Thiết kế bài giảng hình học 10 nâng cao (tập 1) phần 2

Cau 11 .• Cho AABC ndi tiep trong dudng trdn tam O, // la true tam AABC, D la
diem dd'i xiing ciia A qua O. Khi dd HA + WB + ITc bdng
(a)//0

(h)lHO

(c)3H0

(6) 4H0.

Trd ldi: Phuang an dung : (b).

§5. True toa do va he true toa do (tiet 12,13,14)
1. MUC TIEU
1. Kie'n thurc
1. Nhan biet dupe toa dp cua vecto, toa dp ciia diem dd'i vdi true tpa dp va he true
tpa dp.
2. Hpc xong phan nay, hpc sinh cdn xac dinh dupe toa dp ciia vecto va cua diem,
hieu va nhd dupe bieu thifc toa dp cua cac phep toan ve vecto, cac cdng thiic bieu
thi quan he giita eac vecto (cimg phuang).
3. Hpc sinh cung can hieu va nhd dupe cac cdng thiic bieu thi quan he giiia cac
diem : dieu kien de ba die'm thang hang, toa dp trung diem ciia doan thang va toa
dp trpng tam ciia tam giac.
2. KT nang
• Ve kl nang, hpc sinh biet each lua chpn cdng thiic thfch hpp trong giai toan va
tfnh toan chfnh xac.
• Biet phan tfch mdt vecta thanh to hop ciia cac vecto khac bdng toa dp.
• Chiing minh dupe mdt sd bai toan cd lien quan den trung diem, trpng tam nhd
toa dp.
3. Thai dp
• Lien he dupe vdi nhieu va'n de cd trong thuc te'.

• Cd mdi lien he chat che giira vecto va toa dp cua nd.

72


• Viing vang trong tu duy logic.
II. CHUXN BI CUA GV VA M

1. Chuan bi cua GV:
- GV chudn bi sdii hinh ve tii hinh 29 de'n hinh 31 SGK.
- Thudc ke, phab mau,...

2. Chuan bi cua HS :
- HS dpc trudc bai hpc.
m. PHAN PHOI TH6I LUONG

Bai nay gom 3 tiet.
Tiet thur nha't: Tir dau den het muc 3.
Tiet thur hai: Tiep theo den het muc 5.
Tie't thur 3: Phan cdn Iai va hudng ddn bai tap.
IV. TIEN TDINIi DAY HOC
n. KI€M TRR ISni CU
Cau hoi 1.
Cho hinh binh hanh ABCD ; O la tam.
a) Tfnh AB + AD theo AO
b) Hay so sanh hai vecto AB + AD va CO.
Cau hoi 2.
Cho 3 diem A, B, C. Gpi G la trpng tam tam giac ABC, M la diem bat
ki. Tfnh tdng MB + MC + MC

73


Bni MOI

HOAT DONG 1
1. True toa do
a) Mtic dich: Giiip HS hieu dugc tog do eua vecto trin true, do ddi dgi so.
b) Hudng thuc Men
- Ddt vd'n di'.
- Thue Men ^\^ 1.
- Trinh bdy ve do ddi dgi so.
c) Qud trinh thuc Men
• Dat van de
GV ve hinh 27 va thirc hien cac thao tac sau:
- Neu khai niem true toa dp: True va vecta don vi.
- Kf hieu true toa dp.
- Toa dp cua vecto tren true
a

x'

*

0

T

I

.

Hinh 27

• Thuc hien ^ ^ 1
GV thu'c hien thao tac nay trong 5'.

Hoat ddng cua GV
Cdu hdi 1

Hoat ddng ciia HS
Ggi y trd ldi cdu hdi 1

Vecto OA cd toa dp bang bao nhieu? Vecto OA cd toa dp bdng toa dp cua
A va bang a.

74


Ggi y trd ldi edu hdi 2

Cdu hdi 2
Vecto OB ed toa dp bdng bao nhiau?

