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

Chi/ONq III
FHl/CiWCS P I I A P 1X)A 1>0 l l l O K G l i H O ^ G

GlAN

I'huu 1
Gidl THIEU CHLfdNG

1. CAU TAO CHUONG
§1. He toa dp trong khong gian
§ 2. Phuang trinh mat phdng
§3. Phuang trinh dudng thang trong khdng gian
On tap chuang III
On tap cuoi nam

Muc dich ciia chuong
• Chuang III nhdm cung cap cho hpc sinh nhirng kiln thiic co ban vl khai niem
toa dp trong khong gian va nhirng ling dung ciia nd.
Tpa dp vecto va tpa dp dilm.
Bilu thirc tpa dp ciia cac phep toan vecto.
Tich vo hudng ciia hai vecto.
Phuang trinh mat cau.
• Gidi thieu vl phuang trinh mat phdng trong khdng gian.
Vecto phap tuyen ciia mat phdng.
- Piiuong trinh tdng quat ciia mat phdng.
- Dilu kien de hai mat phdng song song, vuong gdc.
Khoang each tir mot dilm din mot mat phdng.
• Phuang trinh dudng thdng trong khong gian:
66

Phuong trinh tham sd ciia dudng thdng.
- Dilu kien dl hai dudng thang song song.
- Dilu kien dl hai dudng thdng cheo nhau.
- Dilu kien dl hai dudng thdng cdt nhau.
II- MUC TIEU
1. Kien thirc
Ndm dupe toan bp kiln thirc co ban trong chuang da neu tren.
= Hieu cac khai niem va tinh chat vecto trong khong gian.
Hiiu va bilt dupe mdi quan he giira vecto phap tuyIn va cap vecto chi phuang
ciia mat phdng.
Hiiu va bilt dupe mdi quan he giira vecto phap tuylh va vecto ciM phuong ciia
dudng thdng.
2. KT nang
Xac dinh dupe cac vecto trong khdng gian.
Van dung dupe eac tinh chat dl giai bai tap
- Chiing minh dupe hai mat phang song song, vuong gdc.
- Lap dupe cac phuong trinh dudng thdng va phuong trinh mat phdng.
- Xac dinh dupe vi trf tuong ddi ciia dudng thdng va mat phdng, giiia hai mat phdng.
3. Thai do
Hpc xong chuang nay hpc sinh se lien he dupe vdi nhilu van de thuc t l sinh dong,
lien he duoc vdi nhirng van dl hinh hpc da hpc d Idp dudi, md ra mot each nhin
mdi vl hinh hpc. Tir dd, cac em cd thi tu minh sang tao ra nhiing bai toan hoac
nhirng dang toan mdi.
Kit luan:
Khi hpc xong chupng nay hpc sinh cdn lam. tdt cac bai tap trong sach giao khoa va
lam dupe cac bai kilm tra trong chuong.

67

P h ^ n 2,
CkC BAI SOAN

§1. He toa do trong khong gian
(tiet 1, 2, 3, 4, 5)
1. MUC TIEU

1. Kien thiic
HS ndm dupfc:
1. Khai niem toa dp vecto trong khdng gian, toa dp dilm va dp dai vecta.
2. Bilu thiie toa dp ciia cac phep toan : cdng, trir vecto; nhan vecto vdi mdt sd
thuc.
3. Bilu thiic toa dp ciia tfch vo hudng ciia hai vecto.
4. Phuang trinh mat cdu.
2. KT nang
•

Thuc hien thanh thao cac phep toan vl vecto, tfnh dp dai vecto,

•

Viet dupe phuong trinh mat cdu.

3. Thai do
•

Lien he dupe vdi nhilu van dl thuc t l trong khdng gian.

•

Cd nhilu sang tao trong hinh hpc.

•

Hiing thii trong hpc tap, tfch cue phat huy tfnh dpc lap trong hoc tap.

