Phuong Phap Giai Toan Vecto

7/31/2019 Phuong Phap Giai Toan Vecto

1/12

CAC DNG BI TP -PHNG PHP GII TON VECTO

CHNG I : VECTO

A. Vecto

: Cho hnh bnh hnh ABCD c tm l O . Tm cc vect t 5 m A, B, C , D , O

a) Bng vect AB ; OB

b) C di bng OB

: Co tm gc C m v n t tng m C C C:

MN BP ; MA PN .

: Co t gic ABCD, gi M, N, P, Q ln t l trng m AB, BC, CD, DA. Chng minh:

MQNPQPMN ; .

i 4: Cho tam gic ABC c trc tm H v O tm ng trn ngoi tip . G m i

xng B qua O . Chng minh : CBAH ' .

i 5: Cho hnh bnh hnh ABCD . Dng BCPQDCNPDAMNBAAM ,,, .

Chng minh OAQ

B. CH

: Co m t Cng mn cc ng tc :

a) PQ NP MN MQ ; b) NP MN QP MQ ;

c) MN PQ MQ PN ;

: Co ng gc C Cng mn ng:

a) 0AD BA BC ED EC ;

b) AD BC EC BD AE

: Co m M, N, P, Q, R, S. Chng minh:

a) PNMQPQMN

. b) RQNPMSRSNQMP

. : Co 7 m A ; B ; C ; D ; E ; F ; G . Chng minh rng :

a) AB + CD + EA = CB + ED

b) AD + BE + CF = AE + BF + CD

c) AB + CD + EF + GA = CB + ED + GF

d) AB - AF + CD - CB + EF - ED = 0

5: Co n n n C c tm O C: 0OA OB OC OD . : Co ng gc u ABCDE tm O Chng minh :


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OOEODOCOBOA

7: Cho lc gc u ABCDEF c tm l O . CMR :

a) OA+OB+OC+OD+OE +OF=0

b) OA+OC+OE = 0

c) AB+ AO+AF

= AD

d) MA+ MC+ ME = MB+ MD+ MF ( M ty ).

i 8: Cho tam gic ABC ; v bn ngoi cc hnh bnh hnh ABIF ; BCPQ ; CARS

Chng minh rng : RF + IQ + PS =0

i 9: co t gc C n t tng m C v tng m C:

0EA EB EC ED .

: Co tm gc C v tng m C C C:

a) 0AN BP CM ; b) AN AM AP ;c) 0AM BN CP .

: Co n tng C n C n g tng m C:

EA EB EC ED DA BC .

: H tc tng m Co m v

a) Co tng m C v m t : 2IA IB IM

b) o co 2NA NB C v t : 2 3IA IB IN

c) o co 3PA PB C v t : 3 2IA IB IP : H tc tng tm Co tm gc C c tng tm :

a) CMR: 0GA GB GC t : 3IA IB IC IG .

b) tc on v 1

4GA . CMR 2 0MA MB MC

c) Co tm gc c tng tm C:

+ 0AD BE CF .

+ T u tm gc c cng tng tm : H tc n n n Co n n n C tm O C:

a) 0OA OB OC OD ;

v t : 4IA IB IC ID IO .

C. I N IN N N I

t t u c 2. T d vet ., CBCABCBA

2 t . 060BAD O 2 . T

| AB AD | ; BA BC ; OB DC .

vu . T


3/12

AC BD ; AB BC CD DA .

t . tru v . t

IB ID JA JC .

D.C

. Cho tam gic ABC v M, N l t tru m AB, AC.a) G P Q tru m MN v BC. CMR : A, P , Q thng hng.

b) Gi E, F tho mn :1

3ME MN ,

1

3BF BC . CMR : A, E, F thng hng.

2. t E tru m AB v F thuc tho mn AF = 2FC.a) G M tru v m tho mn 4EI = 3FI. CMR : A, M, I thng hng.b) Ly N thuc BC sao cho BN = 2 NC v J thuc EF sao cho 2EJ = 3JF. CMR A, J, N thng hng.c) L K tru m EF. Tm P thuc BC sao cho A, K, P thng hng.