Vecto OB c6 toa dp bdng toa dp eua
B va bdng b.
Ggi y trd ldi cdu hdi 3

Cdu hdi 3
Hay bieu dian vecto AB qua hai

jB =

aB-aA^b-a.

vecto OA v a 0 5 .
Cdu hdi 4

Ggi y trd ldi edu hdi 4

Gpi / la trung diem AB. Tim toa
Toa dp cua / bdng
dp cua /.

.

• Do dai dai sd cua vecta tren true
GV nau dinh nghia:
Ni'u hai diem A, B nam trin true Ox thi tog do eua vecto AB dugc ki
Mill la AB vd ggi Id do ddi dgi sd cda vecto AB trin true Ox.
Nhu vdy, AB = AB i .
GV neu hai nhan xet sau:
1) Hai vecto AB va CD bdng nhau khi va chi khi AB = CD (hien
nhien);
2) He thii'c AB + BC = AC tuong duong vdi he thifc AB + BC = AC
(he thiic Sa-lo).
GV cho HS thyc hien thao tac sau trong 3' de cung co kien thCrc.

Hoat-ddng ciia GV
Cdu hdi 1

Cho A cd toa dp la a, B cd toa dp la

Hoat ddng ciia HS
Ggi y trd ldi edu hdi 1
b-a.

B. Tim dp dai dai sd ciia AB.
75


Cdu hdi 2
Tim dp dai dai sd eiia AI (vdi / la
trung diem ciia AB).

HOAT DONG 2
2. He true toa do
a) Muc dich: Giiip HS nam duge hi true tog dgvd van dung trong gidi todn.
b) Hudng thuc Men
- GV niu he true tog do.
c) Qud trinh thuc Men
• GV gidi thieu ve he true toa do thdng qua hinh 28 SGK.

1

_>.
J
0 —>
i

x

Hinh 28

• GV gidi thieu he true toa dp, mdi lien he vdi cac he true da hoc
GV neu cau hoi
HI. Toa dp ciia diem va toa dp cua vecto ed lien quan nhu the' nao?
H2. He true toa dp cd lien quan nhu the' nao vdi true toa dp ?

76


HOAT DONG 3
3. Toa do cua vectd doi v6i he true toa do
a) Muc dich: Giiip HS tim dugc tog do cua vecto, cua diem.
b) Hudng thuc Men
-Thitc Men ^ f 2.
- GV neu dinh nghia trang 27 SGK.
Thuc hiin ? 1
- Niu nhdn xit trang 27 SGK.
c) Qud trinh thitc Men
• Thuc hien ' ^ 2.
- Treo hinh 29 len bang.

yn

^

Hinh 29

GV thirc hien thao tac nay trong 5'.

Hoat ddng cua GV

Hoat ddng cua HS
Ggi y trd ldi cdu hoi 1

Cdu hdi 1
Hay bieu thi vecta a qua i va j .

-

^-

5-

a-ll+—

I

1
Cdu hdi 2
Hay bieu thi vecto b qua i va j .

Ggi y trd ldi cdu hdi 2
b = -3l+0j
Ggi y trd ldi edu hdi 3

Cdu hdi 3

Hay bieu thi vecto u qua / va j .

^^

u = li —

3^

/.

2
Ggi y trd ldi cdu hdi 4

Cdu hdi 4
Hay bieu thi vecto v qua / va j .

v = Oi + - j
1

• Neu dinh nghia
Ddi vdi hi true tog dd(0

; i,j),

ni'u d = xi+yj

thi cap sd

(x ; y) di'or: [joi Id toe do cua vecto a, ki Men Id a = (x ; y) hay
a (x ; y). Sd tint nhdt x ggi Id hodnh do, so thic hai y ggi la tung do ctia

vecto a .

• Thuc hien ? 1

GV thirc hien thao tac nay trong 5'.

78


Hoat ddng ciia HS

Hoat ddng ciia GV

Ggi y trd ldi cdu hdi 1

Cdu hdi 1
Xac dinh toa dp cua cac vecto

a^(l

a, b, U, V trin hinh 29

; -) ; b = (-3 ;0)
1

t

u = (l;--)
Cdu hdi 2

1

; v = (0;-)
1

Ggi y trd ldi edu hdi 2

Dd'i vdi hi tnic tog do (0 ; i ,j),
i = OA, trong dd A la diem (1 ; 0)

hdy chi ra tog do ciia ede vecto
0, i , j , i + j , Ij-i

y = Ofi, trong dd 5(0 ; 1)

;

/ + y = OC, trong ddC(l ; 1).

