L *J 15 52 2 1J Do do dudng thang can tim co phuong trinh:
5b) Toa do giao diem la nghiem cua he:
b) Toa do giao diem la nghiem cua he:
[x + y - 2 z - l l = 0 < 3 x - y + z - 6 = 0 <=>
x + 2y - z - 15 = 0
c) Mp(xOy) : z = 0. Toa do giao diem A la nghiem ciia he
32 x = 4 y = 5 Vay M(4; 5; - 1 ) . z = - 1 2x - y + z = 6 x + 4y - 2z = £ z = 0 <=> y = 10 V a y A » H ; o ) . I 9
Giai tuong tu thi giao diem voi mp(xOz) la B(4; 0; -2) va voi mp(yOz) laC(0; 10; 16).
Vi du 17: Tim k de duong thang d la giao tuyen cua 2 mat phang
(P): 2kx + y - z + l = 0, (Q): x - k y + z - l = 0 nam trong mat phang (Oyz) Giai Giao tuyen d co VTCP: ( 1 - 1 - 1 2k 2k 1 u = K - k 1 1 1 1 - k J = (1 - k; - 1 - 2k; - 2 k - 1) Mp(Oyz) co VTPT i = (1; 0; 0)
Be d nam trong mat phang (Oyz) thi can co:
F u = 0 <=> (1 - k ) l + (-1 - 2k).0 + (-2k2 - 1).0 = 0 <=> k = 1. Thay k = 1 vao phuong trinh ciia 2 mat phang chiia d:
(P): 2x + y - z + l = 0, (Q): x - y + z - l = 0.
Ta co diem M(0; 0; 1) thuoc d va ciing thuoc mat phang (Oyz) nen thoa man.
Vay de d nam trong mat phang (Oyz) thi can va dii la: k = 1. Vi du 18: Trong khong gian co he toa do Oxyz, cho cac diem A(2; 1; 0),
B(l; 2; 2), C( l ; 1; 0) va mat phing (P): x + y + z - 2 0 = 0. Xac dinh toa do diem D thuoc duong thang A B sao cho duong thang CD song song voi mat phang (P).
Giai
x = 2 - t Taco AB = ( - 1 ; 1; 2), phuong trinh AB: j y = 1 + t .
z = 2t
D thuoc duong thing AB => D(2 - t; 1 + t; 2t) => CD = (1 - t; t; 2t) Vecto phap tuyln cua mat phang (P): n = (1; 1; 1)
V i C khong thuoc mat phang (P) nen:
. 1 CD // (P) <=> n. CD = 0 <=> 1.(1 - t) + l . t + 1.2t = 0 <=> t =
Vay D(-; -;-l).
2 2
V i du 19: Chung minh cac mat phang
(Pm): (2 + m)x + (1 + m)y + (1 + m)z + m - 1 = 0 luon di qua mot duong thang co dinh.
Giai (Pm): 2x + y + z - l + m(x + y + z + 1) = 0.
Mat phang (Pm) di qua cac diem M(x; y; z) co toa do khong phu thuoc m . . . , . , , , . f2x + y + z - l = 0
khi va chi khi: {
[x + y + z + l = 0 Cho y = 0 thi x = 2, z = - 3 : A(2; 0; -3) Cho z = 0 thi x = 2, y = - 3 : B(2; - 3 ; 0).
Vay cac mat phang (Pm) di qua duong thang co dinh la giao tuyen ciia 2 mat phang: 2x + y + z - l = 0, x + y + z + l = 0 tiic la duong thang AB co dinh.
V i du 20: Chiing minh cac duong thang dk la giao tuyen ciia 2 mat phang: x + k z - k = 0, ( 1 - k)x - ky = 0, k * 0 luon nam tren mat phang co dinh.
Giai (adsbygoogle = window.adsbygoogle || []).push({});
Giao tuven dk chiia cac diem M(x; y; z) co toa do thoa man he: |X'+kz-k = ° ;k*0.
|Q - k)x - ky = 0
Suy ra: x - (1 - k)x + kz - k + ky = 0.
=> k(x + y + z - l ) = 0 = > x + y + z - l = 0, vi k * 0.
Vay cac during thang dk luon luon nam trong mat phang co dinh (P): x + y + z - l = 0.