. t v M N P m tho mn : 3MB MC O , 3AN NC , PB PA O .CMR : M, N, P thng hng. (

1 1 1,

2 2 4MP C B CA MN CB CA ).

. t v M N t mn 2 ,LB LC 1

2MC MA

, NB NA O . CM : L, M, N

thng hng.

. Cho tam gic ABC vi G l trng tm. I, J tho mn : 2 3IA IC O , 2 5 3JA JB JC O .a) CMR : M, N, J thng hng v M N tru m AB v BC.b) MR tru m BI.

c) G E m thuc AB v tho mn AE kAB . X C, E, J thng hng.

. t . t mn : 2 , 3 2 =IA IB JA JC O . MR ng th qu .

t M tru tue. tru e M v K t e

tren canh AC sao cho AK =3

1AC. Chng minh ba iem B, I, K thang hang

i 8: Cho tam giac ABC. Hai iem M, N c xac nh bi cac he thc

OACNAABOMABC 3; . Chng minh MN // AC.

E.P . X nh v tr m m tho mn mtng th V

m A, B, C. Tm vtr m M sao cho :a) MB MC AB b) 2MA MB MC O

c) 2MA MB MC O d) 2MA MB MC O

e) MA MB MC O f) 2MA MB MC O i 2: Cho tam giac ABC co I, J , K lan lt la trung iem BC , CA , AB . G la trong tam tam

giac ABC . D, E xac nh bi : AD = 2 AB va AE=5

2AC.


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Tnh DEvaDG theo AB va AC. Suy ra 3 iem D,G,E thang hang

888=============================8888888888888===============================888

F.

I.L THUYT:1.TRC T :

Trc t (Trc , hay trc s ) l mt ng th tr t x nh mt m O v mt ve t

v 1ii

O c gi l gc t ve t i c g ve t v ca trc t

2.T c m trn trc:

ve t u nn trn trc (O ; i ) .Do i v u p iau vi a R. S c g d i s ca u hay t ca u i vi trc (O ; i )

m M nm trn (O ; i ) => imOMRm :

S c gi l t c m M.d i s c c :

Trn trc ( O ; i ) 2 m A , B c t v b .d i s c ve t ABhieukyAB

Ta c : abAB .Tnh cht :

ACBCABiOCBACDABCDAB ::);(;; ;3.BI TP

Bi 1:

T d i s c ve t ABtrn trc (O ; i ):

p dng cng thc : AB b a vi a; b l t ca A; B

Th d : Trn trc t (O ; i ) m A ; B ; C c t l t l2 ; 1 v 4.

1.Tnh t ve t CABCAB ;; 2.Ch tru m ca AC.

GII:

1. 1 2 3 3 6

2. 3

AB BC CA

BA BC BA BC

tru m ca ACTng qut :

Cho A ; B trn trc ( O ; i ) c t v b .M tru m ca ABa+b = 2m (m l t ca M)

Bi 2:

Chng minh mt h thc lin quan n cc di i sca cc vec t trn trc (O ; i)

O I


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P Tnh di i sca cc vec t , chng minh h thc i s.Ch .Chn mt trong cc im l im gc ta di i sca cc vec t n gin hn.

Th d :

u ha : Trn trc t (O ; i ) m A ; B ; C ; D c t l t l a ; b ;c ; d

(ABCD) l mt u haCB

CA

DB

DA

.( )( ) ( ) . .

23.