-7-3j,V3 7+0,14).
2j-i = 0D, trong dd D(-l

; 2);

- / - 3 7 = 0 / ^ , trong ddA:(- ;-3)
V3 7+ 0,14y = OM, trong dd
M(S

;0,14).

• Neu nhan xet
Tii dinh nghia toa dp ciia vecta, ta thay hai vecto bdng nhau khi va chi khi chung
cd ciing toa dp, nghia la
d(x; y) = b(x';

y')<^

X = X

y= y

GVnhan manh:
Hai vecto bdng nhau thi cd toa dp bdng nhau;
Hai vecto dd'i nhau thi cd toa dd dd'i nhau.

79


HOAT DONG 4
4. Bieu thurc toa do cua cac phep toan vectd
a) Muc dich: Giup HS lien giifa tog do ciia vecto vd cdc phip todn cua vecto.
b) Hudng thuc Men
-Thuc Men ^ ^ 3.
- GV khdi qudt tinh chdt.
- Thue hiin ? 1
c) Qud trinh thuc Men

• Thuc hien "^t 3.
GV thyc hien thao tac nay trong 5'.
Hoat ddng cua GV

Cdu hdi 1
Hay bieu thi vecto 5 qua hai vecto

Hoat ddng cua HS
Ggi y trd ldi edu hdi 1
a = -3i + lj.

^' j •

Cdu hdi 2
Hay bieu thi vecto b qua hai vecto

b-4i + 5j

i, j .

Ggi y trd ldi cdu hdi 3

Cdu hdi 3
Tim toa dp ciia vecto c =d + b .
Cdu hdi 4
Tim toa dp cua cac vecto 5 = 45 ;
Cdu hdi 5
Tim toa dp cua vecto ii = 4d-b .

80

Ggi y trd ldi edu hdi 2

c =d + b = i + lj

Ggi y trd ldi edu hdi 4

J = 45 = -127+87
Ggi y trd ldi cdu hdi 5
u = 4d-b --I6i

+ 3j.


• Neu khai quat tinh chat
Cho a =(x;y)vd

b =(x'; y'). Khi dd

l)d + b = (x + x'; y + y') ; d-b = (x-x';

y-y');

2) kd = (kx ; ky) vdi k e R ;
3) Vecto b eiing phuong vdi vecto a^O khi vd chi khi cd so k sao
cho x' = kx, y' = ky.
• Thuc hien \?2\ nham ciing cd kien thurc.
GV thyc hien thao tac nay trong 6'.

Hoat ddng ciia GV
Cdu hdi 1

Hoat ddng cua HS
Ggi y trd ldi edu hdi 1

Cap vecta sau cd ciing phuo'ng khdng? Khdng, vi

a^kb.

a ) 5 = f0 ; 5) v a ^ = f - l ; 1) ;
Cdu hdi 2

Ggi y trd ldi edu hdi 2

Cap vecto sau cd cimg phuong CdviH = 2003v
khdng ?
b) M = f2003 ; O; va v = a ; Oj ;
Cdu hdi 3

•

Ggi y trd ldi edu hdi 3

Cap vecto sau cd cung phuong Cdvi e = -Sf.
khdng ?
c ) e = f 4 ; - 8 ) v a 7 = (-0,5 ; Ij;
Cdu hdi 4

Ggi y trd ldi edu hdi 4

Cap vecto sau cd ciing phuong
Khdng vi
khdng ?

m^kn.

6) m = (yll ; 3) van = (3 ; yfl).

6.TKBGHiNHHOC10M{NC)

81


nOATDONG^
.-C

5. Toa do cua diem
a) Muc dich: Giiip HS liin he duge giifa tog do cua vecto vd tog do ciia diem.
b) Hudng thuc Men
- Niu duih nghia
- Thuc hien^^

4.