11. CHUAN DI CUA GV VA HS
1. Chuan bi ciia GV:
• Hinh ve 56 din 62.
• Thudc ke, phan mau,...
68


2. Chuan bi cua HS :
Dpc bai trudc d nha, cd thi lien he vdi phuong phap he tpa dp trong mat phdng.
ID. PHAN PHOI THOI LUONG
Bai dupe chia thanh 5 tilt :
Tilt 1: Ttr dau den hit muc 2
Tilt 2: Tilp theo din hit muc 3
Tilt 3: Tilp theo din hit muc 4
Tilt 4: Tilp theo din hit muc 5
Tilt 5: Tilp theo din hit muc 6
IV. TIENTOINHDAY HOC
n. DRT VAN D€
Cau hoi 1.
Nhdc lai khai niem hinh hop, hinh chdp.
Cau hdi 2.
Cho hinh lap phuong ABCDA'B'CD'
a) Chiing minh eac canh ciia hinh lap phuang xuat phat tir mot dinh
vudng gdc vdi nhau.

b) Cho canh cua hinh lap phuang la a, tfnh dp dai dudng cheo ciia hinh
lap phuong.

a. ani MOI
HOATDONCl
1. He true tpa do trong khdng gian

69


GV mo ta he true tpa dp trong khong gian va neu eau hdi :
HI. Hai vecta i, j co vuong gdc vdi nhau hay khong?
H2. Vecto k co vuong gdc vdi tat ca cac vecta thuoc mat phdng (Oxy) khong?
• GV neu dinh nghia:
He gdm ba true Ox, Oy, Oz doi mdl vudng gdc dugc ggi Id he true tog do
vudng gdc trong khdng gian.
• GV sir dung hinh 56 trong SGK va dat van dl:
H3. Hay dpc ten cac mat phdng tpa dp.
H4. Hay kl ten cac vecta don vi.
H5. Cd the cd them mot gdc tpa dp nira khac O hay khong?
H6. Hay neu cac tfnh chat ciia mat phdng tpa dp, vecta don vi?
-2

—

H7. Tfnh i = i.i

-2

—

-2

j = j.j k = k.k.

H8.Tfnn i.j,j.k, k.i
• Thuc hien | ? l | trong 4 phiit.
Hoat ddng ciia HS

Hoat ddng cua GV
Cdu hdi 1
Tai sao /

Ggi y trd Idi cdu hdi 1
~ j

=k

- I.

Do tinh chat ciia tfch vd hudng ciia
cac vecto cimg phuong va cd dp dai
bdng 1.

Cdu hdi 2
Taisaoi .j ^ i .k = k .i = 0.

Ggi y trd loi cdu hdi 2
Do tfnh chat ciia tfch vd hudng ciia
cac vecto vuong gdc

70


HOATDONC 2
2. Toa do cua vecto
• GV neu dinh nghia :
Trong khdng gian cho vecta a. Bd ba so (x

y

:) thda man

a = x.i + y.j + z.k ggi la tga do ciia vecia a. Ki hieu a(x;y;z) hoac
a = (x;y;z).
H9. Hay tim tpa dp ciia cac vecto i, j , k .
• Thuc hien ?2 trong 4 phiit.
Hoat ddng ciia GV
Cdu hdi 1
Tfnh / .u.
Cdu hdi 2
Tfnh u.j

Hoat ddng cua HS
Ggi y trd Idi cdu hdi 1
((./ = (xi + yj + :k).i ~ xi - x
Ggi y trd Idi cdu hdi 2
HS tu tfnh.

' Thuc hien vf du 1 trong 5'. GV sit dung hinh 57.

Hoat ddng cua GV
Cdu hdi 1

Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1


Bilu diln

OM

theo cac vecto

OM=-(c)J

+ OK)

don vi.
^. 1 -. I T
= 0/+— ; + —A:,
2
Cdu hdi 2
Xac dinh tpa dp ciia OM .
Cdu hdi 3

2

Ggi y trd Idi cdu hdi 2
1 1 ^

OM = 0;
V 2 2ij.
Ggi y trd loi cdu hdi 3

Bilu diln

MG

theo cac vecto

MG

^OG-()M,

don vi.

J___l_
V2)

(I

--0 /+
3
3

Cdu hdi 4

2

J+

k.

Ggi y trd Idi cdu hdi 4

Xac dinh toa do cua MG

HS tu viet.

