62 Cho hinh tru c6 cac day la hinh tron tarn O vh tarn O', ban kinh ddy bang chi^u cao va bang a Tren ducmg tron day tarn O la'y diem A, trdn ducmg tron day tarn O' la'y diem B cho AB = 2a Tinh the tfch kh6'i tir dien OO'AB (Trich de thi Dai hoc - Khoi A - 2006) ChUffng in PHl/ONG P H A P T O A D O T R O N G K H O N G G I A N I Ht TOA DO TRONG KHONG GIAN A L t THUYfeT C A N N H Toa cua diem va cua vecta / He tog do: Trong khong gian CO ba true toa d6 vuong goe vdi doi mot va c6 dinh hudng Ba true nhu vay duac goi la ht toa d6 vuong goc kh6ng gian Didm O goi la g6'c toa d6 - True hoanh, dinh hudmg duang X Ox, c6 vecta don vi i = (l;0;0) o Hinh 34 True tung, dinh hudng duong y'Oy, eo vecta dan vi } =(0; 1;0) f True cao, dinh hudng duang z'Oz, c6 vecta dan vi ^=(0;0;1) - Cac mat phang xOy, yOx, xOz d6i m6t vudng goc vdi duac goi 1^ cac mat phang toa dd Vi i , } , it la cac vecta dan vi tren true, ma cac true vu6ng goc d6i ni6t, nen ta cof = va T.J =0; i.k =0; j k He qua = Trong khdng gian vori he toa dp O x y z dupe gpi la khong gian O x y z Toa dp mot diem Trong khong gian O x y z a, = b,, a2 = b2, a, = bj b) 0) = (0; 0; cho m6t diem M y, m6i diem hoan toan xac dinh boi vecta a) a = h « c) a va b ciing phuong M d) Trong khong OM gian c6 mot so k: a, = kb,, aj = kbj, a, = kbj Oxyz c6 A ( a , ; a2; a,) va B(b,; b2; b,) thi 45 = ( b - a , ; b2-a2; b j - a j ) V i vecto i , j , k la vecta khong dong phang nen c b6 so T i c h v6 hudmg nha't (x, y, z) cho: 1) Bieu thAc toq OM = X i +y.j + z.k D i n h If Trong khong gian O x y z tich v6 hudng cua hai vecto a (aj; a.^, a-,; = OM, +OM3 + OMj Hinh 35 va A (b,; bj; h^) la m Or sty duac xac dinh bcfi cong thiic: Nguoc lai vdfi (x, y, z) ta c6 didm M nhat khong gian thoa man OM = X a b = a,b| + a2b2 + ajbj i + y.j + z.k Bp ba so (x, y, z) gpi la toa dp cua diem M doi v6i he true Oxyz da cho va k i hieu la M ( x , y, z) hoac M = (x, y, z) Toq cua 2) Ifng dung * Do dai cua vecto vecta Trong khong gian Oxyz, cho vecto u , ta luon c6: * Khoang each giua hai diem A AB = M = a,, i + 3.2-j 7(XB - x^f + (ye - yj' + (z^ - z^f +a3 k Bp ba so (a,, a2, a,) xac dinh nhat va gpi la toa dp ciia vecto u, ki * G p i la goc giiia vecta a va b hieu la u (a,, aj, a,) cos = Bieu thirc toa cua cac phep toan vecto S ( b i ; b2; bj) k h i ± b = (a,± b,; a2± bj; a j i b,) b) k a = (ka,; kaj; k-a,) = k(a,; di^; aj) (k la so thuc) a,b, +a2b2 +a3b3 ^/af+af+a^ -^/bf+bf+b => a J b aib] + a2b2 + a3b3 = / Dinh U Trong kldong gian Oxyz cho cac vec ta a (a,; aa; aj) d) a A ( X A ; yA; Z A ) B ( X B ; yg; Zg) la T i c h CO hudng cua hai vecto (hay