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Thiết Kế Bài Giảng Hình Học 12 Tập 2
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TRA N VINH
I hiet ke bai giang H)NHHQC|2
TAP HAI
N H A XUA T BA N H A N0I
TRAN VINH
THIET KE BAI GIANG HINH HOC 12 TAP 2
NHA XUAT BAN HA NOI
Chi/dNq 2
M A T NON, MAT TRU , MAT CAU
Pha n 1
Gidl THlfu CHL/CJNG
I. CAU TAG CHUONG
§ 1. Khai niem ve mat trdn xoay
§ 2. Mat eau
On tap chuang II
I. Muc dich ciia chuong
• Chuang II nham cung ca'p cho hpc sinh nhiing kien thdc co ban ve khai niem cac khd'i trdn xoay trong khdng gian ma chii yeu la mat ndn, mat tru va mat cau. Mat ndn trdn xoay : Day, dudng sinh va dudng trdn day.
Dien tich xung quanh va dien tich toan phan ciia mat ndn.
The tfch ciia mat ndn.
• Mat tru trdn xoay la gi ?
Dien tich xung quanh va dien tich toan phSn cua mat tru.
The tfch ciia mat tru.
• Mat eSu la gi ?
Dien tfch ciia mat cau.
Thi tich ciia mat cau.
II. MUC TIEU
1. Kien thiirc
Nam dupe toan bp kien thiic ca ban trong chuong da neu tren.
Hieu cac khai niem cac mat trdn xoay: Mat ndn, mat tru va mat eau. Nam dupe eac cdng thiic tfnh dien tfch, the tfch cua eac mat trdn xoay. 2. KI nang
Tinh dupe dien tfch xung quanh, dien tfch toan phan ciia cac hinh trdn xoay. Tfnh dupe the tfch cua hinh lang tru, hinh ndn.
3. Thai do
Hpc xong chuong nay hpc sinh se lien he dupe vdi nhieu van de thuc te sinh ddng, lien he dupe vdi nhiing van de hinh hpc da hpc d Idp dudi, md ra mdt each nhin mdi ve hinh hpc. Tii dd, cac em cd the tu minh sang tao ra nhiing bai toan hoac nhiing dang toan mdi.
Ket ludn:
Khi hpc xong chuong nay hpc sinh can lam tdt eac bai tap trong sach giao khoa va lam dupe cac bai kiem tra trong chuong.
Pha n 2
cAc &Al SOAN
{Ti^p theo)
§2. Mat cau
(tiet 6, 7, 8, 9, 10)
I. MUC TIEU
1. Kien thurc
HS nam duoc:
1. Khai niem chung vd mat cau.
2. Diem thudc mat cau , diem d trong va diem d ngoai mat cSu. 3. Giao cua mat cau va mat phang.
4. Giao cua mat cdu va dudng thang.
5. Tie'p tuye'n ciia mat cau.
6. The tfch va dien tich cua mat c^u.
2. KI nang
• Ve thanh thao cac mat cau.
• Xac dinh dupe mdt mat phang la tie'p dien cua mat cau, mot mat phang la tie'p tuye'n cua mat cau.
• Xac dinh dupe vi trf tuong ddi cua mat phang va mat eau.
• Tfnh dupe the tfch va dien tfch cua mat cau.
3. Thai do
• Lien he dupe vdi nhieu van de thuc te' trong khdng gian.
• Cd nhieu sang tao trong hinh hpc.
• Hiing thii trong hpc tap, tfch cue phat huy tfnh dpc lap trong hpc tap.
II. CHUAN BI CUA GV VA HS
1. Chuan bi ciia GV:
• Hinh ve 2.15 de'n 2.26.
• Thudc ke, phan mau,...
2. Chuan bi cua HS :
Dpc bai trudc d nha, cd the lien he cac phep bien hinh da hpc d Idp dudi in. DHAN PHOI TH6I LUONG
Bai dupe chia thanh 5 tiet:
Tie't 1: Tir dau den het phan I
Tie't 2: Tie'p theo de'n het phan II
Tie't 3: Tie'p theo den het phan III
Tie't 4: Tie'p theo den het phan IV
Tie't 5: Hudng din bai tap
IV. TIEN TDlNH DAY HOC
n. DAT VAN D€
Cau hdi 1.
Hinh trdn xoay la gi ? Hay ke mdt vai hinh trdn xoay da hpc. Cau hdi 2.
Hau ndu each tao ra hinh ndn va hinh tru.
C^u hdi 3.
Mat tru trdn xoay va hinh tru gidng va khac nhau d nhihig diem nao?
R. Bni 1V161
HOATDONCl
I. MAT CAU VA KHAI NIEM LI£ N QUAN DEN MAT CAU
1. Mat cau
GV neu cau hdi :
HI. Em hay neu khai niem hinh cau theo each nghi ciia minh.
H2. Qua dia ciu cd phai la hinh ciu hay khdng?
• GV su dung hinh 2.14 trong SGK va dat van de:
H3. Cho hai diem M^ va M2, hay so sanh OM^ va OMj.
H4. Khi cat mdt mat eau va mdt dudng thing di qua O, ta dupe nhiing doan thang. Hay so sanh cac doan thang dd.
H5. Mdt doan thang di qua O eat mat cau tai A va B. Hay neu vi trf ciia O ddi vdi AvaB.
H6. Mdt doan thang di qua O cat mat cau tai A va B. Mdt diem M bat ki tren mat cau dd. Hay do gdc AMB .
H7. Neu dinh nghia mat cau theo y ciia em.
• GV neu dinh nghia :
Tap hgp nhirng diem cdch deu mgt diem O cho trUdc mgt khodng khdng ddi la mgt mat cdu.
H8. Hay neu kf hieu dudng trdn tam O ban kfnh R.
H9. Day cung cua dudng trdn la gi ?
HIO. Hay ndu tfnh chat ciia day cung Idn nhat.
• GV neu dinh nghia dudng kfnh cua dudng trdn:
Dudng kinh cua dudng trdn la ddy cung di qua O.
H I 1. Hay so sanh dp dai eiia dudng kfnh vdi ban kfnh, vdi day cung bat ki?
H12. Khi bie't dudng kfnh ciia mat cau thi cd xac dinh dupe tam va ban kinh cua mat cau hay khdng?
• GV neu su xac dinh cua mat cau:
Mgt mat cdu xdc dinh khi bie't tdm vd bdn kinh hoac dudng kinh ciia nd. 2. Diem nam trong va nam ngoai mat cau. Khd'i cau.
HI3. M thudc mat cau (O ; R). Hay so sanh OM va R.
H14. M d ben trong mat cau (O ; R). Hay so sanh OM va R.
H15. M d ben ngoai mat cau (O ; R). Hay so sanh OM va R.
• GV neu cac khai niem diem nam trong, nam tren va nam ngoai mat cau va cho HS dien vao bang sau:
OM R
Vi trf
3 5
3
Tren
7 6
9 8
• GV neu dinh nghia khd'i eau
Tap hgp tdt cd cdc diem thudc mat cdu S(0; R) vd tdt cd cdc diem d trong mat cdu ggi la khdi cdu.
HI6. Danh dau x vao 6 ma M thuoc khd'i cau
OM
R
Thudc khd'i
cau
3. Bieu di^n mat cau
3 3
3 4
7 6
9 9
• GV neu eac bieu dien khd'i cau theo y sau Ve mdt dudng trdn.
Tam mat cau la tam dudng trdn.
8
- Cac mat phang cdn lai cat mat cau la mdt hinh dupe bieu dien la hinh Elip. Tham khao hinh 2.16
• GV neu qua ve each bieu dien mat cau nhd phep chie'u .
HI7. Hay ve mdt mat cau di qua ba diem A, B va C. Mat cau dd cd duy nhat khdng?
4. Dudng kinh tuye'n va vl tuye'n cua mat cau
HIS. Hay neu khai niem dudng kinh tuye'n va vT tuye'n trong dia If. • GV sii dung hinh 2.17 va dat cac cau hdi:
HI9. Kinh tuyen la ca dudng trdn Idn hay niia dudng trdn Idn?
H20. Hay ve kinh tuye'n va vi tuye'n.
• GV neu dinh nghia kinh tuye'n va vl tuye'n:
Giao CIM mat cdu vd nira mat phdng cd bd la true cua mat cdu ggi la kinh tuye'n. Giao ciia mat cdu vd mat phdng vudng gdc vdi true ciia nd ggi la vT tuye'n. • Thuc hien ^ 1 trong 4 phiit.
Hoat ddng cua GV
Cdu hdi 1
Tam giac ABC cd dac diem gi? Cdu hdi 2
Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1 Tam giac can
Ggi y trd Idi cdu hdi 2
Gpi H la trung diem AB ; OH cd dac diem gi ?
Cdu hoi 3
0 thudc mat phang nao ?
OH 1 AB.
Goi y trd Idi cdu hoi 3 Mat phang trung true ciia AB.
HOATDONC 2
II. GIAO CUA MAT CAU VA MAT PHANG
GV sii dung hinh 2.18.
Dat OH = h ;
1. Trudng hpp h > r
H21.SosanhOMvar.
H22. M nam trong hay ngoai mat cau?
• GV ket luan M nam ngoai mat ciu va giao ciia mat ciu va mat phang la khdng cd. 2. Trudng hpfp h = r
H23. H thuoc S. Diing hay sai ?
H24. M 7^ H thi M khdng thudc S. Diing hay sai.
• GV ket luan :
Giao cua mat cau S va mat phang P: Mat phang (P) tiep xiic vdi mat cau. • Dua vao hinh 2.19 GV dua ra cac cau hdi sau:
H25. Mpi dudng thing thudc (O) khong di qua H cd giao vdi mat ciu hay khdng? • GV neu dinh If:
Dieu kien cdn vd du de mat phang (P) tie'p xuc vdi mat cdu S(0 ; r) tgi H la (P) vudng gdc vdi bdn kinh OH tgi H.
H26. Mpi mat phing vudng gdc vdi OH deu tiep xiic vdi S. Diing hay sai ? H27. Chiing minh khi (P) tie'p xiic vdi S tai H thi OH < OM.
10
2. Trudng hop h < r
Trong hinh 2.20
H28. H thudc mien trong cua S. Diing hay sai ?
H29. M thudc (P), M thuoc S thi OM cd dac diem gi ?
H30. Cho OH = h tfnh HM.
• GV ket luan :
Giao cua mat ciu S va mat phang P la dudng trdn tam H ban kfnh r' - yr -h H31. Khi h = 0, tim tam dudng trdn la giao cua (P) va (S).
• GV neu ket luan
Mat phdng di qua tdm ggi la mat phdng kinh. Giao ciia (P) vd (S) Id dudng trdn vd ggi la dudng trdn Idn.
• Thuc hien A2 trong 5'
cau a
Hoat ddng cua GV
Cdu hdi 1
Cho bie't dp dai OH'.
