Bai 6. Hudng ddn. Dua vao phuong trinh mat cau
1. Dieu kien de hai mat phang song song
H13. Khinao(a)//(a')
• 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 ^ k D ' [ D ^ k D '
, , , „ fn = kir' [(A;B;C) = k ( A ; B ' ; C ) a)= a c^ { <^<
[D = kD' l D = 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 d i 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
• 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).
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
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 cua HS Ggi y trd Idi cdu hdi 1
A B C D
A'~ B'~ C'~ D' Ggi y trd Idi cdu hoi 2
1 -m 4 m 1 - 2 m + 2 -4 Khdng cd m.
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,) ± ( 0 2 ) 0 n,'.n^ = O o A A ' + B B ' + C C = 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.
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 D I £ M DEN M O T MAT P H A N G
• 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 .
ằ Thuc hien vf du 1 trong 4'
Hoat ddng cua GV Hoat ddng cua HS Cdu hdi 1
Tfnh diO, ia))
Cdu hoi 2
TmhdiM,ia))
Gffi y trd Idi cdu hdi 1
|2.0 - 2 . 0 - 0 + 31 diO, ia))
^2'+{-2f+{-lf Ggi y trd Idi cdu hdi 2 diO, ia))
_ | 2 . 1 - 2 . ( - 2 ) - 1 3 + 3| 4
^ 2 2 + ( _ 2 f + ( - 1 ) 2 3
• 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)).
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.
• 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.
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.
TOM TfiT B^l HQC
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:
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
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.
X y z
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 ' ~ f i ' " 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
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 + 3 y - z + 2 = 0 c d vecto phap tuyen la (1 ; 3 ; -1) (b) Mat phing x + 3 y - z + 2 = 0 c d vecto phap tuyen la (1 ; 3 ; 2) (c) Mat phing x 3 y - z + 2 = 0 c d 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.
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 = 0 c d vecto phap tuyen la (1 ; 0 ; -1) (d) Mat phang - x + 3 y - z =:0cd vecto phap tuyen la (-1 ; 3 ; -1) Trd Idi.
a D
b S
c D
d D
D D
•
D
D 0 D D
Cdu 3. Cho hinh lap phuong canh 1 nhu hinh ve.
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.
a D
b D
c D
d D Cdu 4. Dien vao d trdng sau
D D D D
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).
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.
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 c h i n : 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.
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.