tich vecto) /) Dinh nghia: Tich c6 hu6ng (hay tich vecto) cua hai vecta a (ai, a2, a,) Va b (b„ b2, bj) la mot vecto duac k i hieu \di[a,b] d6 duac xac dinh nhu sau: hay a A va c toa f a, a, a, a, Vb,b3 b3b, b,b, ) 2) Tinh chat [a,b]=0 {a,b] o a =k.S l a v a [ a , ] I ( r a , b ] a = 0; r a , b b = 0) a,b = a B Vf DU > sin(a,b) Vi du Cho tii dien ABCD: 3i)Chvtngm\nh: AD+ BC = BD + AC b) Goi M la die'm chia dudng trung tuyen AA, ciia mat phang ABC theo so 3: ( = — ) Chung minh rang: MA, 3)Apdung J Tick dien tick cua hinh binh hanh va thetich khoi hdp • A B C D la hinh binh hanh SABCD = A B A D sin A AB,AD • A B C D A B C ' D ' la hinh hop • VABCD.A'B'C'D'= T A B , A D AA Ba vecta a, c, bdong phang a,b c = Ba vecta a, b, c khdng d6ng phing la phuong trinh mat c^u tarn I(-A;-B;-C) va r = ^A^ + B' + C' - D lai: a) DM = —DA + DB + DC 20 10 20 AD - AC = CD BD - BC = CD AD - AC = BD - BC AD+ BC = BD +AC )Cdchl MA _ MA, ~ ^ D AM =- MA AM = DM - DA MA, = DA, - AM Nensuy DM - DA = - (DA, - AM) DA + — DA, >DM = — 10 10 ' Ma DA, = -(DB + DC) T>oA6 DM = — DA — DB + — DC 10 20 20 X-^'- -V- Cdnh 2: Bien ddi ve phai: = - - b+a — DA ^ — DB ^ — DC 10 20 20 = - — b + a + -{c2 + M^)+ —(Z)M+ MB)+—-{DM = (DM 10 20 + MC) 20 _L DM + — DM +— Z)M +— M4 +— MB +— MC 10 20 20 10 20 = DM + — MA + — (MB + MC 10 20 10 MA + — M A i , MA^^ = MN^= - (a - b +cf= = ld^^MN = MN • 20 2 —(a-b+c) - {a^+ b+c+l{-a.b + a.c-b c) ^ dV2 AB = d Theo c6ng thiic ta c6: M N A B = M N AB VP = VT la dieu phai Chung minh M N a.b = c.a = b.c = a cos 60"= -d\ 1) MN = MA + AB+ BN r - = - - b +a+- BD 2 1BD Vi du U m toa d6 hinh chie'u cua die'm A ( l ; - ; -5) trdn: 1) mp Oxy; 2) mp Oxz 3) mp Oyz; 4) True hoanh; 5) True tung 6) True cao A Giai: Vf du 6: 1) Tren mp Oxy thi cao z = nen toa d6 hinh chie'u cua diim A la Cho a = (3; - ; 5) va = (1; 2; - ) Tim c thoa man cac di6u kidn A,(l;-3;0) sau: c ±Oz, ca = 9, cb = - 2) Tr6n mp Oxz thi tung d6 y = nen toa hinh chieu cua diem A la A j d ; 0; - ) 3) Tren mp Oyz thi hoanh x = nen toa hinh chieu ciia di^m A la Giai: Goi c= (x y, z) V i c Oz o ck = o x + y + z 1= => z = vi c a = « 3x - y = V i cc = X + 2y = - A3(0; -3; -5) 4) Tren true hoanh Ox thi tung y = 0, cao z = nen tea d6 hinh chieucuaAlaA4(l;0;0) 5) Tren true tung Oy thi hoanh x = 0, cao z = nen toa hinli chieu ciia A la A,(0; -3; 0) 6) Tren true cao Oz thi hoanh x = 0, tung y = nen toa hinh chieu ciia A la A^iO; 0; -5) Vi du 4: Cho A(-3; 2; -1) Tim toa diem doi xiing cua A qua gdc toa Giai he: P'^-y^^ ^ • [x + y = - 1^ = [y=-3 c , (2; - ; 0) Vi du Cho tii dien ABCD c6 A ( l ; -2; -1), B(-5; 10; -1), C(4; 