Cdu hdi 2
Cho bie't OM
Hoat ddng cua HS
Ggi y trd Idi cdu hoi 1
OH'= -
2
Ggi y trd Idi cdu hoi 2
OM = r
11
Cdu hdi 3
Tfnh MH'
cau b
Hoat ddng cua GV
Cdu hdi 1
Ta cd OH' = a, OK = b so sanh OH va OK.
Cdu hdi 2
Tfnh H'M va KN.
Cdu hdi 3 So sanh H'M va KN.
• GV ket luan :
Ggi y trd Idi cdu hoi 3 V 4 2
Hoat ddng cua HS
Ggi y trd Idi cdu hdi 1 OH < OK.
Ggi y trd Idi cdu hoi 2 HS tu tfnh
Ggi y trd loi cdu hoi 3 H'M > KN.
Mat phang cang gan tam thi dudng trdn giao cd ban kinh cang Idn 12
HOATDONC3
III. GIAO CUA MAT CAU VA DUON G THANG . TIEP TUYEN CUA MAT CAU
1. Dinh nghia
• GV su dung hinh 2.22 va dat ra cac cau hdi:
GV cho HS chi ra doan thing d, r.
1. Trudng hop d > r
H32. So sanh OM va OH.
H33. M d trong hay ngoai mat cau?
H34. A cd cat mat cau hay khdng?
• GV neu dinh nghia:
Khi d > r thi A vd (S) khdng cdt nhau
2. Trudng hgp d = r
• GV sii dung hinh 2.23 va dat ra eac cau hdi:
13
H35.SosanhOMvaOH.
H33. M ?^ H thi M d trong hay ngoai mat cau?
H34. A cd cat mat cau hay khdng?
• GV neu dinh nghia:
Khi d = r thi A vd (S) tie'p xuc nhau
• GV neu dinh If:
Dieu kien cdn vd du de A tie'p xdc vdi mat cdu S(0 ; r) tgi H la A vudng gdc vdi bdn kinh OH tgi H.
3, Trudng hop d < r
• GV sii dung hinh 2.24 va dat ra eac cau hdi:
H35. So sanh OH va OM.
H36. Mpi diem thudc doan MN cd thudc khd'i cau khdng.
• Ket luan :
Khi d < r thi dudng thdng A cdt (S) tgi hai diem phdn biet.
• GV dat bai toan : Cho biet d, r hay tfnh MN.
KL:MN = 2Vi^
GV cho HS dien vao bang sau
0
14
d
r
MN
3
4 4 7
5 3
• De tdng ket GV nen cho HS dien bang sau :
d
r
Giao cua A va (S)
3 4
5 4
4 4
• GV tdng ket chii y bdi bang sau :
S6 TIEP TUYEN v6l(S)
TJNH CHAT CUA CAC TIEP TUYEN
Qua mdt diem d tren mat cau
Qua mdt diem d ngoai mat cau
• GV neu chii y :
Vd sd Vd sd
Cac tiep tuyen cimg vudng gdc vdi ban kfnh tai tiep diem
Dp dai eac doan thing ndi diem dd va tiep diem bang nhau
Mat cdu ggi Id ndi tie'p da dien ne'u nd tie'p xuc vdi tdl cd cdc mat ciia da dien dd vd khi dd la ciing ndi da dien ngoai tie'p mat cdu.
• Thuc hien ^ 3 trong 5'
cau a
Hoat ddng cua GV
Cdu hdi 1
Gpi O la tam ciia hinh vudng canh a. Chiing minh 0 la giao diem cua cac dudng cheo.
Hoat ddng ciia HS
Gffi y trd Idi cdu hoi 1
HS tu chiing minh.
15
Cdu hdi 2
Tfnh OA.
Cdu hdi 3
Xac dinh tam
mat cau.
caub
va ban kfnh cua
Gffi y trd Idi cdu hdi 2 aS 0 A = ^ ^
2
Ggi y trd Idi cdu hdi 3 HS tu ket luan.
Hoat ddng ciia GV
Cdu hdi 1
Gpi 0 la tam cua hinh vudng canh a. Chiing minh 0 la giao diem ciia cac dudng cheo.
Cdu hdi 2
Tfnh khoang each tii O den cac canh.
Cdu hdi 3
Xac dinh tam va ban kfnh cua mat cau.
cau c
Hoat ddng cua GV
Cdu hdi 1
Gpi O la tam cua hinh vudng canh a. Chiing minh O la giao diem cua cac dudng cheo.
Cdu hdi 2
Tfnh khoang each tii 0 de'n cac mat Cdu hdi 3
Xac dinh tam va ban kfnh cua mat eau.
16
Hoat ddng ciia HS
Ggi y trd Idi cdu hdi 1 HS tu chiing minh.
Ggi y trd Idi cdu hdi 2
a
d = -
2
Gffi y trd Idi cdu hoi 3 HS tu ket luan.
Hoat ddng cua HS
Ggi y trd Idi cdu hdi 1 HS tu chiing minh.
Ggi y trd Idi cdu hdi 2
2
Ggi y trd Idi cdu hdi 3 HS tu ket luan.
HOATDONC 4
IV. CONG THtrC DI$N TICH MAT CAU VA TH^ TICH MAT CAU • GV neu cdng thuc dien tfch mat cau:
S-4m^
• Vdi 71 = 3,14. Hay dien vao bang sau vdi 2 chir sd thap phan
1 2 3 4 r
S
• GV neu cdng thiic the tfch mat ciu:
V = — w
3
• Vdi 71 a: 3,14. Hay dien vao bang sau vdi 2 chvr sd thap phan
r
V
I GV neu chii y
1 2 3 4
a) Dien tich ciia mat cdu bdn kinh r bdng 4 Idn dien tich dudng trdn Idn cua mat cdu dd.
b) The tich V ciia mat cdu bdng the tich ciia khdi chdp cd dien tich ddy bdng dien tich mat cdu vd chieu cao bdng bdn kinh ciia mat cdu.
GV cho HS dien vao bang sau:
r
Dien tfch dudng trdn Idn
1 2 3 4
H.hoc 12/2 17
Thuc hien ^ 4 trong 5'
Hoat ddng cua GV
Cdu hdi 1
Xac dinh tam va ban kfnh mat cau ngoai tiep hinh lap phuong.
Cdu hdi 2
Tfnh the tfch V.
Hoat ddng cua HS
Gffi y trd Idi cdu hdi 1
Tam O la tam hinh lap phuong, ban aV3 kfnh r = .
2
Gffi y trd Idi cdu hoi 2
HS tu tfnh.
HOATDQNG 5
TOM T^T Bfil HQC
1. Tap hgp nhirng diem cdch deu mgt diem O cho trudc mgt khodng khdng ddi la mgt mat cdu.
2. Mgt mat cdu xdc dinh khi bie't tdm vd bdn kinh hoac dudng kinh cda nd.
3. Tap hgp tdt cd cdc diem thudc mat cdu S(0; R) vd tdt cd cdc diem a trong mat cdu ggi Id khdi cdu.
4. Dieu kien cdn vd du de mat phdng (P) tie'p xdc vdi mat cdu S(0 ; r) tgi H Id (P) vudng gdc vdi bdn kinh OH tgi H.
5. Mat phdng di qua tdm ggi la mat phdng kinh. Giao ciia (P) vd (S) la dudng trdn vd ggi la dudng trdn Idn.
6.
d < r
(P) cat (S) theo
mdt dudng trdn
18
Giao ciia (P) va (S)
d > r
Khdng cd
d = r
(P) tie'p xiic (S)
Giao cua Ava (S)
d > r
Khdng cd
d = r
A tie'p xiic (S)
d < r
A cat (S) tai 2 diem
HOATDONC 6
MQT SO Cfia HOI TR6C NGHIEM
Hay dien dung (D) sai (S) vao cac kh^ng djnh sau : Cdu 1.
(a) Mdt mat phing cit hinh cSu thi cat theo mdt dudng trdn. (b) Dudng trdn Idn ed ban kfnh bang ban kfnh mat ciu. (c) Dudng trdn Idn di qua tam O.
(d) Mdt hinh ciu cd vd sd dudng trdn Idn.
Trd Idi.
n D D •
a D
b D
c
D
d D
Cdu 2.
(a) Mdt dudng trdn xoay quanh mdt dudng thing ta dupe mdt liinh cdu [] (b) Mdt mat ciu quay quanh mdt dudng kfnh cua nd ta dupe mdt mat eSu. [] (c) Niia dudng trdn quay quanh mdt dudng kinh ciia nd ta dupe mdt mat ciu. [] (d) Ca ba khang dinh tren deu sai. [1 Trd Idi.
a S
b D
c
d
D
S
19
Cdu 3. Gpi d la khoang each tii O cua mat cau S(0 : r) den mat phing (P) (a) d > r thi (P) cat (S) D (b) d < r thi (P) cit (S) [] (c) d = r thi (P) cit (S) D (d) Ca ba khang dinh tren deu sai U Trd Idi.
a S
h D
c S
d S
Cdu 4. Gpi d la khoang each tii O ciia mat cau S(0 ; r) den mat phang (P) (a) d > r thi (P) tiep xiic vdi (S) • (b) d < r thi (P) tiep xiic vdi (S) [] (c) d = r thi (P) tie'p xiic vdi t (S) [] (d) Ca ba khing dinh tren deu sai |_| Trd Idi.
a S
b S
e
D
d S
Chon khang dinh dung trong cac cau sau:
Cdu 5. Gpi d la khoang each tilt O cua mat ciu S(0 ; r) de'n mat phing (P) Dien vao chd trdng sau :
d
r
Vi tri tuang ddi cua (P) vd (S)
20
3 5
4 4
5 4
5 8
Cdu 6. Gpi d la khoang each tir O ciia mat cau S(0 ; r) den dudng thing A Dien vao chd trdng sau :
d
r
Vi tri tuang ddi cua A vd (S)
3 5
4 4
5 4
5 8
Cdu 7. Cho hinh lap phuang ABCD. A'B'C'D' tam O. Tam cua mat cau ngoai tiep hinh lap phuong la:
(a) O; (b) A;
(c) B; (d) C.
Trd Idi. (a).
Cdu 8. Cho hmh lap phuong ABCD. A'B'C'D' tam O. Tam ciia mat ciu tiep xiic vdi tat ca cac mat ciia hinh vudng hinh lap phuong la :
(a) O; (b) A;
(c) B; (d) C.
Trd Idi. (a).
Cdu 9. Cho hinh lap phuong ABCD. A'B'C'D' canh a. Ban kfnh cua mat cau tiep xiic vdi tat ca cac mat cua hinh lap phuong la :
(a) a;
, ,aV3
(c)--r-
Trdldi. (b).
(b)f. (d)a^/^
Cdu 10. Cho hinh lap phuong ABCD. A'B'C'D' canh a. Ban kfnh cua mat cau ngoai tie'p hinh lap phuong la :
(a) a;
. ,aV3
(O— ;
Trdldi. (c).
(b) ^ 2
(d)aV2
21
Cdu 11. Cho hinh lap phuong ABCD. A'B'C'D' canh a. Dien tfch xung quanh ciia mat cau tie'p xiic vdi tat ca cac mat ciia hinh vudng hinh lap phuong la :
(a) 47ta"; (b)7ia^
(c) 87ra^; (d) 127ia^
Trdldi. (b).