- (JV khdi qudt tinh chdt.
Thuc Men ? 3
- Niu chii y trang 29 SGK.
c) Qud trinh thuc Men
• Neu dinh nghTa toa do ciia diem.
Trong mat phang tog do Oxy, tog do eua vecto OM dugc ggi Id tog do
cua diem M.
Nhdn xet. Neu M = (x: y) thi x^OH

;y=OK.

• Thuc hien j ^ 4.
- GV treo hinh 31 len bang.
ViL

B
C
0

D
Hinh 31

GV thyc hien thao tac nay trong 6'.
82


Hoat ddng cua HS

Hoat ddng cua GV

Gdi y trd ldi edu hdi 1

Cdu hdi 1

a) Toa dp cua mdi diem 0, A, B, C, 0(0 ; 0), A(-4 ; 0), B(0 ; 3), C(3 ; I)
D bdng bao nhiau ?
va D(4; -4).
Cdu hdi 2

Ggi y trd ldi edu hdi 2

b) Hay tim diem E cd toa dp

E trung D.

(4;-4).
Ggi y trd ldi edu hdi 3

Cdu hdi 3

18 = (4; 3).

Tim tpa dp ciia vecto AS.
Khai quat tinh chat

Vdi hai diem M(Xf^/!; yjf^) vd N(x^ ; y^j thi
MN = (X^-XM.

;

y^-yM)

• Thuc hien ? 3
GV thyc hien thao tac nay trong b .
Hoat ddng ciia GV

Hoat ddng ciia HS
•

Cdu hdi 1
Hay tim toa dp OM.

Can hdi 2
Hay tim toa dp 0 N'.
Cdu hdi 3
Tim tpa dp ciia vecta MN .

Ggi y trd ldi cdu hdi 1
OM = (xM;yM).
Ggi y trd ldi edu hdi 2
ON = (xj^;y^)

Ggi y trd ldi edu hdi 3
MN = ON-OM

= (xN-XM;yN--JM)-

• GV neu chii y
83


De thuan tien, ta thudng dung kf hieu (x^ ; yj^) de chi toa dp ciia
diem M.

HOAT DONG 6
6. Toa do trung diem cua doan thing va toa do trong tam cua tam giac
a) Muc dich: Giup HS van dung tog do vecto trong viec tim tog do trung diem, tog
do trgng tdm.
b) Hudng thuc Men
-Th Uc Men "ff^ 5.
- Thitc Men " ^ 6.
- Thifc Men - ^ 7.

- Hudng ddn cho HS ldm vi du trang 30 SGK.
c) Qud trinh thuc Men
• Thuc hien ^

5.

GV thyc hien thao tac nay trong 6'.

Hoat ddng ciia GV
Cdu hdi 1
Hay bieu thi vecto OP qua hai
vecto OM vaON.
Cdu hdi 2
Hay tim toa dp diem P theo toa dp
cua M va N.

Hoat ddng cua HS
Ggi y trd ldi edu hdi 1
1
OP = -(OM + ON).
1
Ggi y trd ldi edu hdi 2
Xp

--(xj^+xj^)

yp=-(yM+yN)

84

GV goi y cho HS neu nhan xet sau:
Ni'u P Id trung diem cua dogn thdng MN thi
• ^ _^M-^^N...

_yM_±M.

• Thuchien^6.
GV thyc hien thao tac nay trong 4'.
Hoat ddng cua GV

Hoat ddng cua HS

Ggi y trd ldi cdu hdi 1
Cdu hdi 1
Gpi M' cd toa dp (x ; y). Hay neu
bieu thu:e lien he giua eac toa dp UM+yM'^^yA'
ciiaM, M'vaA.
Ggi y trd ldi edu hdi 2
Cdu hdi 2
Tim toa dp diem M' dd'i xiing vdi
diem M(l; -3) qua diem A(l ; 1).

• Thuc hien "^^ 7.
GV thyc hien thao tac nay trong 4'.

