JJuang đn gid

Một phần của tài liệu Phương pháp giải toán hình học theo chuyên đề part 3 (Trang 47)

Vi M 6 A r^ M ( t ; 2 t ; l - t ) , A M = ( t - 3 ; 2 t - 2 ; - t + 2), B M - ( t - l ; 2 t + 2 ; - t ) , C M = ( t - 2 ; 2 t - l ; - t - 2 ) B M - ( t - l ; 2 t + 2 ; - t ) , C M = ( t - 2 ; 2 t - l ; - t - 2 ) 1) Taco: M A + MB = V6t^ - 18t +17 + \/6t^ + 6t + 5

St V6 2 /

A p dung bat dSng thuc: Vâ+b^ +ylc^+d^ >^(a + c)^ + (b + d)^ , dSng

thuc xay ra khi — = — / ta c6: b d M A + M B > \ 2 J - l V 2 j = V38 D i n g thuc xay ra « -V6t = V6t + ^<=>t = -^. Vgy M

Phuang phdpgidi Toan Hinh hoc theo chuyen de- Nguyen Phu Khdnh, Nguyen Tat Thu 2)Tac6: A M + CM = Vet^-18t + 17 + ^61^-4t + 9 3V6 -^/6t (Iff St- V6l IN/3. 3V6 76^' ' t 2 - + • + Vl4 ^ 5V3 96 + 5V42

Dang thuc xay ra <» Vay M 714 >/3 2 45-^/42 . 2(45- 7 4 2 ) 4742-35 Ci>t = 3 45 - 742 10 + 3742 • 10 + 3742' 10 + 3742 '10 + 3742

Bdi 3.5.4. Trong khong gian Oxyz cho diem A ( l ; - l ; l ) , dudng thang A c6

X — 1 V z +1

phuang trinh : = y = — va mat phang (P): 2x - y + 2z - 1 = 0 .

1) Viet phuong trinh mat phang (Q) chura duong thang A va khoang each tu A den (Q) Ion nhát.

2) Viet phuong trinh mat phang (R) chiia A va tao vai (P) mpt goc nho nhat. 3) Viet phuang trinh mat phang (a) chua hai diem ]VI(1;1;1),N(-1;2;-1) va tao vai duang thang A mot goc ion nhat.

Jiicong đn giai

Mat ph3ng (P) c6 n ^ = (2; - 1 ; 2) la VTPT

Duong thang A diqua B(l;0;-1) vaco u = ( 2; l ; - l ) laVTCP.

1) Cach 1: Gia su n = (a;b;c) la VTPT cua (Q), suy ra phuang trinh cua (Q) c6 dang: a ( x - l ) + by + c(z + l) = 0<=>ax + by + c z - a + c = 0 (1)

Do A e ( Q ) nen 2a + b - c = 0=>c = 2a + b .

J2c-bl

Do do: d(A,(Q)) = 4a + b 16a2+8ab + b2 Néu b = 0=>d(A,(Q)) =

V a V b ^ T ? V5a2+4ab + 2b2 V5a2+4ab + 2b2 4

Neu b ^ 0 thi ta dat t = ^ , ta c6: ^^â+Sab + b^ ^ 16t^+8t + l _

b 5a2+4ab + 2b2 5 t 2 + 4 t + 2

Xet ham so f(t) vai t e M ta c6: f'(t) = 24t^+54t + 12^ = 0 « t = -2 t = - -

(5t2+4t + 2)2 ' 4

Cty TNHH MTV DWH Khang viea

Suy ra maxf(t) = f(-2) = | , do do m a x d ( A , ( Q ) ) = ^ , dat dupe khi a = -2b

Chon b = - 1 ta tim duQC a = 2,c = 3 . Vay phuang trinh (Q): 2x - y + 3z +1 = 0.

ach 2: Go! K , H Ian luat la hinh chieu ciia A len A va (Q), khi do

d(A,(Q)) = A H < A K , ma A K khong doi nen d(A,(Q)) Ion nhat <=> H = K D i n toi (Q) la mat phang di qua K va nhan AK lam VTPT.

Vi K £ A ^ K ( l + 2 t ; t ; - l - t ) = ^ A K = (2t;t + l ; - t - 2 ) AK 1 A => AK.u = 0«>4t + t + l + t + 2 = 0 1 ^ =>t = - - r ^ K 2 1 1 0 ; - - ; - 5 j . A K = 2' 1) Vay phuang trinh (Q): 2x - y + 3z +1 = 0.