V i du 21: Cho duong thang (d) voi tham so a la giao tuyen ciia 2 mat phang: x.cosa + y.sina + z.sina = 6sina + 5cosa
x.sina — y.cosa + z.cosa = 2cosa - 5sina
a) Chung minh duong thang (d) khong cit mat phing (P): x.sin2a - y.cos2a + z - 1 = 0.
b) Goi (A) la hinh chieu vuong goc cua (d) uen mat phang (Oxy). Chung minh khi a thay doi, duong thang (A) tiep xiic voi mot duong Uon co dinh.
Giai
a) (d) la giao tuyen ciia hai mat phang co cac vecto phap tuyen
n1 = (cosa; sina; sina).
n2 = (sina; -cosa; cosa) nen co vecta chi phuong u = [ n j n2 ] = (sin2a; —cos2a; —1)
(P) co vecto phap tuyen n = (sin2a ; -cos2a; 1).
Ta co u . n = sin22a + cos22a - 1 = 0 nen (d) khong cat (P)
b) Mat phang (Oxy) co phuong trinh z = 0.
Khu z trong PT duong thang (d) duoc PT cua (A) trong mp(Oxy) la:
x.cos2a + ysin2a - 2sin2a - 5 = 0.
Trong mat phang z = 0, khoang each tir diim 1(0; 2; 0) den (A) la:
|0. cos 2a + 2. s.in a - 2 sin 2a - 5|
D(I, (A)) 5: khong doi nen (A) tiep
Vcos2 2a + sin2 2a
xuc voi duong tron tam 1(0; 2; 0), R = 5 tren mp(Oxy). Vidu 22: Tim tam va ban kinh cac duong tron giao tuyen:
a)(S): x2 + y2 + z2 - 6x + 2 y - 2z + 10 = 0 va(P): x + 2 y - 2z + 1 = 0. b) x
2 + y2 + z2 - 12x + 4y - 6z + 24 = 0
2x + 2y + z + l = 0
Giai
a) (S) co tam 1(3;-1; 1), ban kinh R = 1.
Tam H la hinh chieu ciia I len (P). Phuong trinh ciia duong thang d qua I x = 3 + t
va vuong goc voi mat phang x + 2y - 2z + 1 = 0 la: < : - 1 + 2t . (adsbygoogle = window.adsbygoogle || []).push({});
l-2t
Tir do ta suy ra giao diem H ciia d va mat phang img voi t = 0 la H(3; - 1 ; 1). on nen
Vi diem H triing voi I nen (P) la mat kinh cat theo duong tron
ban kinh during tron giao tuyen r = R = 1.
b) (S) co tam 1(6; - 2 ; 3), R = 5.
Phirong trinh duong thang d qua I va vuong goc voi
x = 6 + 2t
mat phang 2x + 2y + z + 1 = 0 la: I y = -2 + 2t
z = 3 + t
Tir do suy ra giao diem H ciia d va mat phang img voi t = ——la tam M 10 14
duong tron giao tuyen H —. 3 '; - 3 ' 3,
Ban kinh r2 = R2 - I H2 = 25 - 16 = 9. Vay r = 3.
Vidu 23: Trong khong gian voi he toa do Oxyz, cho mat phang (P):
2X - 2y - z - 4 = 0 va mat cau (S): x2 + y2 + z2 - 2x - 4y - 6z - 11 = 0. Chung minh rang mat phang (P) cat mat cau (S) theo mot duong tron. Xac dinh toa do va tinh ban kinh ciia duong tron do.
Giai (S) co tam 1(1; 2; 3), ten kinh R = 5. Khoang each tir I den (P): d(I, (P)) = 2
~4 ~3
" 4
suy ra (P) cat mat cau (S) theo mot duong tron.
Goi H va r lan luot la tam va ban kinh ciia duong tron giao tuyen thi H la hinh chieu vuong goc ciia I tren (P):
IH = d(I; (P)) = 3, r = VR2 - I H2 = 4 x = 1 + 2t
Phuong trinh IH: • y = 2 - 2t. T h i x, y, z vao (P) thi duoc t = 1 nen tam [z = 3- t
H(3; 0; 2).