Hoat ddng ciia GV
Cdu hdi 1
Hay vie't he thiic giiia cac vecto

Hoat ddng cua HS
Ggi y trd ldi edu hdi 1
0A+0B+0C=30G.

OA, OB, OC vaOG.
Cdu hdi 2

Ggi y trd ldi edu hdi 2

Tit 66 suy ra toa dp ciia G theo toa
dp eua A, B, C.

85


^G=-(XA+XB+XC)
<

yG=-(yA

+ yB + yc)

GV gdi y HS neu nhan xet sau:
Ni'u G Id trgng tdm eiia tam gide ABC thi

_XA + xg + xc .
-G-

-^

yA+yB + yc

,JG=

^ — •

• Hudng ddn cho HS lam vi du 1 trang 30 SGK
GVhu'dng din hoc sinh lam toan theo cac gdi -

A, B, C khdng thdng hang khi A5 ^ k'SC .

-

Tim toa dp trpng tam tam giac ABC dura vao nhan xet tii •

TON T^T Bfil HOC
1. Tren true toa dp (O ; / ) , neu ii = ai ta ndi u ed toa dp la a, OM = mi ta noi
M ed toa dp m.
1.

Dd'i vdi he true

toa

ddfO ; / , y j ,

ne'u

d = xi+y]

thi

cap sd

(x ; y) dupe gpi la toa dp ciia vecto 5, kf hieu la 5 = (x ; y) hay 5 (x ; y). Sd thii:
nha't X gpi la hoanh dp, sd thii hai y gpi la tung dp eiia vecta 5.
3. Cho 5 = (x ; y)vab=

(x'; y'). Khi dd

l)d + b=(x + x'; y + y') ; d-b=(x-x';

y-y');

2) kd = (kx ; ky)v6i k e R ;
3) Vecta b cimg phuang vdi vecto 5 T^ 0 khi va chi khi cd sd k sao cho
X = kx, y' = ky.
86


4. Trong mat phdng toa dp Oxy, toa dp cua vecta OM dupe gpi la toa dp cua diem M.
5. Vdi hai diem Mix^ ; y^^) va N(xj^ ; yyy j thi
MN = (xf^-XM ;

yN-yM)-

6. Ne'u G la trpng tam cua tam giac ABC thi
^G-

X^+XB

^

+ XC

.
_yA+yB + yc
^yG-

HOAT DONG 7

MCrtfNG D^N Bfll T6P SGK
a) MMC thdng qua gidi bdi tap de eiing ed'li thuyii.
b) Hudng thuc Men
-Chifa cu the mgt sdbdi: 31, 32, 34. Cdc bdi tap cdn lgi hudng ddn ve nhd.
c) Qud trinh thuc Men
Bai 29.
Hudng ddn
HS dn tap lai khai niem :
Hai vecto bdng nhau khi eac toa dp tuofng iing bang nhau.
Hai vecto to dd'i nhau khi cac toa dp tuong iing dd'i nhau.
Cae menh de dung la : b), c), e)
Cae menh de sai la : a), d).
Bai 30.
Hudng ddn-trd ldi.
5 = ( - l ;0), ^ = ( 0 ; 5 ) , c = ( 3 ; - 4 ) , 5 = [ - - ; - \ e =(0,15; 1,3),
V 2 2;
87

/ = (7t; - cos24°).
Bai 31
GV hudng ddn cdu a)
Hoat ddng ciia GV

Hoat ddng ciia HS
Ggi y trd ldi cdu hdi 1

Cdu hdi 1

la=(4

Hay tim toa dp cac vecto la, - 3b

Ggi y trd ldi cdu hdi 2

Cdu hdi 2
Tim

;1), -3b =( - 9;-12).

toa

dp

ciia

u=

(l;-S).

vecto M = 25 - 3^7 + c .
Trd ldi cdc cdu cdn lgi.
h)x + d = b-cox

= -d + b-c.