) Cach 1: Tuang t u nhu tren ta c6 (Q): ax + by + (2a + b)z + a + b = 0 Goi a = ((P),(R)), 0 ° < a < 9 0 ° . 2 a - b + 2(2a + b) Ta c6: cos a = 3Vâ+b2+(2a + b)2, 1 3)1 b^ + 12ba + 36â 2 b 2 + 4 a b + 5a2 Neu a = 0=>cosa = 1 372 „ ^„ b , b2+12ba + 36a2 t2+12t + 36 Neu dat t = — thi ta co: — — = —

a 2b^+4ab + 5â 2t^+4t + 5

7 53

Khao sat ham só f(t) ta tim duoc maxf(t) - f( ) = —

\ . ^' M o 6

Suy ra max|cosaJ dat duoc khi — = , chon b = -7=>a = 10 a 10

Vay phuong trinh (R): lOx - 7y + 13z + 3 = 0.

ach 2: Gpi d la duang thing i qua B va vuong goc voi (P)

Ta CO phuong trinh x = l + 2t d : y - - t z = - l + 2t lay C( 3; - l ; l ) e d , C ^ B = f(t)

Phumtg phdp gidi Toatt Hinh hoc theo chuyen de- Nguyen Phil Khdnh, Nguyen Tat Thu

Gpi H , K Ian lirgt la hinh.chiéu ciia C len (R) va A , khi do a = BCH va Sin a = sin BCH = > .

BC BC BK

Ma khong doi, nen suy ra a nho nhat <=> H = K hay (R) la mat phang di qua A va vuong goc voi mat phang (BCK).

Mat phang (BCK) d i qua A va vuong goc v o i (P) nen n ^ , u ] = (-1;6;4) la VTPT ciia (BCK).

= (10;-7;13) la

H p U

Do (R) d i qua A va vuong goc v o i (BCK) nen n ^ =

VTPT cua (R), suy ra phuong trinh ciia (R): lOx - 7y + 13z f 3 = 0 .

3) Cach 1: Gia su phuong trinh mat ph^ng (a) c6 dang: ax + by + cz + d = 0

X T , X " fa + b + c + d =0 Do M , N e ( a ) nen <^ <=>^ -a + 2 b - c + d = 0. d = - l b 2 1 . c = -a + —b 2

Ta viet lai dang phuong trinh ciia (a) n h u sau: 2ax + 2by + (b - 2a)z - 3b = 0

Suy ra n„ = (2a; 2b; b - 2 a ) la VTPT ciia (a). Goi (p = (AJOLJ)

|4a + 2 b - b + 2a| Ta c6: sincp = n„.u 1 " a u b^ +12ab + 36a2^ V6.V4a^+4b^+(b-2a)^ 5h^ - 4 a b + 8a S h

. Neu a = 0 => sin(p = , voi âO, dat t = — , t e IR

2 a Xet ham so f(t) = ^ +12t + 36 jjy,^,^ maxf(t) = f'^^

5t^ - 4 t + 8

Do do {p„,^,^ <=> sin ^>^^^ f ' chpn b - 5,a = 8

a 8

53 9

Vay phuong trinh ciia (a):16x + 1 0 y - n z - 1 5 = 0.

Cach 2: Ta c6: N M = ( 2 ; - l ; 2 ) la VTCP ciia M N , suy ra p h u o n g trinh duong

x = l + 2t

thang M N :<y = l - t , t e S . Ggi d la duong thang d i qua M , song song voi

z = l + 2t

x = l + 2t

y = l + t , t € R

rz = l - t A . Suy ra phuong trinh d :

308

Cty TNHH MTV DWH Khang Viet

Wkren d ta lay diem Ă3;2;0).

HSoi H , K Ian lugt la hinh chiéu len (a) va M N , khi do

((5^) = A M H -

^ , rrTTj M H M K Ta co: c o s A M H = > ,

M A M A

ma khong doi nen A M H Idn nhat o H = K M A

Hay (a) la mat phang di qua M N va vuong goc voi mat phang (p) = ( M N , d ) Ta c6: np = N M A u = (-1;6;4) la VTPT ciia (p)

Suy ra n„^ = N M A n,^ = ( - 1 6 ; - 1 0 ; l l ) la VTPT ciia (a) Vay p h u o n g trinh ciia (a):16x + 1 0 y - n z - 1 5 = 0.

Bdi 3.5.5. Trong khong gian Oxyz cho mat phSng ( a ) : x + y + z - 3 = 0 va

diem A ( l ; 2 ; 3 ) . Lap phuong trinh d u o n g thSng A nam trong (a) va 1) A d i qua M ( l ; l ; l ) va khoang each t u A den A Ion nhat, nho nhat; 2) A di qua M va khoang each giua A va d : x - 2 _ y _ z Ion nhat.

1 2 - 1

s J{uang dan gidi

Mat phang (a) c6 n = (1;1;1) la VTPT 4

Ggi u = ( a ; b ; c ) la VTCP ciia A , do A c (P) => a + b + c = 0 => c = - a - b (1) 1) Taco: A M = ( 0 ; - l ; - 2 ) =^

Do do:

u , A M l = (c + 2b;2a;-a)

d(A,A) = u , A M (c + 2b)^ +5â (b-a) 2+5â

i â+b^+c^ ^ a 2 + b 2+(a + b)2 ^ll

b ^ - 2 a b + 6â b^+2ab + b^ Neu a = 0 = > d ( A , A ) = 4 - ' v o i a ^^0 dat t - - , t e K