AB

a b c d ab cd I A IB IC ID

AC AD

2 2

1 2 2

1 1 tru m AB

GII:

1. ( )( ) ( )( )

2( ) 2( ) ( ) ( ) 2( ) ( )( )(1)

DA CA a d a ca d b c b d c a ab ac bd cd bc ab cd ad

b d b cDB CB

ab cd ac bc ad bd ab cd c a b d a b ab cd a b c d

2. Ch tru m I ca AB l gc t ta c:

22 2

2

a -b

-a(1)2(ab cd) 0 ab -cd .

cdIA IB IC ID

b cd

3. Chn A l gc t ta c:2 1 1 2 1 1

(1) 2cd bc bdb

hayc d AB AC AD

BI TP:

1.Trn trc t (O; i ) 2 m A v B c t l t a v b .

a)Tm t m M sao cho )( 1 kMBkMA S xM =1

k

akb

b)Tm t tru m I c S2

baxI

c)Tm t m N sao cho NBNA 52 S7

25 abxN

2.Trn trc (O ; i ) m A ; B ; C c t l t ; b ; . T m I sao cho :

0

ICIBIA S 3

cba

xI

3.Trn trc t m A ; B ;C ;D bt k .

a.Chng minh 0 BCADDBACCDAB ... b.Gi I,J ,K ,L l t tru m ca AC ; BD;AB v CD . Ch v K u tru m.

B.H TRC T I.L thuyt :1.T mT .


6/12

);(:;

;:;

yxMjyixOMRyxmpOxyM

aaajaiaaRaampOxya

212121

2.C

Tr p Ox ve t );(;);( 2121 bbbaaa

1 11 1 2 2 1 1 2 2 1 2

2 2

( ; ) ( ; ) ( ; )a ba b a b a b a b a b a b a b pa pa paa b

a p b a pb

3.T mt s t bit :Trong mpOxy cho 2 m A(x1;y1) B(x2;y2) v C(x3;y3)

T vecto 1212 yyxxAB ; T M tru m ca AB

22

2121 yyxxM ;

T trng tm G ca tam gic ABC

33

321321 yyyxxxG ;

II.BI TP:

Bi 1. Chng minh 2 vecto );(;);( 2121 bbvaau cng phng .

P Php:

Gi s2 vet p =>

pba

pbavpu

22

11

Nu h trn c nghi t 2 vet p ; Nu h trn v nghi t 2 vet p.

Ch :Nu b1; b

20 th ;u v p 1 2

1 2

a

b

a

b

Th d :

Cho 2 vecto );(;);( 6231 vu Xt t p a 2 vecto trn.

Gii :

Gi s ;u v p

1

1 2 12

3 6 1 22

pp

u pv p

p p

H c nghim ; vy vu ; p

Th d 2:Tr pOx m A(1;2) B(3 ; 2) v C(4 ;1) , Chng minh ABC l mt tam gic.GII

ACABACAB ;);();( 1

4

5

41544 p => ; ; tng hng .


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V m A ; B ; C to thnh tam gic.Th d 3:

Cho 2 2 ;4u m m ( ;2)v m 2 vet p.GII :

Xt m = 0 => vuvu ;);(;);(

4

2

2

02042 p

Xt 0; ;m u v p

22 2m 2 4

2 2 2 02

1

2

mm m m m m

m

m

m

BI TP:

.Tr pOx m A (1 ;2) B(0 ; 3) C(3; 4) D(1 ; 8) . Bb tr m trn b no thng S ; ;2.Tr pOx m A(1 ;2) B(3 ;1) C(3 ; 5)a.Chng minh ABC l mt tam gic .b.Tm t trng tm ca tam gia1cABC .c)Gi I(0 ; 2) .Chng minh A ; G; M thng hng.d) Gi D(-5;4) .Chng minh ABCD l hnh bnh hnh.

Bi 2:Tm t ca vecto:PP.p dng cc php ton ca vecto:

Th d :

Cho 3 vecto: 525123 ;;; cba

Tm t ca vecto 2 4 2 5u a b c va v a b c

GII

);();();();(

);();();();(

171525105102223

2913208451462

vcba

ucba

Bi tp

1.Cho cc vecto 642

1102 ;;;

cba . Tm t vecto

);(: 3228542 uDScbau

2.Cho tam gic ABC , G l trng tm ca tam gic . Tnh t vecto GBGCGAu 423 S ( ; -14)

Bi 3:P t v t 1 2 1 2 1 2c ( ; ) theo 2 vecto a (a ;a ) va b (b ;b )c c p

P pp

Gi s : 1 1 1

2 2 2

cxa yb c

xa y bxa yb c


8/12

Gii h trn tm x ; y.