Suy ra x = (-6 ; 1)

e) Tim toa dp cua vecto k d + l b rdi so sanh vdi c de dupe he phuong
{lk + 3l = l
trinh <
[k + 4l = l
Giai he ta dupe : k = 4,4 ; / = -0,6.
Bai 32
Hoat ddng ciia GV
Cdu hdi 1

Hoat ddng ciia HS
Ggi y trd ldi edu hdi 1

Hai vecto u va v cimg phuang khi Hai vecto M va v cung phuong khi
nao?
cd mdt sd t sao cho u = tv.
Cdu hdi 2

Ggi y trd ldi edu hdi 2

Tim cac gia tri cua k de hai vecto
Ta cd w =f i
a, V cung phuang.

l2

• - s l ,v

=(k;-4)

J

V ciing phuang vdi d khi va chi khi

88


4
, 2
lk= — suy ra ^ = —.
5
5
Bai 33.
Hudng ddn
HS dn tap lai ndi dung toa dp cua diem.
Cac menh de diing la : a), c), e).
Cac menh de sai la : b), d).
Bai 34.
Hoat ddng cua GV
Cdu hdi 1

Tim toa dp cua AB vaBC .
Cdu hdi 2
Hay cho biet quan he cua AB
vafiC.
Cdu hdi 3
Ket luan ve A, B, C.
Cdu hdi 4
A la trung diem BD khi nao ?
Cdu hdi 5
Hay tim toa dp D.

Hoat ddng ciia HS
Ggi y trd ldi cdu hdi 1
A5=(4;-3), fiC = ( 8 ; - 6 )
Ggi y trd ldi edu hdi 2
= -~BC.
1
Ggi y trd ldi cdu hdi 3
JB

Ba diem A,B,C thdng hang.
Ggi y trd ldi edu hdi 4
AB =

-AD.

Ggi y trd ldi cdu hdi 5
Gia sir D = ( x ; y ). Diem A la trung
diem ciia BD khi va chi khi
^ l+x

-3 =
2

. ^ 1+y
va 4 =
-.
1

vay D = (-7 ; 1).
Bai 35
Hudng ddn
HS dn tap lai phdn dd'i xu:ng true va dd'i xiing tam.

89


DS .•

Mi(x;-y),M2(-x;y),M3(-x;-y).

Bai 36.
Hudng ddn-gidi
HS dn tap lai cdng thirc toa dp trpng tam.
Toa dp hai vecto bdng nhau.
Gidi.
a) Ta ed :
-4 + 2 + 2
3

Trpng tam cua tam giac ABC la G

1 + 4-2
3

hay G(0 ; 1).

b) Gia sii D = (x ; y). Diem C la trpng tam tam giac ABD khi va ehi khi
. -4+1+x
2=
3

.
va

e) Gpi E = (x;y), tacd

1 + 4+ -y
-2=
3 y. Suyra D = (8 ; - 11).
A^ = f6 ; 3j , CE = (x-l

; y + 1).

Tii giac ABCE la hinh binh hanh khi va chi khi AB + CE = 0
6 + x - 2 = 0 va 3 + y - 2 = 0. Vay E = (-4;

hay

-1).

MQT SO Bfil T6P TRfiC NGHIEM
Chpn phuong an tra ldi diing
Cau 1. Cho AABC cd A(l ; 2), B(-l; 1) va C(3 ; 3). Trpng tam G eiia tam giac la
(a)G

(c)G

r?

A
•3

V3

J

("^

u

(b) G

\

•1

(d)G

J

fl

V3

\

2
)

f3

\

U

J

— ;3

Trd ldi: Phuong an diing : (b).
Cau 2. Cho A(-2 ; 1), B(3 ; 1) dp dai vecto A5 la

90


(a) 5;

(b) V26

(c)V27;

(d)V24.

Trd ldi: Phuang an diing : (b).
Cau 3. Trong mat phdng toa dp Oxy cho C(l ; 0). Dung hinh binh hanh 0A6C khi dd :
(a) Tung dp vecto AB bdng 0.
(b) Hoanh dp vecto AB bdng 0.
(C) X^

+ XQ + XC + XO

= 0

(d) A va 5 cd tung dp khac nhau.
Trd ldi: Phuong an dung : (a).
Cau 4. Trong mat phdng toa dp Oxy cho 4 diem : A(0 ; 1), B(l ; 3), C(2 ; 7) va D(0 ; 3).
Ta cd :
(h) AC II AB

(a)AB//CD;
(c) AD HBC;

(6)A0IIBD.