V i du 24: Cho mat cau (S): x2 + y2 + z2 - 2x + 6y - 4z + 13 = 0 va duong thang (d) di qua A(2; 1; 0) co vecto phuong u = ( 1 ; m; - 2 ) . Bien luan theo m so giao diem cua (S) va (d).
Giai
Diem M(x; y, z) thuoc (d) nen x = t + 2, y = mt + 1, z = -2t. Thay vao (S) duoc: (t + 2)2 + (mt + l )2 + 4t2 - 2(t + 2) + 6(mt + 1) + 8t + 13 = 0 <^ (5 + m2) t2 + 2(5 + 4m)t + 20 = 0.
Ta co A' = (5 + 4m)2 - 20(5 + m2) = - 4 m2 + 40m - 75. Bien luan: - Neu A' > 0 o - < m < ~: (d) cat (S) tai hai d i i m phan biet.
z z
- N e u A' = 0 <=> m - hoac m = — : (d) tiep xiic voi (S) tai mot diem. 5 15
Z 2A
5 15 (adsbygoogle = window.adsbygoogle || []).push({});
- Nev A' < 0 <=> m < - hoac m > — : (d) va (S) khong co diem chung.
z z
YJ du 2± Cho tam giac ABC co C(3; 2; 3), duong cao A H nam tren duong
x - 2 y - 3 z - 3 , ,
. duong phan giac trong B M ciia goc B
thang (di):
nam tren duong thang (d2):
-2
x - l y - 4 z - 3 1 -2
Giai 1) Mat phang (P) qua C, 1 (d,) la:
l . ( x - 3) + l . ( y - 2) - 2.(z - 3) = 0 » x + y - 2 z + l = 0.
(P) n (d2) = B ( l ; 4; 3)
Mat phang (Q) qua C, 1 (d2) la:
l.(x - 3) - 2.(y - 2) + l . ( z - 3 ) = 0 <=>x-2y + z - 2 = 0
(Q) n (d2) = 1(2; 2; 4)
K doi xiing voi C qua (d2) thi K nam tren duong thang chiia canh AB. Vi I la trung diem ciia CK nen K ( l ; 2; 5).
x = 1
y = 2 + 2 t . Do do: (A) cat (d,) tai A ( l ; 2; 5). z = 5 - 2t
D A N G 4: G O C V A K H O A N G C A C H Goc giua 2 duong thang: d co VTCP u va d' co VTCP v
cos(d, d') = | cos( u , v ) |
Goc giira duong thang va mat phang: d co VTCP u va (P) co VTPT n sin(d, (P)) = | cos( u,n ) | • Chii y:
- Goc giua 2 vecto tu 0° den 180° va cac goc con lai giua duong thang, mat phang deu tir 0° den 90°
- Goc giira duong thang va mat phang la goc giua duong thang voi hinh chieu ciia no len mat phang.
Khoang each tir Mo(x0, yo, zo) den mat phang:
(P): Ax + By + Cz + D = 0 la: d(M0, (P)) = 1 + By o + C zo + D 1 VA2 + B2 + c2 Khoang each tir mot diem den 1 duong thang:
Cho Mo(x0, yo, zo) va duong thing d qua A va co VTCP u = AB S ( M0A B ) = =• | [ A M0 )A B ] | = - A B . M0H
2 2 [ A M0, u ]
nen: d(Mo, d)
Khoang each giua hai duong thang cheo nhau: Duong thang d) qua M i va co VTCP u i Duong thang d2 qua M2 va co VTCP u 2
l [ u1, u2_ . MXM2 1 u ^ ] . l d(d1,d2) = (adsbygoogle = window.adsbygoogle || []).push({});
• Chii y:
- Khoang each giua cac yeu to song song la khoang each tir 1 diem chpn tren yeu to nay den yeu to kia.
- Khoang each giua 2 duong cheo nhau la khoang each tu duong nay den mat phang chua duong kia va song song voi no.
- Ta co the tim chan hinh chiiu cua mot diim len duong thang, mat phang
de dua ve khoang each giua 2 diem. . - I m g dung toa do Uong khong gian d l giai cac bai toan hinh khong gian co
dien, quan he song song, vuong goc, do dai, goc, khoang each, vi tn tuong doi,...