Cdu 12. Cho hinh lap phuang ABCD. A'B'C'D' tam O. The tfch ciia mat cau tiep xiic vdi tat ca cac mat cua hinh vudng hinh lap phuong la :
(a)-7:a ;
(c) — 7ia ; ,3 Trdldi. (b).
(b) -7ia ^ 6
(d) -7:a ^ 3
Cdu 13. Cho hinh chdp S.ABC ndi tie'p hinh ciu tam O ban kinh r (hinh ve)
Biet tam giac ABC la tam giac deu, SO 1 (ABC). Canh AB bing (a)rV3; (b)2rV3 (c)4rV3; (d)3rV3 Trd Idi. (a).
22
Cdu 14. Cho hinh chdp S.ABC ndi tiep hinh cau tam O ban kfnh r (hinh ve) s
Biet tam giac ABC la tam giac deu, SO 1 (ABC). The tfch hinh chdp bing :
(a) 471 r^ V3 ;
(c)4r'V3;
Trd Idi. (a).
Cdu 15. Cho hinh ciu S(0 ; 4) nhu hinh ve
AB bing :
(a)2V3;
(c)3;
Trdldi. (a).
(b) -71 r^ V3; 3 (d)3r'V3
(b)3V3
(d)6.
Cdu 16. Cho hinh cau S(0 ; 4) nhu hinh ve
Ban kfnh dudng trdn tam O bing :
(a) V7;
(c)3 ;
Trdldi. (a).
HOATDQNC 7
naOTNG D^N Bfil TOP SGK
Bai 1. Hudng ddn. Sii dung dinh nghia hinh eau:
(b)3V7 (d)6.
Ddp sd. Mat ciu tam O (trung diem AB) ban kinh AB 24
Bai 2. Hudng ddn. Dua vao tfnh chat cua mat phang tmng true va dinh nghia mat eau: A
B - - C
Ddp sd
- Dung tam E dudng trdn ngoai tiep tam giac CBD.
Ke Ex L (BCD).
Dung mat phing trung true ciia AB.
I la tam.
HS tu chiing minh.
Bai 3. Hudng ddn. Dura vao tfnh chit ciia giao diem mat cau va mat phang Tap hop la dudng thing di qua tam O' cua dudng trdn va A 1 mp(O')
Bai 4. Hudng ddn. Dua vao dinh nghia mat cau.
A
Goih (O') la dudng trdn ndi tiep tam giac ABC. Sii dung bai 3. Bai 5. Hudng ddn. Dua vao vf trf tuong ddi cua mat phang va mat cau:
cau a.
Chiing minh 4 diem A, B, C va D ciing thudc mdt dudng trdn. cau b. Hudng ddn.
Ta cd MA. MB = MO'^ - r'l Trong dd r' la ban kfnh dudng trdn tam O' MO'2 = MO^ -00' ^ = d^ -00' 2
r'2=r^-00'2
26
Bai 6. Hudng ddn. Dua vao tfnh chat tie'p tuyd'n ciia mat ciu.
- Chiing minh Al va BI la hai tie'p tuyen cua mat ciu.
- Chiing minh AMAB = ALAB.
Bai 7 Hudng ddn. Dua vao dinh nghia hinh tru, tfnh chat ciia hinh tru, dien tfch xung quanh va the tfch hinh tru.
B
\ O
cau a. Chiing minh O la giao diem eac dudng cheo la tam ciia mat ciu. ^ a^+b^+c^
cau b. Tam O' la tam hmh chu nhat ABCD.
Va 2+b2
Bai 8 Hudng ddn. Dua vao tfnh chat ciia tie'p tuye'n mat ciu.
HS tu giai.
Bai 9 Hudng ddn. HS tu giai.
Bai 10 Hudng ddn. HS tu giai.
27
On tap churdng II
(tiet 11,12)
1. MUC TIEU
1. Kien thurc
HS nim duac:
1. Khai niem chung ve mat trdn xoay.
2. Mat tru va cac tfnh chat ciia mat tru.
3. Mat cau va cac tfnh chat cua mat cau.
4. Giao ciia mat cau va dudng thing.
5. Tiep tuyen ciia mat cau.
6. The tfch va dien tfch ciia mat cSu.
2. KI nang
• Giai thanh thao cac bai toan lien quan den mat ciu, mat tru.
• Xac dinh dupe mdt mat phing la tie'p dien cua mat ciu, mdt mat phing la tie'p tuyen ciia mat cau.
3. Thai dp
• Lien he dupe vdi nhieu vin de thuc te' trong khdng gian.
• Cd nhieu sang tao trong hinh hpc.
• Hiing thii trong hpc tap, tfch cue phat huy tfnh dpc lap trong hpc tap.
II. CHUAN DI CUA GV VA HS
1. Chuan bi cua GV:
• Thudc ke, phan mau, ...
2. Chuan bj cua HS :
Dpc bai trudc d nha, cd the lien he cac phep bie'n hinh da hpc d Idp dudi 28
ra. PHAN PHOI THOI LUONG
Bai dupe chia thanh 2 tie't :
Tie't 1: Chiia bai tap va on tap.
Tie't 2: Kiem tra 1 tie't.
IV. TIEN TDINH DAY HOC
HOATDONC 1
I. MO T s d CAU HOI ON TAP
Cdu hdi 1. Neu dinh nghia mat trdn xoay.
Cdu hdi 2. Neu mdt sd hinh trdn xoay trong thuc te ma em biet. Cdu hdi 3. Mdt tam giac thudng quay quanh mdt canh cd dupe mdt hinh ndn khdng? Cdu hdi 4. Mdt hinh ndn dupe tao thanh nhu the nao?
Cdu hdi 5. Mdt hinh binh hanh quay quanh mdt canh cd dupe mot hinh tru khdng? Cdu hdi 6. Mdt hinh tru dupe tao thanh nhu the' nao?
Cdu hdi 7. Neu khai niem true, dudng sinh va dudng cao cua hinh ndn. Cdu hdi 8. Day ciia hinh ndn la hinh gi?
Cdu hdi 9. Neu khai niem true, dudng sinh va dudng cao cua hinh tru. Cdu hdi 10. Neu dinh nghia mat ciu bing cac each khac nhau.
Cdu hdi 11. Hinh cau va mat ciu khac nhau nhu the' nao?
Cdu hdi 12. TCr mdt diem ke hai tie'p tuyen de'n mat cau.
a) Khi nao thi ke dupe?
b) Neu cac tfnh chat cua hai tiep tuyen dd.
Cdu hdi 13. Neu cac vi trf tuong ddi ciia mat phing va mat cau.. 29
Cdu hdi 14. Neu cac vi trf tuong ddi eiia dudng thing va mat ciu. Cdu hdi 15. Neu sU xac dinh mat cau.
Cdu hdi 16. Neu cac cdng thiic tinh dien tfch mat ciu va the tfch khd'i ciu. HOATDONC 2
n. MOT s o CAU HOI TRAC NGHIEM ON TAP CHUONG II Hay chon cau tra Idi dung.
Cdu 1. Cho hinh chdp ndi tiep mdt hinh ndn
(a) Hai hinh chdp va hinh ndn cd dudng cao trdng nhau;
(b) The tfch hinh chdp va the tfch hinh ndn bing nhau;
(c) The tfch hinh chdp Idn hon the tfch hinh ndn;
(d) Ca ba y tren deu diing.
Trd Idi. (a).
Cdu 2. Cho hinh chdp luc giac ddu canh day la 2V3 ndi tiep mdt hinh ndn cd dudng cao la 3
S
30
Dudng cao ke tii S ciia mdi mat bdn ciia hinh chdp la :
(a)2VlO; (b)VlO
To
(c) (d)10
Trdldi. (b).
Cdu 3. Cho hinh chdp luc giac deu canh day la 2v3 ndi tiep mdt hinh ndn cd dudng cao la 3
s
Ban kfnh dudng trdn day la
(a) 2V3;
, , 2V3
Trd Idi. (a).
(b) 2V6 (d) V3.
Cdu 4. Cho hinh chdp luc giac deu canh day la 2v3 ndi tiep mdt hinh ndn cd dudng cao la 3
S
Dudng sinh la
(a) 2^ ;
, , 2V3. (0— , Trdldi. (b).
(b) 2V6
(d) V3.
31
Cdu 5. Cho hinh lang tru luc giac deu canh day la 2\/3 ndi tiep mdt hinh tru cd dudng cao la 3
B
"^ C
A'
, ^-
B'
Dien tfch xung quanh cua hinh lang tru la la :
(a)36V3; (b) 16V3
(c)46V3; (d)26V3. Trd Idi. (a).
Cdu 6. Cho hinh lang tru luc giac deu canh day la 2V3 ndi tiep mot hinh tru cd dudng cao la 3
A B
A' B'
D'
E'
32
Dudng sinh ciia hinh tru la : (a)2V3;
(c)3 ;
Trd Idi. (c).
(b)3V3 (d)6.
Cdu 7. Mdt hinh ciu cd dudng trdn Idn ngoai tiep mdt tam giac deu canh 1 cd ban kfnh la
(a) 27i; (b)37t
(d)6.
Trd Idi. (e).
Cdu 8. Mdt hinh cau cd dudng trdn Idn ngoai tiep mdt tam giac deu canh 1 cd dien tfch toan phan la :
H hoc 12/2 33
(a)2V3;
(c)-i=;
V3
Trd Idi. (c).
(b)3V3 (d) 671.
Cdu 9. Gpi d la khoang each til O ciia mat ciu S(0 ; r) den mat phing (P) Dien vao chd trdng sau :
d
r
Vi tri tuang ddi cda (P) vd (S)
6 5
5 4
4 4
8 8
Cdu 10. Gpi d la khoang each tii O ciia mat cau S(0 ; r) den dudng thing A Dien vao chd trdng sau :
d
r
Vi tri tuang ddi ciia A vd (S)
4 5
4 5
5 7
9 8
Cdu II. Cho hinh lap phuong ABCD. A'B'C'D' tam O. Tam ciia mat cau ngoai tie'p hinh lap phuong la
(a) O; (b) A
(c) B; (d) C.
Trd Idi. (a).
Cdu 12. Cho hinh lap phuang ABCD. A'B'C'D' tam O. Tam ciia mat ciu tiep xuc vdi tat ca cac mat cua hinh vudng hinh lap phuong la :
(a) O; (b) A
(c) B; (d) C.
Trd Idi. (a).
34
HOATDONC 3
m . HUONG DAN BAI TAP SGK
Bai 1. Hudng ddn.
cau a diing vi : Qua ba diem ABC ta cd mdt mat phing. Giao ciia mat ciu va mat phing la dudng trdn.
Bai 2. Hudng ddn.
D
Dudng sinh ciia hinh ndn : BD - av2
Dudng cao ciia hinh ndn AB = a
35
Bai 3. Hudng ddn.
Til tam E cua day hinh ndn ke dudng cao SE.