Trd ldi: Phuong an diing : (a).
Cau 5. Trong mat phdng toa dp Oxy, cho AABC cd trpng tam G va toa dp cac diem
nhu sau :
A(3 ; 2), B(-ll;

0), G(-l ; 2). Toa dp dinh C se la :

(a) C(5 ; 5);

(b) C(4 ; 5);

(c) C(4 ; 4);

(d) C(5 ; 4).

Trd ldi: Phuong an diing : (d).
Cau 6. Cho AABC cd A(l ; -1), B(5 ; -3) dinh C e Oy va trpng tam G e Ox. Toa
dp dinh C se la :
(a) (1 ; 4);

(b) (2 ; 4)

(c) (3 ; 4);

(d) (0 ; 4).

Trd ldi: Phuang an diing : (d).

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Cau 7. Cho A(-2 ; 1), 5(4 ; 5). Dung hinh binh hanh OACB, O la gd'c toa dp. Toa
dp C se la
(a)C(2;6);

(b) C(-2 ; 6)

(c) C(6 ; 2);

(d) C(-6 ; 2)

Trd ldi: Phuong an diing : (d).
Cau 8. Cho A(-l ; 8), B(l ; 6), C(3 ; 4).
(a) A, B, C la ba dinh tam giac;

(b) A, B, C each deu O

(c) A, B, C thdng hang;

(d) AB = AC.

Trd ldi: Phuong an diing : (e).
Cau 9. Cho A(l ; 1), B(3 ; 2), C(m + 4 ; 2m + 1). De A, B, C thang hang thi:
(a) m= I;

(b) m = 2

(e) m = 3;

(d) m = 0.

Trd ldi: Phuong an diing : (a).
Cau 10. Cho tam giac deu ABC canh a. Chpn
he toa dp Oxy nhu hinh ve. Toa dp tam dudng
trdn ngoai tiep AABC la :

aS

(a)(0;^a);

(b)

(c) 0;

(d) 0;

Trd ldi: Phuang an diing : (c).

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;0

aS


On tap chvtdng I (tie1t 13,14)
I. MUC TifiU
1. Kien thiirc
1. Lam cho hpc sinh nhd lai nhflng khai niem co ban nhat da hpc trong chuang :
phep cdng va trii cae vecto, phep nhan vecto vdi mdt sd, toa dp ciia vecto va ciia
diem, cac bieu thiic toa dp ciia cac phep toan vecto.
2. Ve kl nang thuc hanh can lam cho hpc sinh nhd lai nhitng quy tdc da biet: Quy
tdc ba diem, quy tdc hinh binh hanh, quy tdc ve hieu veetp, dieu kien de hai vecto
Cling phuang, ba diem thdng hang,...
2. KT nang
• Cd kl nang tong hpp ve cac phep toan cua vecto, toa dp vecto,...

• Van dung kien thiic de lam cac bai toan.
3. Thai dp
• Lien ha dupe vdi nhieu van de cd trong thuc te.
• Cd the tu sang tao dupe mot so bai toan mdi.
• Virng vang trong tu duy logic.
II. CHUAN DI CUA GV VA HS

1. Chuan bi cua GV:
- GV nhde Iai nhiing kien thiic cdn thie't.
- Chuan bi mdt bai kiem tra 45'.
- Cho hpc sinh lam bai kiem tra 45'.
2. Chudn bi ciia HS :
- De chuan bi cho tie't dn tap, can yeu cau hpc sinh lam viec trudc d nha : tra ldi
eac "cau hdi tu kiem tra" va chuan bi eac bai tap.
m. PHAN PHOI T H 6 I LUONG
Bai nay gom 2 tiet.
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Tiet thir nhat: On tap.
Tiet thii hai : Kiem tra.
IV. TIEN TraNH DAY HOC

HOAT DONG 1
l-T6m tat nhurng kien thurc can nh6
a) Muc dich: hi thd'ng vd dn tap lgi todn bd kii'n thifc co bdn cua chuong 1
b) Hudng thuc Men:
- Tdm tdt todn bg cdc don vi kii'n thicc theo don vi bdi hge.
- Chia ldp thdnh 4 uhdm,mSi nhdm tdm tdt mot muc, sau dd GV nhdn xet vd hifdng
ddn mdi HS tie tdm tdt vdo vd.

c) Qud trinh thuc hiin.
1.