V2 a

Xet ham so f(t) = ^ , khao sat ham so f(t) ta t i m duoc max f(t) = f

t^ + t +1

I 3

= 10, m i n f ( t ) = f(4) = -

• Khoang each t u A den A Ion nhat khi t = c> - = , chon b = -2 a - 3,

3 3 *J

^ = - 1 , suy ra phuong trinh duong thang : A : x - 1 _ y z l = ^

-2 - 1

• Khoang each t u A den A nho nhat khi t = 4 <=> — = 4, chon b = 4 => a a

c = - 5 , suy ra phuong trinh duong thang: A : - ^ ^ ^ ~ '

2) Duong thang d di qua N(2;0;0) va c6 U]' = (l;2;-1) la VTCP

= (2a + b ; - b ; 2 a - b ) : M N = ( l ; - l ; - l ) ,

Do do d(A,d) =

U , U j U , U i . M N = 3b u , U i .MN 3 b

U , U j 7(2a + b)2+b^+(2a-b)^ 4a2+3b2 Dang thuc xay rakhi a = 0=>c = - b ^ u = b(0;l;-l)

x = l Vay phuong trinh A : < y = 1 + 1 .

z = l - t

Bai 3 . 5 . 6 . Lap phuong trinh duong thang d d i qua Ă0; - 1 ; 2) va cat duong

x + l y z - 2

thang d : = — = sao cho: ^ 2 1 - 1

1) Khoang each t u B(2; 1;1) den duong thMng d la Ion nhát, nho nhat. 2) Khoang each giiia d va A: x - 5 y z = — = — la Ion nhát.

2 -2 1

Jiucmg dan gidi

Gia sir d cat d ' tai diem M thi M ( - l + 2t; t; 2 -1), t e M. A M = (2t - 1 ; t +1; -1) la VTCP cua duong thang d .

- ( 1 - t ; l ; 4 - 2 t ) . 1) Taco AB = ( 2 ; 2 ; - 1 ) nen AB, A M

Khoang each tir diem B den duong thang d la

d(B, d) = AB, A M A M 5 t 2- 1 8 t + 18 V 6 t 2- 2 t + 2 = Vf(t) „ , 5 t 2- 1 8 t + 18 . 98t(t-2) T a c o f ( t ) = — nenf'(t)= ^ ' 6t - 2 t + 2 ( 6 t 2- 2 t + 2)2

Tir do ta tim duge max f(t) = f(0) = 18, min f(t) = f(2) = ^ .

310

C t y TNHH MTV DWH Khang Vi$t

mpo do:

^ ,nind(B, d) = dat duoc khi t = 2 => A M = (3; 3; - 2 ) nen phuong trinh , x y + l z - 2

(jtfdng thang can tim d : - = — = " i : ^ -

inaxd(B, d) = 3N/2 dat duoc khi t = 0 => A M = (-1; 1;-1) nen phuong trinh

ctircmg thang can tim d : — = —j- = •

2) A di qua N(5; 0; 0) va c6 vec to chi phuong u ^ = (2; - 2; 1).

Ta CO A M ] - (t - 1 ; 4t - 1 ; 6t), A N = (5; 1; - 2). Khoang each giira hai duong thang la:

f u ^ , A M1 .A N _ 6 _ 3 t

d ( A ; d ) = ^

u ^ , A M ] yjit-lf +{At-lf +i6tf

= 3.

V53t2-10t + 2

Vi f-(t) = nen f'(t) = 0 « t = -2, t =

(53t2-10t + 2 r

Tir do ta tim duoc maxf(t) = f — , khi do A M = - — ( 2 9 ; - 41; 4).

, 3/ / "^•Z

53t^-10t + 2

37'

Vay duong thang d c6 phuong trinh la d : X y + 1 _ z - 2 29 -41

(BMUCIUC

P H U O N G PHAP TOA D O T R O N G M A T PHANG

§ 1. CAC B Al T O AN C O BAN 7

§ 2. X A C D I N H T O A DO C U A M O T D I E M 18 § 3. NH6M CAC BAI TO AN VE HINH BINH HANH 27 § 3. NH6M CAC BAI TO AN VE HINH BINH HANH 27

§ 4. C A C BAI T O A N VE D U O N G T R O N V A C O N I C 40

§ 5. N H 6 M C A C BAI T O A N LIEN Q U A N TAM G I A C 73

H I N H H O C K H O N G GIAN TONG H O P § 1. Q U A N Hfi V U O N G G O C 94 § 1. Q U A N Hfi V U O N G G O C 94 §2. G 6 C 107 § 3. K H O A N G C A C H 129 § 4. T H E T I C H K H O I D A D I E N 149 § 5. M A T C A U - M A T T R U T R O N X O A Y 183

P H U O N G PHAP TOA D O TRONG K H O N G GIAN

§ 1. T I C H C O H U O N G C U A H A I V E C T O V A I J N G D U N G 206

§ 2. LAP P H U O N G T R I N H M A T P H A N G 224

§ 3. P H U O N G T R I N H D U O N G T H A N G 251

§ 4. LAP P H U O N G T R I N H M A T C A U 276

Một phần của tài liệu Phương pháp giải toán hình học theo chuyên đề part 3 (Trang 47)

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