- Mat phing trung true ciia mdt dudng cao bat ki cat SE tai O. Chiing minh O la tam mat cau ngoai tiep hinh chdp.
Bai 4. Hudng ddn.
- Chiing minh tam I dudng trdn ndi tie'p tam giac ABC thudc mat cau dd. - Chiing minh ABC la tam giac deu.
- Chiing minh SI 1 (ABC).
Bai 5. Hudng ddn.
A
36
a) HS tu giai.
Tfnh HB = r = 1V3
- Dudng sinh khdi tru a.
Bai 6. Hudng ddn.
Ke mat phing trung true cua SA cit SO tai I.
Chiing minh I la tam mat ciu.
De tfnh r = SI ta su dung tam giac ddng dang
SI SE „ SE.SA
— = — ^ SI =
SA SO SO
Bai 7. HS tu giai.
Cau hdi trac nghiem :
HS tu giai.
37
HOATDONC 4
MOT SO D^ KI^M TRA CHUONG H
De sd 1
Cdu i. ( 3 d) Cho hinh hop chii nhat ABCDA'B'CD' canh a, b, c. a) Xac dinh tam mat cau ngoai tie'p hinh hop.
b) Tfnh dien tfch mat cau dd.
Cdu 2. (4 d) Cho hinh ndn cd ban kfnh day la 4, dudng cao la 6. a) Tfnh dien tich xung quanh va the tfch hinh ndn.
b) Tfnh ban kfnh hinh cau ngoai tiep hinh ndn dd.
Cdu 3. (3d) Cho hinh tru cd ban kfnh day la 4, chieu cao 3. Cit hinh tru bdi mot mat phing song song vdi true.
a) Hinh thiet dien la hinh gi?
b) Neu cac dung hinh dd biet hinh dd cd dien tfch la 18.
De sd 2
Cdu 1. i 6 d).
Cho hinh cau cd ban kfnh la 3.
a) Tfnh the tfch khd'i cau va dien tfch mat ciu.
b) Tfnh the tfch hinh tru ngoai tiep hinh cau.
Cdu 2. (4 d).
Cho hinh ndn cd ban kfnh day la 6, dudng cao la 8.
a) Tfnh dd dai dudng sinh.
b) Mdt mat phing vudng gdc vdi true cit hinh ndn theo mdt dudng trdn ban kinhd la 2. Tfnh the tfch hinh ndn cut tao thanh.
38
Chi/ofNq 7
PHlTOlVG PHA P IWA 0 6 TROafG HHOIVG GIAIV Pha n 1
Gidl THlfu CHUOHG
I. CAU TAO CHUONG
§ 1. He tpa dp trong khdng gian
§ 2. Phuong trinh mat phing
§3. Phuang trinh dudng thing trong khdng gian
On tap chuang III
On tap cud'i nam
1. Muc dich ciia chuong
• Chuong III nhim cung cap cho hpc sinh nhiing kien thiic co ban ve khai niem ve tpa dp trong khdng gian va nhiing ling dung ciia nd.
Tpa dp vecto va tpa dp diem.
- Bieu thiie tpa dp cua cac phep toan vecto .
Tfch vd hudng cua hai vecto.
Phuang trinh mat cau.
• Gidi thieu vd phuang trinh mat phing trong khong gian.
Vecto phap tuyen cua mat phing.
- Phuong trinh tdng quat ciia mat phing.
- Dieu kien de hai mat phang song song, vudng gdc.
Khoang each tir mot diem den mdt mat phing.
39
• Phuang trinh dudng thing trong khdng gian:
Phuang trinh tham sd cua dudng thing.
- Dieu kien de hai dudng thing song song.
- Dieu kien de hai dudng thing cheo nhau.
- Dieu kien de hai dudng thing cit nhau.
II. MUC TIEU
1. Kien thurc
Nam dupe toan bd kien thiic co ban trong chuong da neu tren.
Hieu cac khai niem va tfnh chat vecto trong khdng gian.
Hieu va bie't dupe mdi quan he giiia vecto phap tuye'n va cap vecto chi phuang ciia mat phing.
Hieu va biet dupe mdi quan he giiia vecto phap tuye'n va vecto chi phuong ciia dudng thing.
2. Kl nang
Xac dinh dupe cac vectp trong khdng gian.
Van dung dupe eac tfnh chit de giai bai tap.
- Chiing minh dupe hai mat phing, song song, vudng gdc.
Lap dupe cac phuang trinh dudng thing va phuang trinh mat phing. Xac dinh dupe vi tri tuong ddi cua dudng thing va dudng thing, dudng thing va mat phing, giiia hai mat phing.
3. Thai dp
Hpc xong chuang nay hpc sinh se lien he dupe vdi nhieu van de thuc te' sinh ddng, lien he dupe vdi nhiing van de hinh hpc da hpc d Idp dudi, md ra mot each nhin mdi ve hinh hpc. Til dd, cac em cd the tu minh sang tao ra nhCrng bai toan hoac nhiing dang toan mdi.
Ket ludn:
Khi hpc xong chuong nay hpc sinh cin lam tdt cac bai tap trong sach giao khoa va lam dupe cac bai kiem tra trong chuong.
40
Pha n 2
cAc BAI SOAN
§1. He toa do trong khong gian
(tiet 1, 2, 3, 4)
1. MUC TIEU
1. Kien thurc
HS nim duoc:
1. Khai niem tao do vecta trong khdng gian, tpa do diem va dp dai vecto. 2. Bieu thiic tpa dp ciia cac phep toan : cdng, trii vecto; nhan vecta vdi mdt so thirc. 3. Bieu thiic tpa dp ciia tfch vd hudng ciia hai vecta.
4. Riuong trinh mat cau.
2. KT nang
• Thuc hien thanh thao cac phep toan ve vecto, tfnh dp dai vecto, • Vie't dupe phuang trinh mat ciu.
3. Thai do
• Lien he dupe vdi nhieu van de thue te' trong khdng gian.
• Cd nhieu sang tao trong hinh hpc.
• Humg thii trong hpc tap, tich cue phat huy tfnh dpc lap trong hpc tap.
II. CHUAN DI CUA GV VA H&
1, Chua n bi cua GV:
• Hinh ve 3.1 den 3.3.
• Thudc ke, phan mau,...
2. Chuan bj cua HS :
Dpc bai trudc d nha, cd the lien he cac phep bie'n hinh da hpc d Idp dudi
III. PHAN PHOI THOI LUONG Bai duoc chia thanh 4 tie't:
Tie't 1 Tie't 2 Tiet 3 Tie't 4
Tii dau de'n het phan I.
Tie'p theo den he't phan II. Tie'p theo de'n het phan III. Phan IV va hudng dan bai tap.
IV. TIEN TDINH DAY HOC
a. DRT VA'N Di
Cau hdi 1.
Nhic lai khai niem hinh hop, hinh chdp.
Cau hdi 2.
Cho hinh lap phuong ABCDA'B'CD'
a) Chiing minh cac canh ciia hinh lap phuong xuit phat tii mdt dinh vudng gdc vdi nhau.
b) Cho canh ciia hinh lap phuong la a, tfnh dp dai dudng cheo ciia hinh lap phuong. n. isni MOI
HOATDONCl
I. TOA DO CUA DIEM VA CUA VECTO
1. He toa do
GV mo ta he true toa do trong khong gian va neu cau hoi : HI. Hai vecto i, j cd vudng gdc vdi nhau hay khdng?
H2. Vecto k cd vudng gdc vdi tat ca cac vecta thudc mat phing (Oxy) khdng? 42
• GV sir dung hinh 3.1 trong SGK va dat van de:
H3. Hay dpc ten cac mat phing tpa dp.
H4. Hay ke' ten cac vecto dan vi.
H5. Cd the ed them mdt gdc tpa dp niia khac O hay khdng?
H6. Hay neu cac tfnh chit ciia mat phing tpa dp, vecto don vi?
-2 — -2 — - 2
H7. Tfnh i = i.i, j = j.j , k = k.k.
H8. Tfnh i.j,j.k, k.i.
• Thue hien ^ 1 trong 4 phiit.
Su dung hinh ve 3.2. GV cho HS len bang ve lai hinh va hudng din HS thuc hien z
N
Hoat ddng cua GV
Cdu hoi 1
Bieu dien OM theo OE va ON. Cdu hdi 2
Bieu dien OE theo OH va OK.
Hoat ddng cua HS
Gffi y trd loi cdu hoi 1
OM = OE + ON
Ggi y trd loi cdu hdi 2
OE=OH+OK.
43
Cdu hoi 3
Ggi y trd Idi cdu hoi 3
= z.k
Tim cac mdi quan he giiia cac vecto ON OH va OK
Cdu hdi 4
Bieu dien OM theo i, j va k.
2. Toa dp cua diem
GV sir dung hinh 3.2 va dat cau hdi:
OK = x.i OH = y.j,ON =
Ggi y trd Idi cdu hdi 4 OM = x.i + y.j + z.k
H9. Cho ba sd thuc x, y va z. Cd bao nhieu diem M thda man OM = x.i + y.j + z.k.
HIO. Cho OM = x.i + y.j + z.k. Cd bao nhieu bd sd sd thuc x, y va z thda man he thiie tren.
• GV tra Idi va neu dinh nghia :
Bd ba sd thuc (x; y z) thda mdn OM = x.i + y.j + z.k ggi la tga do diem M vd ki hieu M (x ; y ; z) hoac M = (x ; y ; z).
HI 1. Cho M (0 ; 0 ; 0) Hay chi ra M.
HI2. Cho M(0 ; 1 ; 2). Hdi M thudc true nao ?
HI3. Cho M(l ; 0 ; 2). Hdi M thudc true nao ?
H14. Cho M(l ; 2 ; 0). Hdi M thudc true nao ?
3. Toa dp vecto
• GV neu dinh nghia :
Trong khdng gian cho vecta a. Bg ba sd (x ; y ; z) thda mdn a = x.i + y.j + z.k ggi la tga do cua vecta a. Ki hieu a(x;y;z) hoac a = (x;y;z).
HI5. Vecta OM va diem M cd ciing tpa dp khdng?
44
• GV neu nhan xet trong SGK:
Tga do cua OM chinh Id tga do cua M.
• Thue hien A2 trong 4 phiit.
Su dung hinh ve 3.2. GV cho HS len bang ve lai hinh va hudng din HS thuc hien
Hoat ddng cua GV
Cdu hoi 1
Tim tpa dp cua AB.
Cdu hoi 2
Tim tpa dp ciia AC.
Cdu hoi 3
Tim tpa dp ciia A C
Cdu hdi 4
Tim tpa dp ciia AM.
Hoat ddng cua HS
Ggi y trd Idi cdu hoi 1 AB(a;0;0)
Ggi y trd Idi cdu hoi 2 AB(a;b;0).
Gffi y trd Idi cdu hdi 3 AB(a;b;c)
Ggi y trd Idi cdu hdi 4 AM(-;b;c ) 2
HOATDQNC 2
II. BIEU THtrC TOA D O CUA CAC PHEP TOAN VECTO
• GV neu dinh If:
Trong khdng gian Oxyz cho ba vecta a(ai;a2;a2) vab(bi ;b2;b3) fa cd : a + b = (aj +bi;a2 + b2;a3+b3)
a - b = (a, -bpa2-b2;a3-b3 )
ka = (kai;ka2;ka3), trong dd k la mgt sdthuc.