1; 11), D(-8; -2; 2) ViS't phuang trinh mat ciu ngoai tie'p tii dien ABCD Giai: Goi I(x; y; z) la tarn mat c&u ngoai tie'p tii dien ABCD, thi ta phai c6: d6, qua eac true toa do, qua cac mat phang toa d6 IA = IB = IC = ID Giai: Qua gdc toa do: toa diem doi xiJng ciia A la: (3; - ; 1) IA=IB Qua true hoanh x' Ox: toa dd di^m ddi xiing ciia A la: (-3; 2; -1) Qua true tung y'Oy: toa di^m ddi xiing etia A la: (3; 2; 1) IA=IC (1) IA=ID Qua true cao z'Oz: Toa di^m ddi xiing cua A la: (3; - ; - ) Qua mat phang Oxy: toa diem ddi xiing ciia A la: (-3; 2; 1) Qua mp Oyz: toa dd diem ddi xiing cua A la: (3; 2; -1) lAJBJC, lA IB IC, ID ,ID lA = ( l - x ; - - y ; - l - ? ) Qua mp Oxz: toa di^m ddi xiing ciia A la: (-3; - ; -1) Vi du 5: => lA = ^il-xr+i2 Cho AB = (2; - ; -1) Tim toa dd diem A, biet B ( l ; - ; 2) IB = (-5-x; 10-y; -1-z) Giai: + yy+il + zr => IB Goi A(x;y;z) ta c6: AB = (1-x; - - y ; 2-z) Suy 2=1-X X -3 = - - y y =2 -l = 2-z z=3 IC= =-1 A ( - l ; 2; 3) (4-x; l - y ; l l - z ) IC = V ( - ; c ) ' + ( ! - > ' ) ' + ( 1 - ^ ) ' A1 ID = ( - - x ; - - y ; 2-z) ID T a c o h a : • 18x - 22y - 5z = ( l - x ) ' + (2 + y ) ' +(1 + zf ( - x ) ' + (2 + y ) ' +(1 + zf x)^+(l-y)'+(ll-z) = (8 + X ) ' + ( + y ) ' + ( - Vidu9 Cho cac die'm A ( ; ; l ) , B ( - l ; ; l ) , C(5;2;3), D(0; - ; ) -2 ^ Chirng minh rling b6'n diem A , B, C, D la dinh ciia hinh t i i dien I (-2; 4; 5) z = I Khoang each l A = = -^{l-xf + ( + >;)' + ( l + z ) ' = Mat e^u ngoai tiep t i i dien A B C D c6 tarn I(-2;4;5) va c6 ban kinh r = l A = Tinh ^AO (O la tam ciia mat BCD ciia hinh t i i dien) G i a i : a Neu bon d i ^ m A , B, C, D la b6n dinh cua m6t hinh tir diSn thi ba vector A B , AC, AD khong ddng phang Vay phuong trinh mat cau la: A B = (-4;4;1), AC = (2;2;2) = ( ; ; ) , A D = ( - ; - ; ) (x+2)' + ( y - ) ' + ( y - ) ' =81 Cdch I: Xet bieu thirc Vidu8 Cho a = (3,;2;2)va = (18;-22;-5) T i m cbie't = 14,c a,c c tao vdfi true tung goc t u ±bva -4 1 -5 + 1 -3 [AC,AD].AB + 1 -3 -5 = -4.8 - 4.6-2 = -58 9^ (dpcm) Giai Cdch Ta khong tim dugc cap so x, y thoa man A B = xAC Goi c = (x; y; z) V i : fl c < < = > y < nen chon y = - Dap so: c = (-4; -6;12) - y + 10 = 3x-z + ll = y = z = -2y y = ±6 V i c tao vdi true tung goc til nen c.j zf ( - x ) ^ + ( + y ) ' + ( l + z)^ = ( + X 3x + 2y + 2z = nghia la he sau v6 nghiem: = 3x + 2y + 2z ^ c o ^ = » =0 18x - 22y - 5z = = c : > x ' + y W = 14' - = X - 3>' = X - 5y ^ 1= X + 3y he v6 nghiem + yAD, 67 Toa d6 trung diem cac canh cua tam giac ABC la (1;3;2), (0;2;0) (2; -2; 4) Tim toa cua cac dinh tam giac ABC b Goi 0(x, y, z) ta c6: AO = AB + BO 