V i du 1: Tinh goc tao bdi dudng thang va mat phang: x = 1 - t a)d: y = 2 ( P) : y- z = 0 (z = t , . , x + 2 y -1 z - 3 /T V> n „ b) d: - = - = — —. (P): x + y - z + 2 = 0 d; 1±1 = Izl = -, (P): 3x - 3y + 2z - 5 = 0. 2 4 3 Giai Goi q> la goc giua d va (P) thi 0° < cp < 90°
a) d co VTCP u = ( - 1 ; 0; 1), (P) co VTPT n = (0; 1; - 1 ) sincp = | cos( u , n)| = ^L^j- => cp = 3 0° b) d co VTCP u = (4; 1; -2), (P) co VTPT n = (1; 1; -1) I / - - x l 14 + 1 + 21 7 sincp = cost u , n ) = . 1 —' = —?=. V16 + 1 + 4.V3 V63 c) d co VTCP u = (2; 4; 3), (P) co VTPT n = (3; - 3 ; 2) I - ^ l 6-12 + 6 0 sincp = cosf u , n ) = , - , = 0 => cp — U V4 + 16 + 9.V9 + 9 + 4 x + 3 v — l z — 2
V i du 2: Tim goc tao bdi dudng thang = = vdi cac true toa do
2 1 1
Giai
Dudng thang da cho co vecto chi phuong u = (2; 1; 1), cac true Ox, Oy, Oz lan luot co vecto chi phuong la T = (1; 0; 0), J = (0; 1; 0), k = (0; 0; 1). Tir do goc tao bdi dudng thang da cho vdi cac true Ox, Oy, Oz co so do sao cho cosin ciia no lan luot bang:
|2| V6 cosepi = I cos(u , i ) | = coscp2 = I cos(u , j ) I = V6.VI 3 111 V6 V6.VT 6 I f- r ^ l I1! ^ coscp3 = I cos(u , k ) | = = -g"
Vj du 3: Tim goc tao boi cac cap duong thang sau day: a)d: b)d: x = 1 + 2t y = - l + t z = 3 + 4t x = 2 + t y = l + t z = 6 d' d' x = 2 - t y = - 1 + 3t z = 4 + 2t x = 2 - 2t1 y = -5 z = 2t Giai
Gpi cp la goc giua 2 duong thang thi 0° < cp < 90 a) d, d' co VTCP u = (2; 1; 4), u ' = ( - 1 ; 3; 2) | - , j 1-2 + 3 + 81 coscp = | cos(u , u ) V4 + 1 + 16.V1 + 9 + 4 ~ V294 b) d, d' co VTCP u = (1; 1; 0), u ' = (-2; 0; 2) |-2| 1 coscp = | cos(u , u ') cp = 60° V2.V8 2
Vi du 4: Cho tu dien ABCD co A(3; 2; 6), B(3; - 1 ; 0), C(0; - 7 ; 3) va D(-2; 1; -1).
a) Tinh goc giua cac cap canh doi.
b) Tinh goc giua duong thang A D voi mp(ABC). Giai
A B. CD = 0 do do a) Ta co AB = (0; -3; -6), CD = (-2; 8; -4)
g(AB, CD) = 90°
Tuong tu AC . BD = 0, AD . BC = 0 nen goc giua cac cap d6i deu bing 90°. b) Duong thing A D co VTCP u = DA = (5; 1; 7)
Mp(ABC) co VTPT n = [ A B , AC ] = (5; - 2 ; 1)
Gpi cp la goc giua A D vdi mp(ABC) thi: sincp= | cos(DA, n ) | = ^H^. 5 Vi du 5: Cho dudng thang d la giao tuyen cua 2 mat phang (P):
x - z. sina + cosa = 0, (Q); y - z.cosa - sina = 0. Chung minh dudng thang d tao vdi true Oz mpt goc khong doi.