• GV neu eac tfnh chat ciia tpa dp vecto :
Cho

cdc

vecta

i(| = ( X | ; y, ; 2,),

/i, =(A-2 ; ^2 ' - 2 )

'''^' ^^'•'' ^ '">' ^''

ta cd :
1) ii, = fh <=> xj = X2, yi = y2, 'i = '2
2) » , + / Y 2 = ( x , + J r 2 ; y | + y 2 ; 2 , + Z 2 )
ij

"i-»2=(-^i--^2;>'i->'2;zi--2)

4) kfi^ = ikxi; kyi; kz^)

5) 11^ .112 =X^X2+

>•]^2 + z,Z2

6ihhv^=V^f+>-?+zf
7) cos(/<, /I2) •

X i X 2 + y i y 2 + ZiZ2
-2 , -.2 , .2

^Xf+yf+zJ-

/ 2 , ,.2 , „2

v<7/ »| 9t 0; M2 '^ 0

^^2+^2+22

5j ((, J. »2 ci> ii|.i(2 = 0 C5> X|X2 +)']y2 +Z|Z2 = 0

72


HOATDONC 3
3. Toa dd ciia diem
HIO. Cho OM = x.i + y.j + z.k. Cd bao nhieu bd sd thuc x, y va z thda man he thirc
tren.
• GV tra Idi va neu dinh nghia :
Bg ba sd thuc (x; y ; z) thda mdn OM = x.i + y.j + z.k ggi Id Iga do diem
M vd ki hieu M (x ; y zj hoac M = (x : y . z).

Hll. Cho M (0 ; 0 ; 0). Hay chi ra M tren he true tpa dp.
H12. Cho M(0 ; 1 ; 2). Hdi M thudc true nao ?
H13. Cho M(l ; 0 ; 2). Hdi M thuoc true nao ?
H14. Cho M(l ; 2 ; 0). Hdi M thudc true nao ?
• Thuc hien |?3| trong 4 phiit
Hoat ddng ciia GV
Cdu hdi 1
Tai sao M{0\ 0 ; 0 ) .
Cdu hdi 2

1

Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1
Vi 0 0 = 6 = (0 ; 0 ; 0)
Ggi y trd Idi cdu hdi 2

Tai sao M e (Oxy) o z = 0, tire

M e (Oxy) c^OM Ik c:> ()M.k = 0

la M = (x ; y ; 0).

<^ z = 0 tuc la M = (A- ; y ; 0).

Cdu hdi 3
Hay giai thich cac y cdn lai.

Ggi y trd Idi cdu hdi 3
HS tu lam tuong tu.

• Thuc hien [?4j trong 4 phut
Hoat ddng ciia GV
Cdu hdi 1
M e Ox khi nao?

Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1
M (x ,• y : z) G Ox <=> X = z = 0.

73


Cdu hdi 2

Ggi y trd Idi cdu hdi 2

M e Oy, M e Oz khi nao?

HStu tra Idi.

• Thuc hien - ^ 1 trong 4 phiit.
Sir dung hinh ve 59. GV cho HS len bang ve lai hinh va hudng ddn HS thuc hien
1k

o,

[

i j __ic__i

•

:

:

! ^-':

;

^-"T

:

\

!

r -"i-^Tr 1 i 1
,.-i k

/^

_..;

' 1

i

1

i

J \

\\

:

i

;

•--I^J

rr*

r^

:•>

7*

7

^

...-::l.^;;::...^:::l...;::;.fi^;:::.

Hoat ddng cua HS

Hoat ddng cua GV

Ggi y trd Idi cdu hdi 1

Cdu hdi 1
Tim toa dp eiia A, B, C, D va E.

Cdu hdi 2
Xac dinh P.

A = (2 ; 0; 0). Cac dilm khac HS tu
lam.
Ggi y trd Idi cdu hoi 2
HS tu xac dinh.

HOATDONC 4
4. Lien he gi&a toa do cua vecto va toa do cua hai diem mut
• GV neu dinh nghia :
Cho hai diem A(x^ ; JA ' ^/l) "^^ ^i^B > >'B' ^B)74


1) AB = (xg - x ^ ; J5-^'/i; 2fi-z^)
2) AB ^ Mx, -x^f

' Thuc hien f \

+{yB-yA)'+[-B-^^A)

2 trong 4 phiit.

a)
Hoat ddng ciia GV
Cdu hdi 1

Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1

Neu cdng thiic vecto vl trung

07 = ^ ( 0 A + 0fi)

dilm I ciia AB.

Ggi y trd Idi cdu hdi 2

Cdu hdi 2
Tim tpa dp ciia I.

y/ = 2\yA

+ yB)^

2/ = 2 ^ - ^ ^

- " . ) •

b)
Hoat ddng cua GV

Cdu hdi 1
Neu cdng thii'c vecta vl trpng tam

Hoat ddng cua HS
Ggi y trd loi cdu hdi 1
OG = ^{OA + '0B + 0C^

G ciia tam giac ABC.
Cdu hoi 2

Ggi y trd Idi cdu hdi 2

Tim tpa dp ciia G.
-VG = 3(>'-4 +yB +

yc);

c)

Hoat ddng ciia GV
Cdu hdi 1
Neu cong thiic vecto vl trpng tam

Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1
~0E ^ ^(7)A+ '0B+ ~0C+ ~0D^

75

E ciia tir dien ABCD.
Cdu hdi 2
Tim toa dp eiia E.

