn: 1 1 1 1

m n p

.

D l trc tm ca MNP

. 0 03

. 0 03

( ) ( ) 1 1 1 1

DP NM DP NM m nm

DN PM DN PM m pn p

D P D Pm n p

Kt lun, phng trnh ca mt phng : 13 3 3

x y zP

.

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

v

d: 1 21 2 1

x y z . Vit phng trnh 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 = 0 Mt phng (P) ct Oz ti A(0; 0; -c), ct d ti B(1- c; -2c; -2-c)

Theo gi thit 2

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



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

Vit phng trnh mt phng (P) qua M, song song v khong cch gia ng thng v mt phng (P) bng 3. HD: Gi s phng trnh mt phng (P) c dng Ax + By + Cz + D = 0

T gi thit ta c h

2 2 2

3 2 04 0

3

B C DA B C

C D

A B C

28

B CB C

TH 1: 2B C chn 1, 2 2, 8C B A D TH 2: 8B 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 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

: 1x ty t t Rz t

, 21:

1 1 1x y z

. Vit phng trnh 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 , 3I 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; 3B M . Vit phng trnh mt phng ( )P cha ,B M v ct cc trc ,Ox Oz ln lt ti cc im A v C sao cho th tch khi t din OABC bng 3 ( O l gc to ). Gii: Gi ,a c ln lt l honh , cao ca cc im ,A C .

V 0;3;0B Oy nn : 13

x y zPa c .

4 34;0; 3 1 4 3M 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 h 46 6 234 3 6 4 3 6 32

aac ac ac a c a cc

Vy 1 22: 1; : 1

4 3 3 2 3 3x y z x y zP P



Bi 20: Cho im M(1; 2; 3). Lp phng trnh mt phng (P) bit rng (P) ct ba trc Ox, Oy, Oz ln lt ti A, B, C sao cho M l trng tm ca tam gic ABC. Gii: Do A, B, C ln lt thuc Ox, Oy, Oz nn ta gs A(xA ; 0 ; 0), B(0 ; yB ; 0), C(0 ; 0 ; zC).

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

MCBA

MCBA

MCBA

zzzzyyyyxxxx

333

963

C

B

A

zyx

.

Mt phng (P) i qua A(3 ; 0 ; 0), B(0 ; 6 ; 0), C(0 ; 0 ; 9)

Nn (P) c phng trnh l 1963

zyx .018236:)( zyxP

(phng trnh mt phng theo on chn) Bi 21: Trong khng gian vi h trc ta Oxyz cho hai im (0; 1;2)M v ( 1;1;3)N . Vit phng trnh 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 0A B C l mt vect php tuyn ca mt phng (P).

Phng trnh mt phng (P) c dng; 1 2 0 2 0Ax B y C z Ax By Cz B C

1;1;3 3 2 0 2N 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

BK P

B C BCd

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

- Nu 0B th 2 2 2

1 1,24 2 4

2 1 2

Bd K P

B C BC CB

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

0 02 0 2

D DA B A B

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

Theo gi thit th: 2 2 2 2

4 3, ,

5 5

B Cd B P d C P

B C B C

34 3 4 34

BB C B CC

Chn C = 4 suy ra B = 3 Vy 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 trnh 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,C l im nm trn Ox, Oy, Oz tng ng c honh , tung v cao dng v 4OA = 2OB = OC suy ra A(a;0;0) , B(0;2a;0) v C(0;0;4a) vi 0a

Phng trnh (ABC) 1 4 2 4 02 4

x y z x y z aa a a

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

a

Vy phng trnh (ABC) l 4 2 17 0.x y z Bi 24: Trong khng gian vi h ta Oxyz. Cho mt cu (S) : 921 222 zyx .