Giai (P) co VTPT n = G; 0; -sina) (Q) co VTPT m = (0; 1; -cosa) Do do d co VTCP u = [ n , m ] = (sina; cosa; 1) True Oz co VTCP k = (0; 0; 1) Ta co | cos(d; Oz) |
Vay dudng thang d hop Oz goc 45° khong doi. cosiu , k ) | = - =
V i du 6: Cho M ( l ; 0; 2), N ( l ; 1; 0), P(0; 1; 2). Goi A, B, C lan luat la giao diem mp(MNP) voi Ox, Oy, Oz.
a) Chung minh ba duong thang AP, B M , CN dong qui tai G.
b) Goi cp!, 97, cp3 la cac goc tao boi OG voi OA , OB, OC . Chung minh
2 2 2 (adsbygoogle = window.adsbygoogle || []).push({});
COS cpi + COS CP2 + COS 93 = 1.
Giai
a) M N = (0; 1; - 2 ) , NP = ( - 1 ; 0; 2) nen tim duoc mat phang (MNP):
2x + 2y + z - 4 = 0.
Cho y = z = 0 ^ > x= 2 = > A(2; 0; 0) z = x = 0 ^ y = 2^> B(0; 2; 0) x = y = 0 ^ > z = 4=> C(0; 0; 4)
Ta co N A = N B , PB = PC, M A = MC nen trong tam giac ABC thi 3
2 2 4
trung tuyen AP, B M , CN dong qui tai trong tam G( — ; — ; —).
3 3 3
b) Taco OG = ( - ; - ; - ) , OA = (2; 0; 0) nen
' 3 3 3
coscp i = cos( OG. O A ) = —j= v6
Tuong tu coscp2 = -j^, coscp3 = -JL => dpcm.
Vi du 7: Tinh khoang each tir diem M den duong thang A: \ »«vr. o * x + 2 y - 1 z + 1
a) M(2; 3; 1), A: = =
7 1 2 -2
1 3
b) M(2; 3; -1), A di qua diem M0(—; 0; — ) va co vecto chi phuonu 2 4
u = (-4; 2; - 1 ) .
Giai
a) Duong thang A di qua diem Mo(-2; 1; -1) co vecto chi phuong u = (1; 2; -2).
Taco M„M = ( 4 ; 2; 2), [ u , MnM ] = ( 8 ; - 1 0 ; - 6 ) . |P..M^M]| Vs2 + 102 + 62 10V2
Vay khoang each can tim la: d =
22 + 22
Cach khac: Goi (a) la mat phang di qua M(2; 3; 1) va vuong goc voi A thi (a) co vecto phap tuyen (1; 2; - 2 ) , (a) co phuong trinh:
x - 2 + 2(y - 3) - 2(z - 1) = 0 hay x + 2y - 2z - 6 = 0. Tir do tim duoc toa do giao diem H cua (a) va A la:
T Tr 14 17 17 . 10V2
b) Taco M0M = f | ; 3 ; - - M ; [ u , M0M ] ; - l 7
Vay khoang each can tim la: d u , , M0M
2 2 _ V2870 (adsbygoogle = window.adsbygoogle || []).push({});
14
Vi du 8: Tim khoang each tir diem A(2; 3; -1) den duong thang d la giao tuyen ciia 2 mat phang (P): x + y - 2 z - 1 = 0, (Q): x + 3y + 2z + 2 = 0.
Giai (P): x + y - 2 z - 1 = 0, (Q): x + 3y + 2z + 2 = 0. 5 Cho z = 0 thi x + y - 1 = 0 x + 3y + 2 = 0 x = 5 3
Do do giao tuyen d di qua M0( — ; — ; 0) va co VTCP
2 2
u= [ n 7 n X ] = ( 8 ; - 4 ; 2 ) h a y ( 4 ; - 2 ; D.
Do do d(A, d) A M c " [205
14
Vi du 9: Tim khoang each giua hai duong thang sau: x = l + t a) d : y = - 1 - t d' : z = 1 b) d x _ y - 4 _ z + l ^ T " 1 -2 d' : Giai x = 2 - 3t y = -2 + 3t' z = 3 x = - t ' y = 2 + 3t' z = -4 + 3t
a) Duong thang d di qua diem M i ( l ; - 1 ; 1), co vecto chi phuong ux = (1; - 1 ; 0). Duong thang d' di qua diem M2(2; -2; 3) co vecto chi phuong u2 = ( - 1 ; 1; 0). V i ux va u2 cimg phuong nhung ux u2 khong cimg phucmg voi