Ggi y trd Idi cdu hdi 2
^E =J{XA

+ ^B + ^C + ^D)

yE =j{yA+yB
^E =j(^A

+ yc +

;

yD);

+^B +^C +Zfl)-

• Thuc hien vi du 2 trong 5'
a)
Hoat ddng cua GV
Cdu hdi 1
Tim toa dp ciia A, B, C, D va E.

Cdu hdi 2
Xac dinh P.

Hoat ddng cua HS

Ggi y trd Idi cdu hdi 1
A = (2 ; 0; 0). Cac dilm khac HS tu
lam.
Ggi y trd Idi cdu hoi 2
HS tu xac dinh.

• Thuc hien vf du 2 trong 5'
a)
Hoat ddng ciia GV
Cdu hdi 1

Hoat ddng ciia HS
Ggi y trd Idi cdu hoi 1

Khi nao 4 dilm khdng ddng Khi ba vecto AB, AC,AD
phang.
Cdu hdi 2
Hay chiing minh cau a)

ddng phang.
Gin y trd Idi cdu hoi 2
HS tu giai.

b)
Hoat ddng ciia GV
Cdu hdi 1

76

Hoat ddng cua HS

Ggi y trd Idi cdu hoi 1

khdng


Khi nao hai dudng thdng vudng Khi tich ciia hai vecto nhan hai dudng
thdng dd lam gia bdng 0.
gdc ?
Ggi y trd Idi cdu hdi 2
Cdu hdi 2

HS tu giai.

Hay chiing minh cao b)
c)
Hoat ddng ciia HS

Hoat ddng cua GV

Ggi y trd Idi cdu hdi 1

Cdu hdi 1
Neu khai niem hinh chdp diu.
Cdu hdi 2

HS tu neu.
Ggi y trd Idi cdu hdi 2

Hay chiing minh cau c)

Chiing minh : DA = DB = DC, tam
giac ABC diu.

2)
Hoat ddng ciia GV

Hoat ddng cua HS

Cdu hoi 1

Ggi y trd Idi cdu hdi 1

H cd tfnh chat gi ?

H la trpng tam tam giac ABC.

Cdu hdi 2

Ggi y trd Idi cdu hdi 2

TimH.

US tu giai.

HOATDONC 5
5. Tich cd hudng ciia hai vecto
HI5. Nhdc lai tfch vo hudng ciia hai vecto.
HI6. Neu bilu thirc tpa dp vl tfch vo hudng eiia hai vecto.
• GV neu dinh nghia 2

77


Tich cd hirdng (hay lich vecta) cua hai vecta uia; b; c) vd via'; b [c')
Id mdl vecta dugc ki hieu Id \u, v] (hoac « A v ) vd cd tga do dugc xdc
dinh nhu sau :

[«-v] =

b

c

b'

c'

c
c'

a
a'

a

b

a'

b'

(bc'-b'c;ca'-c'a;ab'-a'bj.

H17. Tim tfch cd hudng cua u = (l;2;3) va v = ( - 3 ; - 2 ; l ) .
• Thuc hien (?>, 3 trong 4 phiit.
HoaFddng ciia GV
Cdu hdi 1
Tfnh

Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1
-- -;'. /

iJ

/ 0
V1

= iO;0;l)

0

0 1
5

0 0

0 '

1 0\

0 1/

=k

Ggi y trd Idi cdu hdi 2

Cdu hdi 2
Tfnh cac bilu thiic cdn lai.

HS tu tfnh.

• GV neu tfnh chat ciia tfch cd hudng:
1. [u, v] = 0 khi vd chi khi hai vecta u vd v ciing phuang.
2. Vecta \ii, v]

vudng gdc vdi cd hai vecta

U vd

v

lire Id

[it, v].u-[u, v'].i^ = 0.
3. |[H, V]| = |ii|.|v|.sin(i<, v).
HI8. Hay chiing minh cac tfnh chat tren.
9 GV neu chii y trong SGK:
Ta ve cdc vecta OA = /7 OB ^ v. Ne'u hai vecta u vd v khdng cimg
phucmg, ggi S Id dien lich hinh binh hdnh cd hai cgnh la OA vd OB, khi
do |(7|.|v|sin((7, v) = OA.OB.sinAOB = S.