• GV hudng din HS chiing minh dinh If tren.
H16. Hay so sanh cac tpa dp ciia a va b khi a = b .
• GV neu he qua 1:
Hai vecta bdng nhau thi cdc tga do tuang img bdng nhau.
HI7. Hay viet cac bieu thiic tpa dp cua he qua 1.
• GV neu he qua 2:
Vecta 0 cd cdc tga do bdng 0.
HI8. Hay viet cac bieu thiic tpa dp ciia he qua 1.
• GV neu he qua 3:
Hai vec ta cdng phuang thi mdi tga do cua vec ta ndy bdng k Idn tga do tuang Umg cda vec ta kia..
HI9. Hay vie't cac bieu thiic tpa dp cua he qua 3.
• GV neu he qua 4:
Khi bie't tga do ciia AvdB ta cd tga do ciia AB bdng cdch lay mdi tga do tuang img cda B trit di tga do tuang img ciia A..
H20. Hay viet cac bieu thiic tpa dp cua he qua 4.
46
• Trong sach GK khdng cd vf du nhung GV nen lay vf du minh hpa cho dinh If va he qua nay. Vi day la kien thiic rat quan trpng..
Vfdu.ChoA(l ;1 ; 1),B(-1 ; 2; 3) va C (0 ; 4 ;-2).
a) Hay tim cac tpa dp ciia AB va AC .
b) Tim tpa dp ciia vec to 3AB .
c) Tfnh AC + 3AB.
• GV gpi mdt HS giai cau a.
Hoat ddng ciia GV
Cdu hoi 1
Tim AB.
Cdu hoi 2
Tim AC
Hoat ddng ciia HS
Ggi y trd Idi cdu hoi. 1 HS tu giai.
Ggi y trd Idi cdu hoi 2 HS tu giai.
• GV gpi tie'p HS thii 2 tra Idi cau b va sau dd gpi HS thii ba lam cau c. HOATDONC 3
III. TICH VO HUONG
1. Bieu thurc toa do cua tich vd hudng
• GV ndu dinh If
Trong khdng gian Oxyz cho ba vecta aia^;32;a2) vab(b, ;h2;h-^) ta cd: a.b = ajb2 + a2b2 + a3b3.
• GV hudng din HS chiing minh dinh If tren.
2. tTng dung
a) Do ddi cua vectff
• GV neu cau hdi sau :
H21. Trong boat ddng 2, hay tfnh dp dai AC
47
GV neu dinh nghia :
Cho a(a| ;a2 ;a3) khi do do ddi ciia vecta aki hieu va:
3.1 ~r 3,T ~r 3.-5
b) Khodng cdch gida hai diem
H22. Cho A (XA ; yA ;ZA) va B (Xg ; y^ ;z^).
Xac dinh AB.
Tfnh AB.
GV neu ket qua 2:
Khodng cdch gida hai diem AB Id
AB = AB = 7(XB-XA)^+(yB-yA)^+(ZB-ZA)^
c) Gdc giita hai vectff
• GV neu cdng thiic tfnh gdc giiia hai vecto :
Cho cdc vecta iij = (x,; yj; Zj), M2 = (-^2 ! }'2 > ^2) ^^ ^^ ^ ^^y >*'
^1-^2+yi>'2+^1^2
ta cd COS(M] , M2) = ^x\+y1 + z\ yjxj+yj+zl
H23. Khi nao hai vecto vudng gdc vdi nhau ?
• GV neu he qua :
M] J. U2 o iii.ii2 = 0 <^ x^X2 + yxy2 + 21Z2 = 0.
H24. Vecto to 0 vudng gdc vdi mpi vecto.
• Thuc hien ^ 3 trong 4 phiit.
Hoat ddng cua GV
Cdu hoi 1
Tfnh b + c.
48
Hoat ddng ciia HS
Gffi y trd Idi cdu hoi 1 b + c = (3;0;-3)
Cdu hdi 2
Tfnh a(b + cl.
Cdu hoi 3
Tfnh a + b .
Cdu hoi 4
Tfnh a + b
Ggi y trd Idi cdu hoi 2
a(b + c) = 3.3 + 0.0 + (-3).l = 6.
Ggi y trd Idi cdu hoi 3
a + b = (4;-l;-l) .
Ggi y trd Idi cdu hoi 4
a + b -Vl8
HOATDONC 4
IV. PHUONG TRINH MAT CAU
• GV neu each chia mdt sd khd'i da dien va dat cau hdi: H25. Tfnh khoang each giiia hai diem M(x ; y ; z) va I (a ; b ; e). H26. Bie't khoang each dd la r, hay lap bieu thiie mdi quan he dd. • GV neu dinh If
Mat cdu tdm I(a ; b ; c), bdn kinh r cd phuang trinh (x-a)^ +iy-b)'^ + iz-c)^ =r^
• GV hudng din HS chiing minh dinh If tren.
• Thuc hien ^ 4 trong 4 phiit.
Hoat ddng ciia GV
Cdu hoi 1
Hay xac dinh a, b va c
Cdu hdi 2
Xac dinh r.
Cdu hoi 3
Viet phuong trinh mat ciu.
Hoat ddng cua HS
Ggi y trd Idi cdu hoi I
a= 1, b = -2 vac = 3.
Ggi y trd loi cdu hdi 2
r = 5.
Ggi y trd Idi cdu hdi 3
(x-l)2+( y + 2)2+(z-3)^=25 .
H.hoc 12/2 49
H27. Hay neu mdt dang khac ciia phuong trinh mat ciu.
• GV neu nhan xet:
Phuang trinh x + y^ + z + 2ax + 2b\ + 2cz + d = 0 la phuang trinh ciia mat cdu khi vd chi khi d^ + b^ + c' > d. Khi dd tdm mat cdu la diem I(-a ; -b ; -c )vd bdn kinh mat cdu la
r = yja^ +b' +c^ -d.
H28. d phai thoa man deu kien gi de x^ + y" + z^ + 2ax + 2by + 2cz + d = 0 la phuang trinh ciia mat ciu ?
• GV cho HS thuc hien vi du trong SGK.
HOATDONCL
TOM T ^ Bfit HOC
1. Cho cac vecto u^ = (x,; yi; z,), ^2 = (A^ ; y2 : '-2) va sd k tuy y, ta cd : 1) »i = fh <=> xi = X2, yi = y2. zi = ^2
2) », - U2 = (AI + x, : y, + y, ; ?i + -2)
3) ;ii - i?2 = (-^'i - ^'2; >'i - >''2; ^i - ^2)
4) kn^ = (A-A'i; ky^ ; .fe,)
5) N|.U2 = .Vj.vo + 3'iy2 + z,Z2
6) K| = V"i" = v-^T +>'i" +^\
7) cos((7, iin) = "•"'"- —'•'- ''^ vdi ii, ?t 0; U2 ^ 0 V-T +.^T +-r \-^'2 +-^': +- :
8) M, -L ih <=> i A', AT + y,y2 + -1-2 = ^ •
2. M =(A- ; y ; z) <=> 0/if = xi + yj + zk
50
3. Cho hai diem Aix^ ; y^ ; z^) va Bixg; VB ' ^fi)-
1) AB = ixg-X/^;yg-y^;zB-z^)
i 1 1 9
2) A5-^(A:g-x^ ) +{yB-yA) +{^B-^A)
4. Mat ciu tam I(a ; b ; c), ban kfnh r cd phuang trinh
(x-fl)2+(y-fc)^+(z-c)2=r2
2 2 2 •) - Phuong trinh x + y + z + 2ax + 2by + 2cz + d = 0 la phuofng trinh cua mat cau 2 2 2 •'
khi va chi khi a + b + c > d. Khi dd tam mat cau la diem I(- a ; - b ; - c) va
ban kfnh mat cau la
•I? + b^+c-d.
HOATDONC 6
MOT SO C^U HOI TR^C NQHim
Hay dien dung (D) sai (S) vao cac khing dinh sau :
Cdul. Cho a = (l;2;3), b = (-2;3;-l). Khi dd a + b cd toa dp la (a) a + b cd toa dp la (-1 : 5 ; 2) Lj (b) a- b cdtoaddla(3;-l ;4) • (c) b- a cdtoaddla(3 ;-l :4) [] (d) Ca ba khing dinh tren dSv; sai \_\ Trd Idi.
a D
b D
c
d
S
S
51
CAM 2. Cho a = (l;2;3), b = (-2;3;-l). Khi dd a + b cdtoaddia (a)3a + b cdtoaddIa(l ;9;8 ) • (b) a-2 b cdtoaddla(5;-4;5 ) • (c)2b- a cdtoaddIa(5;-4;5 ) [] (d) Ca ba khing dinh tren deu sai [ j Trd Idi.
a D
b D
c
s
d
s
Cdu5. Cho a = (l;2;3), b = (-2;3;-l). Khi dd a + b cdtoaddia
(a) a.b = 1 • (b)a.b=-l • (c)2b.a=2 [] (d) Ca ba khing dinh tren deu sai Q Trd Idi.
a
D
b S
e
D
d S
Cdu 4. Cho hinh cau cd phuong trinh : (x -1)^ + (y + 2f + (z + 3)^ = 2
(a) Tam ciia hinh cau la 1(1 ; -2 ; -3) (b) Tam cua hinh cau la I(-l ; 2 ; 3) (c) Ban kfnh ciia hinh cau la 2 (d) Ban kfnh ciia hinh cau la yf2 Trd Idi.
U U D D
a
D
52
b S
c S
d D
Chon khang djnh diing trong cac cau sau:
Cdu 5. Trong cac cap vecto sau, cap vecto ddi nhau la (a)a = (l;2;-l) , b = (-l;-2;l) ;
(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);
Trd Idi. (a).
Cdu 6. Cho hinh ve :
z
..-;•" "
L
.---i"" i ,/f"""
...j-' " i _,J--'''' i ,.<-'' i
„.••!'••' i ...]••''•' i ,i''' 1
Ic 1 i
,.k--i---,-j--'-'- _....f i ..i--^
-TE^I
^ 1 i l
.i---i ]i:r(
J.--' ' ..•••'•' ..-• •
'! -4 ^
:::::...^:': ^-"B,--'
Diem D cd toa dp la
(a) (5 ; 1 ; 0); (c) (1 ; 5 ; 0); Trd Idi. (d).
(b) (0 ; 1 ; 5);
(d)(l;0;5).
53
Cdu 7. Cho hinh ve :
Diem C cd toa dp la
(a) (4 ; 4 ; 0); (b) (4 ; 0 ; 4)
Trdldi. (b).
Cdu 8. Cho hinh ve :
Dl-
•"" \ y"\ r
(c) (0 ; 4 ; 4) (d) (0 ; 0 ; 4)
.--1" \L^o -••*" --'"'' .--•'' -••"' ---•"' -•'''
^ ;
Diem A ed toa dd la
(a) (0 ; 2 ; 0); (b) (2 ; 0 ; 2) Trd Idi. (c).