Th d :

Cho 525123 ;;; cba .

1.Chng minh ;a b p

2. P t v t c te 2 ve t a v b Gii:

1.3 2

;-1 5

a b p

2. Gi s

15

3 2 15 1117c

2 5 5 11 17 17

17

xx y

xa y b c a bx y

y

BI TP

1.Cho 1; 2 3;1 4; 2 .a b c

P t ve t a theo 2 vecto b ; c p s:3 7

5 10a b c

2.Cho 5; 2 4;1 2; 7a b c

a.Chng minh b;a ng cng png B.Phn tch vecto

theo 2 vecto ; : 2 3c a b DS c a b

Bi 4:

Tm t nh tht ca hnh bnh hnh ABCD khi bit A (x1;y1); B (x2y2 ) ;C(x3;y3)ng pp :Cch 1

Gi D (x;y). Tnh ;DA BC.

ABCD l hnh bnh hnh1 3 2

1 3 2

ADx x x x

BCy y y y

-Gii h trn tm D(x ; y)Cch 2:

-Tmtng m I ca AC

-Tm D bit tng m ca BD

Th d :Cho tam gic ABC vi A(1;2) B(3 ;1) C(3 ; 5) .Tm D sao cho ABCD l hnh bnh hnh .GII :

G tng m ca AC =>I(1 ;2

3)


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C n n n > tng m ca BD => );(Dy

x45

31

23

Bi tp:Co m A(2;1) B(2;1) C(2 ;3) .a.Chng minh A,B,C khng thng hng . Tm D sao cho ABCD l hnh bnh hnh . S:

2;1)2.Cho tam gic ABC vi A(1;2) B(3;2) C(4 ; -1) .

Tm tng m I ca AC .b.Tm D sao cho ABC n n n S: );(D; 502

3

2

3

Tong mpOx co m M(-4 ; 1) N(2;4) P(2 ;2) ln t tng m ca 3 cnh BC ; CAv AB ca tam gic ABC.Tm ; ;C S: 8; -4;-5) C(-4;7)b.Chng minh 2 tam gic ABC v MNP c cng trng tm.4.Cho tam gic ABC vi A(3;6) B(9;10) C(-5;4) .a.Tm t trng tm G c tm gc C S:b.Tm D sao cho BGCD l hnh bnh hnh.

5Co m A(-2 ; -3) B(3;7) C(0;3) v D(-4 ; -5) .a.Chng minh AB //CD Tm go m I c v C S -12;-13)

5: Tm go m c on thng AB v CD vi A(x1;y1) ; B(x1;y2) ; C(x3;y3) ;D(x4;y4)

Cch gii:G x; go m c ng thng AB v CD

; cung phuong

; cung phuong

AI AB

CI CD

Gii tm I(x;y)

go m c on AB v CD;

;

IA IB nguoc huong

IC ID nguoc huong

Th d 1:Tong mpOx co m A(0 ; 1) B(1; 3) C(2 ;7 ) v D(0;3).Tm go m c on thng AC v BD

GII:

Gi; (1)

; cung phuong (2)

AI AC cung phuongI AC BD

BI BD

1( ; 1) ; (2;6) (1) 6 2 2

2 6

( 1; 3) ( 1;0) (2) 3

x yAI x y AC x y

BI x y BD y

2 2 2 4;3 ; 2) ;4 2

3 3 3 3x I IA IC IA

I thu n AC


10/12

1 2 1;0 ;0) 2 ;0 2

3 3 3IB ID IB

I thu n BD

Vy2

I ;33

l giao c on AC v BD

Bi tp :1. Tong mpOx co m ; ; C ;7 v ;Tm go m c on thng

v C S: on AD khng ct BC)2. Trong mpOx co m A(0 ; 1) B(-1; -2) C(1 ;5 ) v D(-1;-1).Tm go m c on thng C v Tm go m ca BD v AC