78


Vdy dd ddi ciia vecta [u, v] bang sddo dien lich hinh binh hdnh ndi tren.
Ifng dung tfch cd hudng
a) Tinh dien lich hinh binh hdnh
Ne'u ABCD Id hinh binh hdnh thi dien tich S ciia nd Id
S = AB.ADsinA = IA^I . |AD| . sin(A6, A D ) = | [ A 5 , A D ]
H19. Cho A(l ; 2 ; 3), B(-l ; 2 ; 0). Tfnh dien tfch hinh binh hanh OABC, AOCB.
b) Tinh the tich hinh hop
Ne'u ABCD.A'BC'D' la hinh hop vdi dien lich ddy ABCD la S, chieu cao
la h = AH, cp la gdc hgp bdi hai vecta AA' vd [Afi, AD] (h.61) thi the
lich cita hinh hop dd Id : [AB, AD] . AA'
• Thuc hien ,Q, 4 trong 4 phut.
Hoat ddng ciia GV
Cdu hdi 1

Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1

Gia sir ba vecto u, v. vv ddng

Gia sir ba vecto u, v, w ddng phdng.

phdng, chiing minh [/7, v].w = 0.

Khi dd :
Nlu
U,

V

U, V

u, V

cimg

phuang

thi

va

do

dd

=0

.w = 0 w -0.

Nlu u, V khdng ciing phuong thi
w = pu + qv nen :
II, V

=P

.w =

U, V

u, V

.11 + q

.(pu + qv)
ll, y

.v = 0.

79


Cdu hdi 2

Ggi y trd Idi cdu hdi 2

Gia sir [/7, vj.w = 0chiing minh

Ngupe lai gia sir

u, V

.M^ = 0. Nlu

ba vecto ddng phdng.
u, V

0 thi ;(, V ciing phuong va

do dd u, V, w ddng phdng.
Neu

u,v

^Q

thV ca ba vecto

/(, V, w diu vudng gdc vdi vecta
u, \

^ 0

nen ba vecto dd dong

phdng.
• GV neu tfnh chat:
;7 _L V c^ ».v = 0.

u vd V ciing phuang <=> [», v J = 0.
II, V, w ddng phdng <^ \u, vj.vT' = 0.
• Thuc hien vf du 4 trong 6'
a)
Hoat ddng eiia GV
Cdu hdi 1

Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1

Khi nao 4 dilm khong ddng HS tu tra Idi
phdng.
Cdu hdi 2
Chiing minh 4 dilm dd khong
ddng phdng.

Ggi y trd Idi cdu hdi 2
Chirng minh \JA,

'BC'\

'BD +Q.

b)
Hoat ddng cua GV

80

Hoat ddng ciia HS


Ggi y trd Idi cdu hdi 1

Cdu hdi 1
Tfnh dp dai dudng cao ke tir A.

AH = —ABC_ HS tu tfnh tiep.
BC
Ggi y trd Idi cdu hdi 2

Cdu hdi 2
Tfnh ban kfnh dudng trdn ngoai
tilp tam giac ABC.

SABC

= P - ' ' ^ ' ' =

-••

P

c)

Hoat ddng cua GV

Hoat ddng cua HS

Cdu hdi 1

Ggi y trd Idi cdu hdi 1
HS sir dung true tiep cdng thiic.

Tfnh COSCBD

Ggi y trd Idi cdu hdi 2

Cdu hdi 2

HS tu tfnh.

Tfnh cosa.
d)
Hoat ddng cua GV

Hoat ddng ciia HS

Cdu hdi 1

Ggi y trd Idi cdu hdi 1

Tfnh thi tfch tii dien.

V

- ^ \BA,BC\BD

^ ABCD -

Cdu hdi 2

r

Ggi y trd Idi cdu hdi 2

Tfnh chilu cao cua tii dien.

HS tu tfnh.

HOATDONC

6

6. Phuomg trinh mat cau
• GV neu each chia mdt sd khd'i da dien va dat cau hdi:
H20. Tfnh khoang each giiia hai dilm M(x ; y ; z) va I (a ; b ; c).
H21. Bilt khoang each dd la r, hay lap bilu thirc mdi quan he do.
• GV neu dinh li

_5
~6


Mdl cdu tdm I(XQ ; yg ; ZQ), bdn kinh R cd phuang trinh
(x-Xo)2+(y-yo)2+(z-Zo)^=/?^
• GV hudng ddn HS chimg minh dinh If tren.
• Thuc hien trong ^

5 phiit.