54
(c) (2 ; 0 ; 0) (d) (0 ; 0 ; 2)
Cdu 9. Cho hinh ve
.-••'• I Of--
-r--.-3-^--
.y"'\ k if:;
ic
^ :^ y' •
.y'\ ji^o.-' •••••[ .-•-• .---..---.--- " V
..y...y.B^:::i...-"'
Diem B cd toa dd la
(a) (4 ; 4 ; 0);
Trd Idi. (a).
Cdu 10. Cho hinh ve
(b) (4 ; 0 ; 4)
(c) (0 ; 4 ; 4);
(d) (0 ; 0 ; 4)
z
A
L
f-
..-••<"] A""' i .-'!'' i .-•i''"'
--
.
--f
_: ./ i
i i i c i i
£T[ M 1 1 1 ; ; ; ; ; 1 i i 1 i i
')..
i .i---"i >i - :.--': A--^ ...--'
j^..^::l..^:l..^:;:
• ;..-• • ....-•' ...-• • .,.- • y -:::;£L::::1..--""
X
Diem E cd toa dp la (a) (3 ; 4 ; 3);
Trd Idi. (a).
...^
(b) (4 ; 3 ; 4)
(c) (3 ; 4 ; 4);
(d) (3 ; 0 ; 4) 55
HOATDONC 7
HOG^NG D^N Bfil T6P SGK
Bai 1, Hudng ddn. Dua vao tfnh chat cua eac phep toan vecto
Caua.Tacd 4a = {8;-20;12) ; — b ^ ' 3 0;- 2 I 3' 3
Tii dd ta cd ke't qua.
; 3c-(3;21;6).
caub. Tacd ; 4b = (0;8;-4); -2c = (-2;-14;-4) .
Tilr dd ta cd ket qua.
Bai 2. Hirdng ddn. Dua vao tfnh chat chit XQ - -(x ^ + Xg + x^ );
yG=^(yA+yB+yc); ZG= gl^A+ZB+zc)
Bai 3. Hudng ddn. Dua vao tfnh chat cua phep toan toa dp. Hai vecto bing nhau.
A(1 ;0;1)
D(1 ; -1 ; 1
B(0 ; 1; 2)
C'(4 ; 5 ; -5
Tfnh toa dp diem C bang each gpi C(x ; y ; z) va DC-(x-l; y + l;z-l) AB = (-1;2;1) va DC = AB Ta cd C(2; 0; 2)
Tfnh toa dd A' bang each : AA' = DD' ta ed A'(3 ; 5 ; 6)
Tuong tu ta cd B' (4 ; 6 ; -5), D'(3 ; 4 ; -6).
56
Bai 4. Hudng ddn. Dua vao tfnh chat ciia tfch vo hudng hai vecto U\.U2 =x\X2+yiy2+h^2
a) a.b = 6 .
b) c.d = -21 .
Bai 5. Hirdng ddn. Dua vao phuong trinh mat cau. a) Phuang trinh mat cau dupe vie't dudi dang :
(x-4f+(y-l) 2 + z2=16.
Tir dd ta ed tam va ban kfnh mat cau.
b) Phuang trinh mat ciu dupe vie't dudi dang :
^2 :^2
(x-1)^ + 4 y + -3
z + - 19^
Tii dd ta cd tam va ban kfnh mat cau.
Bai 6. Hudng ddn. Dua vao phuong trinh mat cau. a) Xac dinh tam mat cau : I = (3;-1 ;5), ban kfnh mat ciu r = 3. Tii dd ta cd tam va ban kfnh mat ciu
(x-3f+( y + l)'+(z-5 f =9 .
b) Xac dinh tarfi mat cau : C - (3;-3;l), ban kfnh mat ciu r = v5 Tii dd ta cd tam va ban kfnh mat ciu
(x-3f+( y + 3)^(z-lf=5 .
§2. phtiTdng trinh mat phang
(tiet 5, 6, 7, 8, 9)
I. MUC TIEU
1. Kien thiirc
HS nim ducfc:
1. Vecto phap tuye'n ciia mdt mat phing, cap vecto chi prfiuong ciia mat phing. 2. Sir xac dinh mdt mat phing.
3. Biet duoc phuong trinh tdng quat va phuong trinh tham so cua mat phang. 4. Xac dinh dupe didu kien de hai mat piling song scHig va hai mat piling vudng gdc.
2. Kl nang
• Lap dupe phuong trinh mat phing khi biet mdt diem va vecto phap tuyen, khi bie't mot diem va cap vecto chi phuong.
• Xac dinh dupe vi trf tuong ddi eiia hai mat phing, hai mat phing song song, hai mat phing vuong gdc.
• Tim dupe khoang each tii mdt diem den mdt mat phing.
3. Thai dp
• Lien he dupe vdi nhieu van de cd trong thuc te' ve mat phing trong khdng gian. • Cd nhieu sang tao trong hinh hpc.
• Hiing thii trong hpc tap, tfch cue phat huy tfnh dpc lap trong hpc tap. n. CHUAN BI CUA GV VA H6
1. Chua n bi cua GV:
• Hinh ve 3.4 den 3.8 trong SGK.
58
• Thudc ke, phan mau,...
• Chuan bi sin mdt vai hinh anh thuc te' trong trudng ve hai mat phang vudng gdc , hai mat phing song song.
2. Chuan bj cua HS :
• Dpc bai trudc d nha, dn tap lai mdt sd kien thiic da hpc.
• Chuan bi thudc ke, biit chi, biit mau de ve hinh.
m. DHAN PHOI TH6 I LUONG
Bai nay chia thanh 5 tiet:
Tie't 1: tii diu den het dinh nghia phan I.
Tie't 2 : tie'p theo den het muc 1 phin II.
Tie't 3: tie'p theo den het phan II.
Tie't 4 : tie'p theo den het muc 1 phan III.
Tie't 5: tie'p theo den het phin IV.
IV. TIEN TDINH DAY HOC
n. DRT VAN ff>€
Cau hdi 1.
Cho hinh lap phuong ABCD.A'B'CD' cd A trung vdi gdc toa dp. AB triing vdi Ox, AD triing vdi Oy, AA' triing vdi Oz
a) Tim toa dp tat ca cac dinh cua hinh vuong.
b) Tim toa dp vecto AM vdi M la trung diem CC
Cau hdi 2.
Neu mdt so tfnh chit co ban cua phep toan ve vecto.
59
B. Bni MOI
HOATDONCl
I. VECTO PHAP TUYEN CUA MAT PHANG
GV neu mot so eau hdi sau day:
HI. cd bao nhieu dudng thing vudng gdc vdi mat phang
H2. Mot mat phing xac dinh khi nao ?
• GV neu dinh nghia :
Cho mat phdng (a). Ne'u vecta n khdc vecta 0 cd gid vudng gdc vdi mat phdng (a) ggi la vecta phdp tuyen cua mat phdng (a)
H3. Cho n la vecto phap tuyen ciia (a), hdi k n cd la vecto phap tuyen cua (a) khdng?
• GV ndu chii y :
Neu n la vecta phdp tuyen cua mp{a) thi kn {k^Q) cdng la vecta phdp tuyen cda mp(^a).
• GV neu va hudng din HS giai bai toan 1. ( Su dung hinh 3.4). 60
Hoat ddng ciia GV
Cdu hdi 1
De chiing minh n la vec to phap tuyen ciia (a) ta can chiing minh va'n de gi ?
Cdu hoi 2
Hay chiing minh nhan dinh tren. GV neu nhan xet:
Hoat ddng cua HS
Ggi y trd Idi cdu hdi 1 Ta chiing minh n.a = n.b = 0.
Ggi y trd Idi cdu hdi 2 HS tu chiing minh.
Vecta n thod mdn n.a = n.b - 0 ggi Id tich cd hudng cua hai vecta
a va b , ki hieu n = a A b hoac n = Thuc hien A l trong 5 phiit.
Hoat ddng cua GV
a,b
Hoat ddng cua HS
Cdu hoi 1
Tim toa dp vecto AB
Cdu hoi 2
Tim toa dd vecto AC
Cdu hoi 3
Tim vec to phap tuyen ciia mat phing (ABC)
Ggi y trd Idi cdu hoi 1
AB = (2;l;-2 )
Gffi y trd Idi cdu hdi 2
AC = (-12;6;0)
Gffi y trd Idi cdu hdi 3
n = (l;2;2).
61
HOATDONC 2
II. PHUONG TRINH TONG QUAT CUA MAT PHANG • GV neu va cho HS giai bai toan 1. (Sii dung hinh 3.5)
M(x ;y ; x)
Hoat ddng cua GV
Cdu hdi I
Tim toa dp vecto MQ M
Cdu hdi 2
M Q M va n cd quan he vdi nhau nnirthe'nao ?
Cdu hoi 3
Hay chiing minh nhan dinh" ciia bai toan.
Hoat ddng cua HS
Gffi y trd Idi cdu hdi 1 HS tu tfnh.
Gffi y trd Idi cdu hdi 2 MoM.n=0.
Gffi y trd Idi cdu hdi 2 HS tu chiing minh.
• GV neu va cho HS giai bai toan 2. (Sir d ling hinh 3.5)
Hoat ddng cua GV r Hoat ddng cua HS
Cdu hdi I Ggi y trd Idi cdu hdi I
Hay chpn diem MQ thoa man phuong trinh da cho.
Cdu hdi 2
Gpi (a) la mat phang qua M„ va nhan n(A;B;C) lam vecto p\rd~i tuye'n. Chiing min-^ M ihup'? (a) khi Ax + By + Cz +D = 0. 1
HS tu chon.
Ggi y trd Idi cdu hdi 2 HS tu chiing minh.
62
1. Djnh nghia
• GV neu dinh nghia sau :
Ax + By + Cz + D = 0 trong dd A^ + B^ + C^ > 0 ggi la phuang trinh tdng qudi ciia mat phdng (or).
H4. Tim vecto phap tuyen eiia mat phing : Ax +By + Cz + D-0 • GV neu nhan xet a) :
(a) cd phuang trinh Ax + By + Cz + D = 0 thi vecta phdp tuye'n cua (a) la n = (A;B;C)
H5. Lap phuang trinh mat phing di qua Mo(x,); yo; Z;,) va nhan n = (A;B;C ) la vecto phap tuyen.
• GV neu nhan xet b) :
Mat phdng (a) di qua diem MQ (XO,>'O,ZO) ^''^' "^^ vecta phdp tuye'n n (A ; B Old A(x-Xo) +B(y-yo) + C(z-Zo) = 0.
• Tliuc hien ^ 2 trong 5 phut.