68 Tim tren true hoanh mot diem each deu hai diem A(l; -3;7) va B(5;7;- -5) AO = AC + CO 69 AABC CO A(l;2; -1), B(2; -1;3), C(-4;7;5) Tim d6 dai ducmg phan AO=AD+DO giac BD 3AO = AB + AC + AD-(OB + OC + OD) Vi O la irong tarn tarn giac BCD ntn OB + OD + OC = 1 Suyra: AO =-{AB + AC + AD) =-(-5;l;5) 70 AABC CO A(-4; -1;2), B(3;5; -10) Tim toa d6 dinh C bie't trung di^m canh AC thu6c true tung,^ trung diem canh BC thu6c mpOxz 71 AABC CO A(6;2;3), goc toa d6 la trung diem canh AC Trong tam G ciia AABC thu6e true tung Tim toa B, C VsT A0\ ^^i-5y+l'+5' 72 AABC CO A(-l;2;3), tam G trung vdi g6c toa do, Be Ox, = ^ C e mpOyz Tim toa d6 B, C 73 Tim the tich tii dien ABCD biet toa d6 cac dinh A(2; -1;1), B(5;5;4), C BAI TAP C(3;2;-1),D(4;1;3) 74 Cho hinh hop ABCD.A'B'C'D', biet A(-1,0,1), B(2;l;2), D(l;l;2), 63 Cho tii dien ABCD Tim diem O cho: C'(4,-5;l) OA + OB + OC + OD = a) Tim toa d6 cac di^m eon lai ciia hinh h6p Chiing minh di^m O la dilm nha't b) Tim the tich hinh hop tren 64 Cho tii dien ABCD Goi A', B', C , D' la cac di^m theo thir tu chia cac doan thang AB, BC, CD, DA theo ty s6 k: 75 Cho lang tru diing ABCA,B,C, c6 day ABC la tam giac vudng AB = A C = a, AA, = aV2 Goi M, N 1^ lugt la trung diem ciia doan A'A BB CC DD AB BC CD DA AA, va BC, Chiing minh MN la dirdng vu6ng goc chung cua cac ducmg =k CMR vdfi moi di^m O bat ky kh6ng gian, ta lu6n c6: OA + OB + OC + OD = OA' +OB+OC' +OD' Vol gia tri nao ciia k thi b6'n diem A',B',C,D' d6ng phing? 65 Cho a = S,b =l,(a,6) = 30" Tinh goc tao bdi tdng va hifu hai thing AA,va BQ.Tinh VMA,BC, • 76 Trong khong gian toa d6 Oxyz cho 0(0; 0; 0), B(a; 0; 0), D(0; 1; 0), O' (0,0,a) la bon dinh ciia hinh hop chu nhat OBCD.O'B'CD' Tim add 'BD L^T: (Trich d6 thi DHXD, 1999) 77 Trong kh6ng gian toa d6 Oxyz cho hinh tii dien ABCD, bie't toa d6 cac vecta a,b 66 Tarn giac ABC c6 toa d6 cac dinh A(3; -1;6); B(-l;7; -2), C(l; -3;2) Chiing minh tam giac ABC la tam gidc vu6ng dinh A(2; 3; 1), B(4; 1; -2) C(6; 3; 7), D(-5; -4; 8) Tinh dai ducmg cao ciia tii dien xua't phat tur A (Trieh de thi DH Duoc, 1999) II M A T P H A N G A THUYfeT CAN NH6 Phuofng trinh long quat cua mat phdng * Vector phap tuyen: Vecta n^O vu6ng goc vol mat phang («) goi la vecta phap tuyen cua mat phang (a) Dinh ly: Trong kh6ng gian Oxyz ne'u m|t phang (a) c6 cap vecto chi phaang la a (a,; a^, a,) va b (b,; h^, b,) thi ( a ) c6 m6t vecta phap tuy^fn CO toa n = V > J Kh6ng CO mat x va y (A = 0, B = 0) thi mat phing song song hoac Itrung voi mat phang Oxy Tuong tu mat phang Ax +D = song song hoac ttrung vdi mat phang Oyz, mat phang By + D = song song hoac