Cdch I
Hoat ddng ciia HS

Hoat ddng cua GV

Ggi y trd Idi cdu hdi 1

Cdu hdi 1
Tam I cua mat cau d dau ?

Cdu hdi 2

I la trung dilm. AjAj .
Ggi y trd led cdu hoi 2

Tim tpa dp I.

Cdu hdi 3

fa^+a2 h+b2,c^+C2
/= ^ 2 ' 2 ' 2

]
j

Ggi y trd Idi cdu hdi 3

Vilt phuong trinh mat cau.
R = \A,A2
1
i(ai-a2f+(bi-b2f+(ci-C2f
~2
viet phuong trinh mat cau.

HS ty

Cdch 2
Hoat ddng cua GV
Cdu hoi 1
Gia sir M = (x ; y ; z) tim tpa dp

Hoat ddng cua HS
Ggi y trd Idi cdu hoi 1
A,M = (x - a, ;y - b^ ;z -Cj),

A[M va A2M.
A2M = (x - ^2 ;>' - ^ 2 ' 2 ~ '^2)-

Cdu hdi 2
Vilt phuong trinh mat cdu.

Ggi y trd Idi cdu hoi 2
{x-a^){x-a2)

+

+ (z-c,)(z-c-2) = 0
• Thuc hien ^ C 6 trong 5 phiit.
82

{y-b^){y-b2)


Cdch 1
Hoat ddng cua HS

Hoat ddng ciia GV
Cdu hdi 1

Ggi y trd Idi cdu hdi 1

Gpi phuang trinh mat cau cd d^O
dang x' + y" + 7^ + 2ax + 2by
+ 2c: + d - 0, hay tim mdi
quan he khi mat cau di qua A.
Cdu hoi 2

Goi y trd Idi cdu hdi 2
l+2a = 0;l+2b

= 0;\+2c

=0

Tim mdi quan he khi mat cdu
di qua B, C va d.

Goi y trd Idi cdu hdi 3

Cdu hdi 3
Vilt phuong trinh mat cdu.

2

2 ' ^

X +y +z~-x-y-z

= 0.

Cdch 2
Hoat ddng ciia GV
Cdu hdi 1

Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1

Gpi / (x; y; z) la tam mat cdu. Em 1A=^IB = IC = ID.
cd nhan xet gi vl : IA, IB, IC va
ID.

Ggi y trd Idi cdu hdi 2

Cdu hdi 2
Tfnh IA, IB,IC va ID va tim cac
mdi quan he ciia x, y va z.

x^ + y^ + z^ = (x - 1)2 + y^ + z^
' x^ + y^ +z^ =x^ +(y-

1)2 + z^

^2 + y2 + z^ = X^ + y2 + ^. _ lj2

Cdu hdi 3
Vilt phuang trinh mat cdu.

Ggi y trd Idi cdu hdi 3
HS tu vilt.

H22. Hay neu mdt dang khac ciia phuang trinh mat cdu.
• GV neu nhari x l t :

83


Phuang trinh x^ + y' + z^ + 2ax + 2by + 2cz + d = 0 Id phuang
trinh ciia mat cdu khi vd chi khi a^ + b^ + c^ > d. Khi dd lam mat
can Id diem l(-a

: -b ; -c) vd bdn kinh mdl cdu Id
r = ^a^ +
>

7

2

b^+c^-d.
2

H23. d phai thoa man diu kien gi de x + y + z + 2ax + 2by + 2cz + d- = 0 la
phuang trinh ciia mat cdu ?
7 trong 5 phiit.
• Thuc hien ^
Hoat ddng cua HS

Hoat ddng ciia GV

Ggi y trd Idi cdu hdi 1

Cdu hdi 1

,
, , 2 2
'
Phuang trinh a) cd la phuang
Khdng phai, vi he sd cua x y va z'
trinh mat cdu hay khdng ?
khdng bdng nhau.
Cdu hdi 2
Grn y trd Idi cdu hdi 2

Phuang trinh b) cd la phuang La phuang trinh mat cau cd tam (1; 0; 0),
ed ban kfnh bdng 1.
trinh mat cdu hay khdng ?
Ggi y trd Idi cdu hdi 3

Cdu hdi 3

Phuang trinh c) cd la phuang Khong phai, vi phuong trinh sau khi nit
gpn vdn cdn chiia so hang -2xy.
trinh mat cau hay khdng ?
Ggi y trd Idi cdu hdi 4

Cdu hdi 4
Phuang trinh d) cd la phuong
trinh mat cdu hay khdng ?