GV gpi ba HS len bang dien vecta phap tuye'n vao d trdng sau:
HSl:
(a)
VTPT:n
HS2:
(a)
VTPTin
4x - 2y - 6z + 7 = 0 4x +2y-6 z + 7 = 0
4x + 2y -+6z + 7 = 0 4x +2y-6z- 7 = 0 63
HS3:
(a)
VTPT:n
4x - 2y - 6z - 7 = 0 4x +2y + 6z + 7 = 0
• Thuc hien ^ 3 trong 5 phiit. Hoat ddng ciia GV
Cdu hdi 1
Xac dinh MN
Cdu hdi 2
Xac dinh MP
Cdu hdi 3
Xac dinh VTPT cua (MNP) Cdu hdi 5
Lap PTTQ ciia mat phing. 2. Cac trudng hop rieng
Hoat ddng cua HS
Ggi y trd Idi cdu hoi 1 MN = (3 ; 2 ; I)
Ggi y trd loi cdu hoi 2 MP = (4 ; 1 ; 0).
Ggi y trd loi cdu hdi 3 n = (-1 ; 4 ; -5)
Ggi y trd Idi cdu hdi 4 X - 4y + 5z -2 = 0.
a) Mat phdng di qua gdc tog dg:Sit dung hinh 3.6.
H6. Diem O (0 ; 0 ; 0) thudc mat phing (a). Tim D.
• GV ke't luan :
Mat phdng (a) di qua gdc tog do O khi vd chi khi D = 0.
b) Mol Irong cdc he sd: A, B hgc C bang 0 ; Sii dung hinh 3. 7 H7. A = 0, mat phing (a) va Ox cd quan he nhu the' nao ? • GV ke't luan :
Mat phdng (a) song song (hoac chira) true tog do Ox khi vd chi khi A = 0.
H8. Phat bieu trong trudng hpp : B = 0 hac C= 0.
64
GV neu tdng quat:
Mat phdng (a) song song vdi true tog do ndo dd khi vd chi khi he sd tuang img cda bie'n sd'bdng 0.
• Thuc hien A4 trong 5 phiit.
Hoat ddng ciia GV
Cdu hoi 1
B = 0 thi-(a) cd dac diem gi ? Cdu hdi 2
C = 0 thi (a) cd dac diem gi ? GV neu tdng quat:
Hoat ddng cua HS
Ggi y trd Idi cdu hoi 1 (a)//Oy.
Ggi y trd loi cdu hoi 2 (a) // Oz.
Mat phdng (a) song song vdi true tog do ndo dd khi vd chi khi he sd tuang img cua bie'n sd'bdng 0.
c) Mat phdng (a) triing vdi mot trong cdc mat phdng tog do: Su dung hinh 3. 8 H9. A = 0, B = 0 mat phing (a) va mp(Oxy) cd quan he nhu the' nao ? • GV ke't luan :
Mat phdng (a) song song hoac triing vdi mat phdng (Oxy) khi vd chi khi A = B = 0.
HIO. Phat bieu trong trudng hop : B = 0, C= 0 hoac C = 0, A = 0. • Thuc hien ^ 5 trong 5 phiit.
Hoat ddng cua GV
Cdu hdi 1
A = C = 0 thi (a) cd dac diem gi? Cdu hdi 2
B = C = 0 thi (a) cd dac diem gi?
Hoat ddng ciia HS
Ggi y trd Idi cdu hoi I (a) // Oxz.
Gffi y trd Idi cdu hoi 2 (a) // Oyz.
H.hoc 12/2 65
GV neu tdng quat:
Mat phdng (a) song song vdi mat phdng tog do ndo dd khi vd chi khi he sd tuang img cda cdc bie'n sd'bdng 0.
• GV neu cac cau hdi
Hll. Tii phucmg trinh : Ax + By + Cz +D = 0 <=>- + - + - = 1 bing each nao? a b c
GV nhan xet: Ax + By + Cz + D = 0 vdi cac he sd A, B, C, D deu khac 0.
D , D £> ^ u V .. . .
Khi dd bang each dat a = ; b = ; c = . ta dua phuong tnnh tren ve
dang : — + - + - = 1
a b c
• GV neu nhan xet:
X y z ' ., '
Phuang trinh — + — + - = 1 goi la phuang trinh dogn chdn cua mat phdng.abc ' HI2. Phuong trinh doan chin ciia (a) cit cac true theo diem nao? GV nhan xet:
Rd rdng mat phdng cd phuang trinh (2) cdt cdc true Ox, Oy, Oz Idn lU0 tgi cdc diem M(a ; 0 ; 0), N(0 ; b ; 0) vd P(0 ; 0 ; c).
• GV neu vf du trong SGK va giai.
HOATDONC 3
m. Difiu KlfeN DE HAI MAT PHANG SONG SONG, HAI MAT PHANG VUONG GOC.
• Thuc hien ^ 6 trong 5 phiit.
1 Hoat ddng ciia GV Cdu hoi 1
Xac dinh vecto phap tuyen ciia (a)? 66
Hoat ddng cua HS
Ggi y trd Idi cdu hdi 1 n = (l;-2;3)
Cdu hoi 2
Xac dinh vecto phap tuye'n ciia (P)? Cdu hoi 3
Hai vecto tren quan he nhu the' nao? • GV dat vin de :
Ggi y trd Idi cdu hoi 2 ir-(2;-4;6)
Gffi y trd loi cdu hoi 3 Hai vecto tren cdng tuyd'n.
Trong khdng gian toa dp Oxyz, cho hai mat phing (a) va (a') lin luprt cd phuong trinh :
ia):Ax + By + Cz + D = 0
ia') :A'x + B'y + C'z + D' = Q;
chiing lan lupt cd eac vecto phap tuyen la «(A ; 5 ; C) va n\A ;B'; C). Khi nao (a) va (a') song song ?
Khi nao (a) va (a') vudng gdc ?
1. Dieu kien de hai mat phang song song
H13. Khinao(a)//(a')
67
• GV ke't luan : (a) // (a') c^ hai vecto phap mye'n ciia hai mat phing dd cdng tuye'n. H14. Hay vie't bieu thiic toan hpc de hai mat phing (a) va (a') song song. • GV ke't luan :
, ^„ , ,^ fn = ki? f(A;B;C) = k(A;B';C) a// a' o \ o ^
[D^kD ' [D^kD '
, , , „ fn = kir' [(A;B;C) = k(A;B';C) a)= a c^ { <^<
[D = kD' lD = kD'
• GV cd the neu each khac cua ke't luan tren cho de hieu hon: Cho hai mat phdng (a) vd (a') Idn lu0 cd phuang trinh :
(a) :Ax + By + Cz + D = 0
(a') :A'x + B'y + C'z + D' = 0
a) Hai mat phdng dd song song khi vd chi khi
A__B__C_ D_
A'~ B'" C'^ D'
b) Hai mat phdng dd trdng nhau khi vd chi khi
A^_B__^_D_
A'~ B'~ C'~ D'
• GV neu chii y :
Hai mat phdng (a) vd (a') cdt nhau o(A ; B ; C) i^ k(A ; B'; C)
GV cd the neu each khac cho di nhd:
Cho hai mat phdng (a) vd (a') Idn lu0 cd phuang trinh :
(a) :Ax + By + Cz + D^O
(a') A'x + B'y + C'z + D' = 0
Hai mat phdng dd cdt nhau khi vd chi khi A . B • C ^A': B': C 68
• GV neu vf du trong SGK va giai. GV cd the neu vf du khac. • Sau day la vf du khac :
Cho hai mat phing (a) : x -my + 4z + m = 0
i^):x-2y + (m + 2)z-4 = 0.
Hay tim gia tri ciia m de :
a) Hai mat phing dd song song.
b) Hai mat phing dd trdng nhau.
c) Hai mat phing dd cit nhau.
Cau a.
Hoat ddng cua GV
Cdu hoi 1
Neu dieu kien de (a) // (P).
Cdu hdi 2
Xac dinh m de (a) // (P).
Caub.
Hoat ddng ciia GV
Cdu hdi 1
Neu dieu kien de (a) = (P).
Cdu hoi 2
Xac dinh m de (a) = (P).
Hoat ddng eiia HS
Ggi y trd loi cdu hdi 1
A _ B _C D
A'' B'" C'^ D'
Ggi y trd Idi cdu hoi 2
1 -m 4 m
I - 2 m + 2 -4
m-2
Hoat ddng cua HS
Ggi y trd Idi cdu hdi 1
ABC D
A'~ B'~ C'~ D'
Ggi y trd Idi cdu hoi 2
1 -m 4 m
1 - 2 m + 2 -4
Khdng cd m.
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cau c.
Hoat ddng cua GV
Cdu hdi 1
Neu dieu kien de (a) cit (P). Cdu hdi 2
Xac dinh m de (a) cit (P).
Hoat ddng cua HS
Ggi y trd Idi cdu hdi 1 A:B :C ^A': B': C Ggi y trd loi cdu hoi 2 m ^ 2.
2. Dieu kien de hai mat phang vudng gdc
GV sii dung hinh 3.12 va dat cac cau hdi:
H15. Nhan xet ve hai vecto Uj va n2 .
• GV neu dieu kien :
(a,) ±(02) 0 n,'.n^ = OoAA'+BB'+CC = 0
• GV neu vf du trong SGK va giai :
Hoat ddng cua GV Hoat ddng ciia HS
Cdu hoi 1
Mat phing da cho cd cap vecto chi phuong nao ?
Cdu hdi 2
Xac dinh vecto phap tuye'n ciia mat phing cin lap.
Cdu hdi 3
Xac dinh mat phing can lap.
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Gffi y trd Idi cdu hdi 1
AB va n.
Ggi y trd Idi cdu hoi 2
Mat phing cd vecto phap tuyen la n = AB, n = (-I;13;5)
Ggi y trd loi cdu hdi 3
X - 13y-5z+5 = 0.
HOATDONC 4
IV. KHOANG CACH TtTMOT DI£ M DEN MO T MAT PHANG • GV neu dinh If:
Trong khdng gian Oxyz cho mat phdng (a) cd phuang trinh :
Ax + By + Cz + D = 0 vd diem M^fx^ ; y^ ,• ZQ).
Khdng cdch tirM„ de'n (a) ki hieu d^M^, (o.)) vd dugc tinh theo cdng thitc:
' AJCQ + ByQ + CZQ + D\
dfMo,fa))=-I lA^ + B^ +C^
• De chiing minh dinh If tren, GV can dua ra cac budc sau
Gpi M,(x, ; y, ; z,) la hinh chieu ciia M,, tren (a).
Tfnh dd dai MJMQ.U
Tfnh dp dai : M,Mo .
» Thuc hien vf du 1 trong 4'
Hoat ddng cua GV Hoat ddng cua HS
Cdu hdi 1
Gffi y trd Idi cdu hdi 1
Tfnh diO, ia))
diO, ia))
|2.0 -2.0- 0 + 31 ^2'+{-2f+{-lf
Cdu hoi 2
TmhdiM,ia))
Ggi y trd Idi cdu hdi 2
diO, ia))
_|2.1-2.(-2)-13 + 3| 4
^22+(_2f+(-1) 2 3
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• Thuc hien vf du 2 trong 4'
Hoat ddng ciia GV
Cdu hdi 1
Chpn mdt diem M bat ki thudc (a). Cdu hdi 2
Tfnh diM, ia)).