Bi 6: Tm ta im trong mt phng ta : tm t m M(x ; y) trong mp Oxy , ta dng ng vung gc MA1v Ox v MA2 vi Oy

Ta c x = 21 OAy;OA

Th d : Cho hnh bnh hnh ABCD c AD = 4 v chiu cao ng vi cnh AD = 3, BAD=600

.Chn h trc t n n v . Tm t cc vecto ACva;CD;BC;AB

Bi tp:

Co tm gc u ABC c cnh l a . Chn h trc t Ox n : O tng m BC , trcon cng ng vi tia OC , trc tng cng ng vi tia OA.a.Tm t cc nh ca tam gic ABC.b.Tm t tng m I ca AC.c.Tm t tm ng trn ni tip tam gic ABC : Co tm gc C Cc m M(1; 0) , N(2; 2) , p(-1;3) ln t tng m cc cnh BC,CA, AB. Tm t cc nh ca tam gic

: Co ; ; ; ; Cm+; m+ Tm m m A, B, C thng hng

HA x

y

D

B CK BH AD =>BH=3 ;AB=2 ; AH =


11/12

: Co tm gc u ABC cnh a . Chn h trc t (O; i ; j tong O tng

m BC, i cng ng vi OC, j cng ng OA .

a) Tnh t c cc nh ca tam gic ABCb) Tm t tng m E ca ACc) Tm t tm ng trn ngoi tip tam gic ABC

Bi 4 : Cho lc gc u ABCDEF. Chn h trc t (O; i ; j tong O tm lc gc u ,

i cng ng vi OD , j cng ng EC.

Tnh t cc nh lc gc u , bit cnh ca lc gic l 6 .

Bi 5:Cho A(-1; 2), B (3; -4), C(5; 0). Tm t m D nu bit:

a) AD 2 BD + 3CD = 0

b) AD 2 AB = 2 BD + BC c) ABCD hnh bnh hnhd) C n tng c C vi BC = 2AD

Bi 6 :Co m I(1; -3), J(-; c n tn an bng nhau AI = IJ = JBa) Tm t ca A, Bb) Tm t c m i xng vi I qua Bc) Tm t ca C, D bit ABCD hnh bnh hnh tm K(5, -6)

Bi 7: Cho a =(2; 1) ; b =( 3 ; 4) v c=(7; 2)

a) Tm t c vect u = 2a - 3b + c

b) Tm t c vect x tha x + a = b - c Tm cc s m ; n tha c = ma + nb

i 8 : Trong mt phng t Oxy cho A(4 ; 0), B(8 ; 0), C(0 ; 4), D(0 ; 6), M(2 ; 3).a/ Chng minh rng: B, C, M thng hng v A, D, M thng hng.b/ Gi P, Q, R l t tru n thng OM, AC v BD. Chng minh r m P, Q, Rthng hng.i 9. Trong mt phng t Ox m A(1 ; 3), B(-2 ; 2). ng th qu t Ox ti Mv ct Oy ti N. Tnh din tch tam gic OMN.i 10. Trong mt phng t Oxy cho G(1 ; 2). Tm t m A thuc Ox v B thuc Oy sao cho G ltrng tm tam gic OAB.

i 11. Trong mt phng t Oxy cho A(-4 ; 1), B(2 ; 4), C(2 ; -2).a/ Ch nh ca mt tam gic.b/ Tnh chu vi ca tam gic ABC./ X nh t trng tm G v trc tm H.i 12. Cho tam gic ABC vi A(1 ; 2), B(5 ; 2), C(1 ; -3)./ X nh t m D sao cho ABCD l hnh bnh hnh.b/ X nh t E i xng vi A qua B.c/ Tm t trng tm G ca tam gic ABC.


12/12

i 13. Cho A(1 ; 3), B(5 ; 1).

a/ Tm t m I tha .0 IBIAIO b/ Tm trn tr m D sao cho gc ADB vung.*************************************************************************************

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