7

2

'

Phuong trinh nit gpn thanh x +y + z" = 1.
Dd la phuong trinh mat cau vdi tam
(0; 0; 0) va ban kfnh bdng 1.

HOATDONC 7
TOM TfiT Bfil HOC
1. Cho cac

vecto

w, =(X|; y, ; z,),

ta cd :
1) (7, = 1/2 <=> xj = X2, y, = y2, z^ = z^

84

"3 ^ (-^2 ' 3'2 ' ^2)

^^ so k tuy y,


2)

u^+^2={x^+x2•,y^+y2\Zl+h)

3) «, - « 2 = ( x , - X 2 ; y, -^2= Zi -Z2)
4) kui = (kxi ; ky^ ; fe,)
5) Mj.r<2=x,X2+y|y2+z,Z2

6)\u^\ = ^|ii^ = ^lx[Vy^+zf
x,X2+y,y2 + ZiZ2

7) COS(H, , i<2):

..2 , ,.2 , _2

vdi i(| ?i 0; t72 ^ 0

/ 2 , ..2 . _2

^xi^+y^+zf

V4+3'2+22

8) H, ± M2 " ^ "l-"2 ~ 0 *^ -^1^2 +}'l}'2 +^1^2 "*-*•
2. M =(x ; y ; z) <» OM = xi + yj + zlc
3. Cho hai dilm A(x^ ^ )'/i ^ 2^) va fi(xg ; yg ; zg).
1) AB = ( x g - x ^ ; y g - y ^ ;

Zg-Zj^)

2) A5 = V(xg - x^ f + (yg - y^ )^ + (Zfi - z ^ )^
4. Mat cau tam I(a ; b ; c), ban kfnh r cd phuang trinh

ix-af+iy-bf+iz-cf^r'^
Phuang trinh x + y + z + 2ax + 2by + 2cz + d = 0 la phuang trinh ciia mat cau
2 2 2
^
^
^
khi va chi khi a + b + e > d. Khi dd tam mat cau la diem I ( - a ; - b ; - c ) va
ban kfnh mat cdu la
r = ^ a^+b^

5. [;7, v] =

b

c

c

a

a

b \

b

c'

c'

a

a'

b'

+c^-d.

= (bc'-b'c;ca'-c'a;ab'-a'b).

6.
I. \u, v] = 0 khi vd chi khi hai vecto iJ vd v cung

phuang.

85


2.Vecia [u, v] vudng gdc vdi cd hai vecta u vd v ,tircld [u, v].»=[/7, v].v=0.
3. |[;7, v^]| = |;7|.|v|.sin(i7, v).
7. 5 = AB.ADsmA = [AB\
VABCD..VB'CD'=|[^,AD]

I A 5 | . sin (AS, Td^) = |[ Afi, A B ] |
JA'.

HOATDONC

8

MOT SO Cfia HOI TR^C NGHIEM
Hay diln diing (D) sai (S) vao cac khang dinh sau :
Cdul.

Cho a = (l;2;3), b = ( - 2 ; 3 ; - l ) . Khi dd a + b cd toa dp la

(a)a + b c d t o a d d l a ( - l ; 5 ;2)

[]

(b) a - b c d t o a d p l a ( 3 ; - l ;4)

[]

(c) b - a c d t o a d p l a ( 3 ; - l ;4)

[]

(d) Ca ba khdng dinh tren diu sai

[]

Trd Idi.
a

b

c

d

D

D

S

S

Cdu2

Cho a = (l;2;3), b = ( - 2 ; 3 ; - l ) . Khi dd a + b cd toa dp la

(a)3a + b c d t o a d p l a ( l ; 9 ; 8)

[]

(b) a - 2 b c d t o a d d l a ( 5 ; - 4 ; 5)

[]

( c ) 2 b - a cdtoaddla(5 ;-4;5)

[]

(d) Ca ba khdng dinh tren diu sai

Q

Trd Idi.