• Thuc hien A 7 trong 5 phiit.
Hoat ddng cua GV
Cdu hdi 1
Hai mat phing nay cdng song song vdi mat phing nao?
Cdu hdi 2
Tfnh khoang each giiia hai mat phing dd.
TOM TfiT B^l HQC
Hoat ddng ciia HS
Ggi y trd Idi cdu hoi 1 GV cho HS chpn diem bat ki. Gffi y trd Idi cdu hdi 2 diM, ia))
diM, ia)) = 3.
Hoat ddng cua HS
Ggi y trd Idi cdu hdi 1 MP(Oyz)
Gffi y trd Idi cdu hdi 2 d((a), (p)) =1 -8 -(-2)1 = 6.
1. Vecto n^O gpi la vecta phdp tuye'n ciia mat phing (a ) neu gia eiia vecta n vudng gdc vdi mat phing (or).
2. Trong khdng gian Oxyz, cho mat phing id) di qua diem A/Q(XQ,>^0'^O) ^^ ^^ vecto phap tuyen niA; B ;C):
ia) : Aix - XQ) + Biy- yo) + Ciz - Zg) = 0.
Dat D = -iAxQ + ByQ + CZQ) thi phuong trinh ciia mat phing (or) dupe viet dudi dang :Ax + fiy + Cz + D = 0 trong dd A^ + B^ + C^ > 0.
3. Cac trudng hop rieng:
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Phuang trinh cua (a)
By + Cz +D = 0
Ax + By +D = 0
Ax + Cz +D = 0
Cz +D = 0
By +D = 0
Ax+D = 0
X y z
Dgc diem cua {a)
(a) song song hoac chira Ox. (a) song song hoac chira Oz. (a) song song hoac chira Oy.
(a) song song hoac triing Oxy. (a) song song hoac trimg Oxz. (a) song song hoac Irimg Oyz.
4. Phuang trinh doan chan : —I- — + — = 1.
a b c
5. Cho hai mat phing (a) va (a') lin lupt cd phuang trinh :
(a): Ax + By + Cz + D = 0
(a'): A'x + B'y + Cz + D' = 0
a) Hai mat phing dd cit nhau khi va chi khi A : B : C T^ A' : B' : C. b) Hai mat phing dd song song khi va chi khi
A__B__C_ D_
A'~fi'"C' D'
c) Hai mat phing dd triing nhau khi va chi khi
A__B__£^_D_
A'~ B'~ C'~ D'
6. Khoang each tii mdt diem den mat phing :
\AXQ + ByQ + CZQ + D\ diMQ,ia)) = ^- i A^ + B^ + C
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HOATDONC 5
MQT SO C6U HOI TRAC NGHI|M
Cdu 1. Hay dien diing, sai vao cac d trdng sau day:
(a) Mat phing x + 3y- z + 2 = 0cd vecto phap tuyen la (1 ; 3 ; -1) (b) Mat phing x + 3y- z + 2 = 0cd vecto phap tuyen la (1 ; 3 ; 2) (c) Mat phing x 3y- z + 2 = 0cd vecto phap tuyen la (1 ; -3 ; -1)
(d) Mat phing -x + 3y - z + 2 == 0 cd vecto phap tuye'n la (-1 ; 3 ; -1) Trd Idi.
D D • D
a D
b S
c
D
d
D
Cdu 2. Hay dien diing, sai vao cac d trdng sau day: (a) Mat phing 3y - z + 2 = 0 cd vecto phap tuyen la (0 ; 3 ; -1) (b) Mat phing x + 3y + 2 = 0 cd vecto phap tuyd'n la (1 ; 3 ; 2) (c) Mat phing x- z + 2 = 0cd vecto phap tuyen la (1 ; 0 ; -1) (d) Mat phang -x + 3y- z =:0cd vecto phap tuyen la (-1 ; 3 ; -1) Trd Idi.
D 0 D D
a D
b S
c
D
d
D
Cdu 3. Cho hinh lap phuong canh 1 nhu hinh ve. 74
Hay dien dung, sai vao cac d trdng sau day: (a) Mat phing ABCD cd phuang trinh la z = 0 (b) Mat phing BB'CC cd phuong trinh la x = 1 (c) Mat phing A'B'C'D' cd phuong trmh la z = 1 (d) Mat phing CCD'D cd phuang trinh la y = 1 Trd Idi.
D D D D
a D
b D
c
D
d D
Cdu 4. Dien vao d trdng sau
Phuong trinh ciia (a) Dgc diem ciia (a) (a) song song hoac chiia Ox.
(a) song song hoac chira Oz.
(a) song song hoac chda Oy.
(a) song song hoac triing Oxy.
(a) song \i.ng hoac triing Oxz.
(a) song song hoac triing Oyz.
Chgn cdu trd Idi dung trong cdc bdi tap sau:
Cdu 5 Cho hinh ve. Hinh lap phuang ABCD.A'B'CD' cd canh 1.
Mat phing (A'BD cd phuang trinh nao sau:
(a)x + y+l=0 ; (b)x + y + z= l (c) x + z = 1 ; (d) y + z + 1 = 0 Trdldi (b).
Cdu 6. Cho hinh ve. Hinh lap phuang ABCD.A'B'CD' cd canh 1. , z
Mat phing (CBD) cd phuang trinh nao sau:
(a) -X + y +z + 1 = 0 ; (b) -x + y + z = 1 (c) X + z = 1 ; (d) y + z + 1 = 0 Trd Idi . (b).
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Cdu 7. Cho mat phing cd phuang trinh (P) : x + 2y + 3z - 1 = 0. Mat phing nao sau day song song vdi (P)
(a) 2x +4y + 6x 1=0 ; (b) 2x +4y - 6z -2 = 0;
(c) 2x - 4y + 6z -2 = 0; (d) - 2x +4y + 6z -2 = 0.
Trd Idi (a).
Cdu 8. Cho mat phing cd phuong trinh (P): x + 2y + 3z - 1 = 0. Mat phing nao sau day triing vdi (P)
(a) 2x +4y + 6z -2 = 0; (b) 2x +4y - 6z -2 = 0;
(c) 2x - 4y + 6z -2 = 0; (d) - 2x +4y + 6z -2 = 0.
Trd Idi (a).
Cdu 9. Cho mat phing cd phuong trinh (P): x + 2y + 3z - 1 = 0. Mat phang nao sau day vudng gdc vdi (P)
(a) 2x +4y + 6x -2 = 0; (b) 2x +4y - 6x -2 = 0;
(c) 2x - 4y + 6x -2 = 0; (d) -3x + z -2 = 0.
Trd Idi. (d).
Cdu 10. Cho mat phing cd phuong trinh (P) : x + 2y + 3z - 1 = 0. Khoang each tilr M(l, 2, -1) den (P) la :
(a)^ ; (b)i-; (c) ^ ; (d) ^ Vl4 14 6 7
Trd Idi (a).
HOATDONC 6
HaQNG DfiN GIfil Bfil TfiP SfiCH GIfiO KHOfi
Bai 1. Sii dung phuong trinh ciia mat phing.
a) Hudng ddn. Six dung cong thiic : Aix -XQ) + Biy -yo) + Ciz - ZQ) = 0. Ddp sd. 2x + 3y + 5z -16 = 0.
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b) Hudng ddn. Vecta phap tuye'n cua mat phang dd la : [u, vl = (2; 6; 6)
Su dung cdng thiie : Aix -XQ) + Biy -yo) + Ciz - ZQ) = 0..\
Ddp sd. X - 3y + 3z - 9 = 0.
- - X V 7
c) Hudng ddn. Sit dung phuang trinh doan chin : h -2— -\ = 1 - 3 - 2 - 1
Ddp so. 2x + 3y + 6z + 6 - 0.
Bai 2. Sii dung tfnh chit cua trung diem va mat phang trung true. •Trung diem eiia AB la M = (3 ; 2 ; -5).
. AB = (2;-2;-4 )
Ddp so. (a) : X - y - 2z + 9 = 0.
Bai 3. Sii dung cac trudng hpp rieng cua mat phang.
a) Hudng ddn. Six dung hoat ddng 4.
Ddp sd. mp(Oxy): z = 0, mp(Oxz): zy= 0, mp(Oyz): x = 0.
b) mp(a) //(Oxy) nhan k(0;0;l) lam vecto phap tuye'n.
Ddp sd. (a): z + 3 = 0.
Tuong tu : (p) //(Oyz): x -2 = 0 ; (y) //(Oxz): x - 6 =0.
Bai 4. Su dung phuong trinh tdng quat cua mat phing.
a) Hudng ddn. mp(a) chiia true Ox va di qua P se nhan i va OP lira cap vecto chi phuang.
i ^ = [i,OP] = (0;-2;l).
(a): 2y + z = 0.
b) Hudng ddn. mp(P) chiia true Oy va di qua P se nhan j va OQ lam cap vecto chi phuang.
i^ = p,OQ] = (-3;0;-l).
(P) : 3x + z = 0.
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c) Hudng ddn. mpiy) chiia true Oz va di qua R se nhan k va OR lam cap vecto chi phuang.
" Y = k,ORj = (4;3;0).
(y): 4x + 3y = 0.
Bai 5. Sii dung phuang trinh tdng quat ciia mat phing.
Nen cd hinh ve de HS trung binli de tudng tupng.
A(5; 1;3)
D(4 ; 0 ; 6)
B(1 ; 6 ; 2)
C(5 ; 0 ; 4)
a) Hudng ddn. mp(ACD) se nhan AC va AD lam cap vecto chi phuang. r ^ = [AC,AD] = (-2;-l;-l).
(ACD):2x + y + z-14 = 0
Tuong tu : (BCD): 6x + 5y + 3z -42 = 0.
b) Hudng ddn. mp(a) nhan AB va CD lam cap vecto chi phuong. n^ = [AB,CD] = (lO;9;5).
(p): 10z + 9y + 5z-74 = 0.
Bai 6. Sii dung phuong trinh tdng quat ciia mat phing.
Nen cd hinh ve de HS trung binh de tudng tupng.
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Hudng ddn. mp(a) se nhan xxn = (2;-l;3) lam vecto phap tuye'n.
(a):2x- y + 3z-ll=0 .
Bai 7. Su dung phuang trinh tdng quat ciia mat phing.
Mat phing (a) can lap nhan: AB va nn n lam cap vecto chi phuong . Do dd
%=(l;0;-2) .
(a): X - 2z + 1 = 0.
Bai 8. Sii dung tfnh chat: Hai mat phing song song khi va chi khi:
A^_B__C_ D_
A'~ B'~ C D'
. 2 m 3 -5 ^, ., a) — = — = — yt. — Tu do ta CO n = -4, m = 4. /? -8 -6 2
. 3 -5 m - 3 ^ . ^. . 10 9
b) — = — = — ^— Iudotacon = ,rn= —
2 « - 3 1 3 2
Bai 9. Sii dung true tiep cdng thiic tfnh khoang each
IAJTQ + ByQ + CZQ + D
diMQ,ia)) = ^- V ^ B^ +C'
a) diMQ, (a)) = 5
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