Lp phng trnh mt phng (P) vung gc vi ng thng 1:1 2 2x y zd

v ct mt cu (S) theo

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

+ ng thng d c vtcp (1,2 , 2 )u

, (P) vung gc vi t a nn (P) nhn u lm vtpt

Phng trnh mt phng (P) c dng : 2 2 0x y z D

+ (P) ct (S) theo ng trn 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 zyx v (P2) : 053522 zyx Bi 25: Trong khng gian vi h to Oxyz, cho mt cu (S) c phng trnh 011642222 zyxzyx v mt phng () c phng trnh 2x + 2y z + 17 = 0. Vit phng trnh mt phng () song song vi () v ct (S) theo giao tuyn l ng trn c chu vi bng 6. Gii: Do () // () nn () c phng trnh 2x + 2y z + D = 0 (D 17) Mt cu (S) c tm I(1; 2; 3), bn knh R = 5 ng trn 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 22.1 2( 2) 3 74 5 12

172 2 ( 1)

Vy () c phng trnh 2x + 2y z 7 = 0 Bi 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 trnh mt phng (P). Bit rng mt 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 = 2 Mt 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 cb

a b c



2 2 2 2 2 2

22 6 22 6 2 6

(I)2 59 4 4 4 3 10 05 11

2

a ca c b b ca 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 trnh mt phng : 2 1 2 1 2 0P c x c y c z

2 2 6 0x 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 hp t gii:

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

x y zd Vit phng trnh mt phng

cha d sao cho khong cch t A n ln nht. s: 4 3 0x y z Bi 2: Trong khng gian vi h ta Oxyz cho hai ng thng d v d ln lt c phng trnh :

2:1

yd x z

v 2 5 : 32 1

x zd y

. Vit phng trnh mt phng )( i qua d v to vi d

mt gc 030 s: 2 4 0 ; 2 2 0x 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 trnh mt phng (P) qua hai im A v B, ng thi khong cch t C ti mt phng (P) bng

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

Bi 5: Cho ng thng d c phng trnh: 2

: 22 2

x td y t

z t

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

song song vi d v I(-2;0;2) l hnh chiu vung gc ca A trn d. Trong cc mt phng qua , hy vit phng trnh ca mt phng c khong cch n d. 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 ty tz t

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

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

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

x y zd Vit phng trnh mt phng

cha d sao cho khong cch t A n ln nht. s: 2 2 15 0x 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 trnh mt phng (P) i qua im M song song vi ng thng ng thi khong cch gia ng thng v mt phng (P) bng 4. s: 4 8 16 0x y z hay 2 2 4 0.x y z Bi 9: Vit phng trnh mt phng cch u hai ng thng d1 v d2 bit:

1

2: 2

3

x td y t

z t

v 21 2 1:

2 1 5x y zd

s: 3 4 7 0x y z

Bi 10: Trong khng gian vi h ta Oxyz, cho ng thng 2 0

:2 6 0x y

dx y

v

mt cu 2 2 2: 2 2 2 1 0S x y z x y z . Vit phng trnh mt phng (P) cha d sao cho giao tuyn ca mt phng (P) v mt cu (S) l ng trn 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 yx z

, 21:

1 1 1x y z

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

Bi 12: Lp phng trnh mt phng cha ng thng 8 11 8 30 0

2 0x y z

x y z

v tip 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 ty tz t

v 1

1 1

7: 1 2

8

x ty tz

Vit phng trnh )( tip xc mt cu (S) v song song vi 1 v 2 Bi 14: Lp phng trnh mt phng (P) i qua hai im 1;2;3A v 2;3; 4B v ct mt cu 2 2 2: 2 6 4 15 0S x y z x y z theo giao tuyn l mt ng trn c chu vi 8

Bi 15: Lp phng trnh mt phng (P) song song vi hai ng thng 15 1 13:

2 3 2x y zd

27 1 8:

3 2 0x y zd

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 trnh mt phng (P) i qua im 2,3,1A 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 trnh mt phng (P) i qua im 2,1, 1A 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 zx y z

Bi 18: Lp phng trnh 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 trnh mt phng i qua hai im 0;0;1 ; 3;0;0A B ng thi a. To vi mt phng Oxy mt gc 60o b. Vung gc vi mt phng : 2 3 1 0P x y z



Bi 20: Lp phng trnh 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 a. i qua A (1,3,-2) b. Vung gc vi mt phng: : 2 4 1 0x 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 gc Oxyz, cho 2;5;3A v ng thng 1 2:

2 1 2x y zd

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

s: : 13 6 12x y z

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

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

OA OB OC nh nht

s: : 2 3 14 0x y z Bi 25: Trong khng gian vi h ta Oxyz, vit phng trnh mt phng i qua im 2;5;3M v ct chiu dng ca cc trc ta Ox, Oy, Oz ln lt A, B, C sao cho OA 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;0A v song song vi ng thng 1 2 1:

1 1 1x y zd

. Vit phng trnh mt phng to vi mt phng xOy mt gc nh nht

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

s :

: 05 1: 3 02 2

y z

x y z

CHUYN : VIT PHNG TRNH NG THNG TRONG

KHNG GIAN

1. Vit phng trnh tham s hoc chnh tc



- 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 trnh tham s 0

0

0

x x aty y btz z ct

vi t R l tham s

phng trnh chnh tc 0 0 0x x y y z za b c

(khi , , 0a b c )

2. Vit phng trnh tng 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 hay

1 1 1 1

2 2 2 2

0:

0a x b y c z d

da x b y c z d

- Tm vecto ch phng u ca ng thng d Cch 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 vtcp u cng phng vi vecto MN

hay vecto MN

chnh l vtcp ca d Cch 2: T phng trnh chnh tc hoc tham s chuyn v phng trnh tng qut Cc dng bi tp Dng 1: Vit phng trnh ng thng i qua im 0 0 0 0; ;M x y z v tha mn iu kin cho trc 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 du 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 phng trnh tham s ng cao tng ng 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; 3AB

, 1;1; 5AC

, 2; 2; 2BC

v , 18;8; 2AB 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



Vy phng trnh ng thng 1

: 2 43 5

x td y t

z t

Bi 2: Trong khng gian vi h trc ta cc vung gc Oxyz, cho 3 im A(0,0,1); B(-1,-2,0) ; C(2,1,-1). 1. Vit phng trnh ca mt phng (P) i qua 3 im A,B ,C 2. Vit phng trnh tham s ca ng thng i qua trng tm ca tam gic ABC v vung gc vi mt phng (P). 3. Xc nh chn ng cao h t A xung ng thng BC Gii: 1. Phng trnh mt phng (P) i qua A,B,C. Ta c VTP (P) l: , (5, 4,3)Pn AB AC

Phng trnh 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)Pda n

Phng trnh 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 trnh tham s ca BC l: 1 32 3

x ty tz 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 v hai im A(-3;0;1), B(1;-1;3). Trong cc ng thng i qua A v song song vi (P), hy vit phng trnh ng thng m khong cch t B n ng thng l nh nht Gii: Ta c (4; 1;2); (1; 2; 2)PAB n

Pt mt phng (Q) qua A v // (P) : 1(x + 3) 2(y 0) + 2(z 1) = 0 x 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.

Pt tham s 1

: 1 23 2

x tBH y t

z t

Ta H = BH (Q) tha h phng trnh :



1 , 1 2 , 3 2 10 1 11 7; ;2 2 1 0 9 9 9 9

x t y t z tt H

x y z

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

a AH

Phng trnh 3 0 1:26 11 2

x y z

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

x y z

Bi tp t gii: Bi 1: Vit phng trnh chnh tc ca ng thng i qua im 1;1;2M 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 5x y z

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

p s:

1: 1 2

1 3

x ty tz t

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

p s:

1 3: 2

2

x td y

z

Bi 4: Vit phng trnh ng thng d i qua im 1;2; 1A v song song vi ng thng giao tuyn ca hai mt phng : 3 0x y z v : 2 5 4 0x y z

p s:

1 4: 2 7

1 3

x td y t

z t

Loi 2: C mt cp vecto khng cng phng cho trc, khi iu kin l - C mt cp vecto ch phng a 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

- 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 bit v A B u AB

Ch : - Nu gi thit l vung gc vi mt vecto c bt k th hiu c l vtcp, - Nu gi thit l song song vi mt vecto d

th hiu d

l vtpt



Bi tp gii mu: Bi 1: Vit phng trnh tham s ca ng thng i qua im 1;1; 2A song song vi mt phng