86


a

b

c

d

D

D

s

S

Cdu 3. Cho a = (l;2;3), b = ( - 2 ; 3 ; - l ) . Khi dd a + b cd toa dp la
(a) a.b = 1

•

(b)a.b=-l

•

(c)2b.a=2

\2

(d) Ca ba khdng dinh tren diu sai

[]

Trd Idi.
a

b

c

d

D

s

D

S

Cdu 4. Cho hinh cau cd phuang trinh : (x -1)^ + (y + 2)^ + (z + 3)^ = 2
(a) Tam ciia hinh cdu la 1(1 ; -2 ; -3)

D

(b) Tam cua hinh eau la I(-l ; 2 ; 3)

•

(c) Ban kfnh ciia hinh cdu la 2

D

(d) Ban kfnh ciia hinh cau la \/2

•

Trd Idi.
a

b

c

d

D

S

S

D

Chon khang djnh diing trong cac cau sau:
Cdu 5. Trong cac cap vecta sau, cap vecto ddi nhau la :
(a) a = ( l ; 2 ; - l ) , b = ( - l ; - 2 ; l ) ;

87


(b)a = ( l ; 2 ; - l ) , b - ( l ; 2 ; - l ) ;
(c) a - ( - l ; - 2 ; l ) , b = ( - l ; - 2 ; l )
(d) a = ( l ; 2 ; - l ) , b = ( - l ; - 2 ; 0 ) ;
Trdldi.

ia).

•

Cdu 6. Cho hinh ve :
]•••"

i

i

i

:

i

,','

i

i

i

~L

-rr'

1

i

j

i

i

i

i

•

,^f""

t''

i

1

I

i

.-i''''
,-i''

1 ^Ti

y" '• r

\

ic 1

•;..•••

\ . . . - ' ' _ » ; ; :

.-':"'

1

1

'\TJ;T—^—pj—^J

1

i

1^—^J—^J—•

Diem D cd toa dp la
(a) (5 ; 1 ; 0 ) ;

(b) (0 ; 1 ; 5)

(c)(l;5;0);

(d) (1 ; 0 ; 5)

Trdldi. (d).
Cdu 7. Cho hinh ve :

t

V

..,•••" i Dy-

•«—

..;-"• i k

!

i

ic i

"E^

\

\

1

J

-i.

\j

^

-4^
J»

X

88

M

^

j„.

_^

^

^,H«

,*

•


Dilm C cd toa dp la
(a) (4 ; 4 ; 0 ) ;

(b) (4 ; 0 ; 4)

(c) (0 ; 4 ; 4 ) ;

(d) (0 ; 0 ; 4)

Trdldi. (b).
Cdu 8. Cho hinh ve :

Dl

\c

i £:

! /(.-

./ !

-U^

:^..^:::1.^:::1.^::1.^::::'B..:;;1
Dilm A cd toa dp la
(a) (0 ; 2 ; 0 ) ;

(b) (2 ; 0 ; 2)

(c) (2 ; 0 ; 0 ) ;

(d) (0 ; 0 ; 2)

Trd Idi. (c).
Cdu 9. Cho hinh ve :

..-• ••• i D l

I E\

^rn• « . - • • •

89


Dilm B cd toa dp la

(a) (4 ; 4 ; 0 ) ;

(b) (4 ; 0 ; 4)

(c) (0 ; 4 ; 4 ) ;

(d) (0 ; 0 ; 4)

Trd Idi. (a).
Cdu 10. Cho hinh ve :
L

i Dy

1 i
i

',

"\ ..i--"11 --bTr
'—ill
./lii

hr-^

V

..--•

.--"B.---

i
^'^

ic i
''

'•

\ 1i
i

i

^'^

y'

i

i

i

i

7-^—7^

1

"• Ai

Diem E cd toa dp la
(a) (3 ; 4 ; 3 ) ;

(b) (4 ; 3 ; 4)

(c)(3;4;4);

(d) (3 ; 0 ; 4)

Trd Idi. (a).
Cdu 11. Cho A (1 ; 2 ; 3), B(-l ; 0 ; 0)
Dilm C ma OABC la hinh binh hanh la
(a) (0 ; 2 ; 3) ;

(b) (4 ; 3 ; 4)

(c) (3 ; 4 ; 4 ) ;

(d) (3 ; 0 ; 4)

Trd Idi. (a).
Cdu 12. Cho A (1 ; 2 ; 3), B(-l ; 0 ; 0)
Dien tfch hinh binh hanh OABC la

90