triing v6i lat phang Oxz f * Ne'u A,B,C,D khac 0, bang each dat a = - — , b = - — , c = - — ta c6 A B C \ dua (1) \i dang: - + ^ + - = 1(2) a b e Khi mat phang (a) cat cac true Ox, Oy, Oz \in luat tai cac di^m: (a;0;0); (0;b;0); (0;0;c) Phuomg trinh (2) la phuang trinh ciia mat phang theo doan chan Vi trf tuong doi cua hai mat ph^ng Cho hai mat phang (or,): A,x + B,y + C,z + D, = = (ajbj - ajbj-, ajb, - a,b3;a,b2 - a^hi) • Nhuvay n = Dinh nghia: Phuong trinh c6 dang Ax + By + Cz + D = d6 A, B, C («2):A2X + B2y +C2Z + D2 = khong d6ng thcri bang (A^+ B^+ 0) dugrc goi la phuong trinh t6ng quat cua mat phang ( a , ) n ( a ) ? i ^ o A,:B,:C, ^ A2:B2:C2 * Nhan xet B, C, A * Neu mat phang ( a ) c6 phirong trinh t6ng quat laAx + By + Cz + D = ti (a,)songsong{a.^) \ k = (0;0;1) lam vecta phap tuydn Vay phuong trinh mat phang phai tim la: 0(x - 2) + 0(y + 1) + l(z + 1) = « z + = ^1 - -1 = (0; 1; 1) Mat phang (R) di qua A(-l;2;3) nen phuong trinh mat phang (R) la: V i du 2: 0(x + l ) + ( y - ) + ( z - ) = Viet phuong trinh mat phang qua ba diem c:>y + z - = M(3;-1;2);N(4;-1;-1);Q(2;0;2) Vi du 4: Giai: Trong kh6ng gian vdi he true toa dd Oxyz, cho tii dien ABCD vdi Veeto phap tuyen ciia mp (MNQ) la: — -3 (2; -1; 6); B(-3; -1; -4); C(5; -1; 0), va D ( l ; 2; 1) -3 - • n= V 1 Chitng minh rang tam giae ABC vudng Tinh ban kinh va dudng tron = > « = (3;3;1) Phuomg trinh mp (MNQ): 3x + 3y + z + D = Mp di qua Q(2; 0; 2) suy D = -(3.2 + 1.2) = -8 Vay pt mp (MNQ): 3x + 3y + z - = , jndi tie'p tam giae ABC Tinh the tich ciia tii dien ABCD (Trich de thi D H Thuy san, 1999) Giai: V i du 3: Trong khdng gian Oxyz cho di^m A ( - l ; 2; 3) va cac mat phang =>C5.C4 = (-3).8 + 0.0 + 6.4 =0 (P):x-2 = , ( Q ) : y - z - l =0 C4 Cfi nen A ABC vudng tai C Viet phuong trinh mat phang (R) di qua di^m A va vuong goc vdi hai mat phang (?) va (Q) (Trich de thi vao D H LuSt Ha Noi, 1999) Giai: Veeto phap tuye'n ciia (P): Veeto phap tuye'n cua (Q): = (-5; 0; -10); CA = (-3; 0; 6);Cfl = (8; 0; 4) = (1; 0; 0) CA = 7(-3)'+0'+6' =3V5 CB = V ' + ' + ' = 4V5, AB = 5V5 vay SABG = ^ 3V5 4V5 = 30 (dvdt) Goi P la nita chu vi A A B C va r la = (1; 1; -1) ban kinh dudng trdn ndi tiep tam giae A B C en ^ABC - pr=>r - ——— Khoang each + Zo = — ^ — «49(5 + z„)= (6zo-9)^ 6V5 Mat phang (ABC) c6 cap vecta chi phuong: o Zo^ + 46zo + 88 = CA =(-3;0;6) =3(-l;0;2) C5 =(8;0;4) =4(2;0;1) Nen vecta phap tuyen cua mat phang(ABC) la: 2 -1 -1 nABC = 1 2 = (0;5;0) = 5(0;1;0) Vay mp(ABC) c6 phuang trinh y + = Khoang each tir D den mp (ABC) la: +1 d(D, (ABC)) = Vo'+i'+o' Dap so: Vi du 6: z„ = - z„ = - 4 M,(0;0;-2) va