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

x y zd

Gii: - Mt phng (P) c vtpt 1; 1; 1Pn

v ng thng d c vtcp 2;1;3du

- ng thng song song vi mt phng (P) v vung gc vi ng thng d nn c vtcp , 2;5; 3P du n u

- ng thng i qua 1;1; 2A v c vtcp u

c phng trnh l 1 2

: 1 52 3

x ty tz t

Ch : Nu bi yu cu vit phng trnh ng thng dng tng qut Cch 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 trnh tham s chuyn v phng trnh tng qut Bi 2: Vit phng trnh tham s ca ng thng i qua im 1;1;1M ng thi vung gc vi hai

ng thng 11 2:

8 1 1x y zd v 2

2 3 0:

1 0x y z

dx y z

Gii: - ng thng 1d c vtcp 1 8;1;1u

v 2d c vtcp 2 2; 3;1u

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

- ng thng i qua 1;1;1M v c vtcp u

c phng trnh l 1 4

: 1 61 26

x ty tz t

Ch : Nu bi yu cu vit phng trnh ng thng dng tng qut Cch 1: ng thng chnh l giao tuyn ca hai mt phng - 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 trnh tham s chuyn v phng trnh tng qut Bi tp t gii: Bi 1: Vit phng trnh chnh tc ng thng i qua 1;2;5M ng thi song song vi hai mt phng : 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

Ch : Nu bi yu cu vit phng trnh ng thng dng tng qut Cch 1: ng thng chnh l giao tuyn ca hai mt phng



- 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 trnh tham s chuyn v phng trnh tng qut Bi 2: Vit phng trnh ng thng d i qua im 2; 1;1A v vung gc vi hai ng thng ln lc c vtcp l 1 1;1; 2u

v 2 1; 2;0u

p s: 2 4

: 1 21

x td y t

z t

Bi 3: (H TCKT 1999) Vit phng trnh chnh tc ca ng thng i qua im 1;1; 2A song song 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 zd

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

x y z

Dng 2: Vit phng trnh ng thng i qua mt im 0 0 0 0; ;M x y z ct ng thng d v tha mn iu kin cho trc iu kin cho trc l - Vung gc vi ng thng 1 cho trc - Song song vi mt mt phng (P) cho trc Ch : - Nu gi thit l vung gc vi mt vecto c bt k th hiu c l vtcp, - Nu gi thit l song song vi mt vecto d

th hiu d

l vtpt

Phng php chung: Trng hp 1: Nu bi yu cu vit phng trnh ng thng dng tng qut Cch 1: Xc nh cc vtcp v vtpt ng thng chnh l giao tuyn ca hai mt phng - Mt phng i qua im 0M v cha ng thng d - Mt phng i qua im 0M v tha mn iu kin cho trc Kt lun : - Nu th bi ton c v s nghim - Nu th / /d th bi ton v nghim, ct d th l ng thng cn dng Trng hp 2: Nu bi yu cu vit phng trnh ng thng dng tham s hoc chnh tc Cch 2: Xc nh cc vtcp v vtpt - Mt phng i qua im M0 v tha mn iu kin cho trc (hin nhin mt phng cha ng thng ) - Gi M d , ta M l nghim ca h + Nu khng tn ti giao im th bi ton v nghim + Nu c v s nghim (tc l d ) th bi ton c v s nghim + Nu c nghim duy nht th tnh vecto 0M M

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

Cch 3: Xc nh cc vtcp v vtpt



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

- Tnh vecto 0M M

- T iu kin cho trc ta dn n mt phng trnh bc nht theo tham s t, tm t 0M M

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

Ch : - Vi cch ny th ng thng d phi dng tham s (nu d dng tng qut hay chnh tc th chuyn v dng tham s) - Ta im M phi theo tham s t - tm ta im M ta c th lm nh sau

chuyÊn ĐÊ viẾt phƯƠng trÌnh mẶt phẲng - ĐƯỜng thẲng - mẶt cẦu.pdf

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