Vecta

GV cho HS tra Idi cac cau hoi sau de on tap khai niem vectd.
HI. Vecto la gi?
H2. Vecto khac doan thdng d nhiing diem nao?
H3, Qua hai diem phan biet cd the cd bao nhiau vecto khac vecto-khdng?
H4. Neu khai niem vecto-khdng.
H5. Nau khai nieip '^ ^^ vecto cung phuang, cimg hudng, ngupc hudng.
2.

Tong va hieu tc^ V (jf

GV cho HS tra Idi cac cau hoi sau de 6' lap cac quy tac.
Cho ba diem A, B, C. Hay xac dinh

HI. T^ + W.
H2. A 5 - 5 C .
H3. Vdi O bat ky. Hay xac dinh ~OA-CB.
Cho ABCD la hinh binh hanh.

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H4. Xac dinh AB + AD.
3.

Tich ciia mdt vecta vdi mot sd

GV cho HS tra Idi cac cau hoi sau de on tich cua mot vectd vdi mot so.
HI. Cho b = kd.
a) So sanh dp dai cua a va b.
b) Khi nao hai vecto a va b cimg hudng? ngupc hudng?
H2. Hay nau cac tfnh chat cua phep nhan vecto vdi mdt sd.
H3. Cho I la trung diem cua AB. Hay dien so thfch hop vao chd trdng
07 = ....f0A + 0 5 j .
H4. Diem G la trpng tam tam giac. Hay dien so thfch hpp vao chd trdng
0G = ....(OA + OB + OC).
4,

Toa dp ciia vecta va ciia diem

GV cho HS tra Idi cac cau hoi sau de on tap toa do cua vectd va cua diem.
Hay dien vao chd trdng

HI. Dd'i vdi he true (O; /, j ) hay Oxy thi
1) u = (a; b) -^ u = ...i +...J ;
2) M = (x; y)<^OM = (...;

:..).

H2.Ne'uA = fx; y), B = (x'; y') thi JB = (X'-...;
H3. Neu il = (x; y) vav=(x';

...-y').

y') thi

1) u + V = (... + x'; y + ...);
2) kU = (...X; k...).

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HOAT DONG 2
Cau hoi tir kiem tra
a) Muc dich: Giup HS tu ddnh gid dugc muc do tii'p thu kii'n thde eua minh sau
khi hge xong chuong 1.
b) Hudng thuc hien:
- Hudng ddn HS trd ldi cdc edu hdi.
c) Qud trinh thuc Men hogt dgng.
Cau 1
GV ggi mot HS trd ldi
Hudng ddn. Vecto cd hudng, doan thdng khdng cd hudng. Mdt doan thang, vai
trd hai ddu miit nhu nhau, mdt vecto thi vai trd hai dau miit khac nhau.
Cau 2.
GV ggi mot HS trd ldi
Hudng ddn. La mdt hinh binh hanh.
Cau 3.
GV ggi mot HS trd ldi
Hudng ddn. Ta cdng hai vecto dau tien, cdng ldn luot cac vecto tie'p theo.
Cau 4.
GV ggi mgt HS trd ldi
Hcdng ddn. Xem lai dinh nghia hieu hai vecto.
Cau 5.
GV cho 4 nhdm HS thdo ludn 4 cdu. Moi nhdm cd 1 HS trd l&i.
Hudng ddn. HS xem lai khai niem tong, hieu hai vecto.
Trd ldi. a) Sai, b) diing, c) sai, d) diing.

Cau 6.
GV ggi mgt HS trd ldi
Hudng ddn. Cd, khi ta nhan vecto dd vdi -1 ta dupe vecto dd'i.
96