M2 (0;0;-44) Tim tap hop nhung di^m M(x,y,z) each mp (P): 4x - 4y - 2z + = m6t khoang bang Giai: = The tfch tir didn DABC la: j3.30 = 30(dvtt) Vr du 5: Tim mot diem \xtn true cao each d6u diem A(l; -2;0) va mp (P): 3x - 2y + 6z - = Giai: Goi M € Oz CO toa dp M(0; 0; z^) Ta c6: MA = (1; -2; -Zo) M de'n mp (P) la: 6z„-9 V3'+(-2)'+6^ 6z„-9 OZ " V a y r = ^ = V5 V=^h.SABc= di^m MA = ^1^ + (-2)'+ (-z^) Khoang each tijr M d6n mp (P) duac tinh theo cong thiic: 4A: - 4;; - 2z + = V4'+(-4)^+(-2)^ ±12 = x - y - z + Vay tap hop cac di^m M g6m hai mat phang: 4x - 4y - 2z - = va 4x - 4y - 2z + 15 = C BAI T A P 78 Viet phuang trinh mat phang di qua diem (3;-2.-7) va song song mp x - z + =0 79 Viet phuang trinh mat phang qua hai diem A(l;-l;-2) va B(3;l;l) va vu6ng goc vdi mp (P): x - 2y + 3z -5 = 80 Viet phuang trinh mat phang qua true tung va di^m A(l;4;-3) 81 Viet phuong trinh mat phang qua hai diem A(7;2;-3), B(5;6;-4) va song song vdi true hoanh + 4t ^ , ITe "Tg" 120 (A) y = - 6t , khu t ta co: z = -l-8t Do chinh la phuang trinh chinh tac cua ducmg thang (A) Dap.anBdung 121 M(3; 6; -7), N(-5; 2; 3), Q la trung di^m cua MN thi toa d6 cua Q = (-l;4;-2) QM = (4;2;-5) Dudng thang QM di qua diem Q va M, nen phuofng trinh ducmg thang QM la: x-3 y-t6 z + x + y-2 z = —f— = va — — = — — = —r•4 f ^ - X = Dap an D dung j 122 Ducmg thang A i di qua diem M(l; 7; 3) va c6 vecta chi phuofng M=(2;l;4) Ducmg thang A' di qua diem M'(6; - ; -2) va c6 vecta chi phuong MM' = (5;-8;-: 5) u = (3;-2; 1) Bieuthiic [u , u'] MM' = (9; 10;-7).(5;-8;-5) = 9.5+10.(-8)+7.5 = => A va A' dong phang, matkhac [ Z , u ' ] = (9;10;-7) Vay A va A' cat DapanAdiing ' 123.a)Tac6: IA + L ? = 2IE C + ID = 2IF T^O b)C6: IA+ IB +IC +ID =2(IE + IF)=0 MA +MB =2 M^ MC+ MD = MF MA+ MB +MC +MD = 2(ME + 124 1) Do G, la tam cua A BCD nen MF)^4.MI (1) G,B +G,C +G,D = 2) Ta da bid't trpng tam ciia tii dien chia m6i trung tuyen ciia tii dien theo ti s6' 3:1 (trung tuy6h ciia tii dien la doan thing n6'i dinh va tam cua mat d6'i dien) nen GA =30^ (2) Tiir(l)tac6: G.G + GB +G,G +GC +G,G +G,D = r = - , vay I = ( - ; - ; - ) 3 3 1 3) Dudng thang vuong goc vdi mp(ABC) c6 vecta chi phuong (1; —; - ) vay ducmg thang di qua I vuong goc vdi mp(ABC) c6 pt: 127 1) Di^m A e O x , B e O y => mp O A B chinh la mat x=- phang Oxy nen c6 pt z = Tucmg tu pt mat phang =0, y Mat 1 z = - + -t 3 =0 phang (ABC) chan tren ba true tea d6 tai ba Tat nhien diem J doi xiing vdi I qua mp(ABC) ciing thuoc ducmg thang diem A, B, C ndn pt long quat cua mp(ABC): + t y ^ l t (OBC), ( O C A ) ^ lucrt la: X , ,,1 Gia su J ( - + 1 1 ; - + - / o ; - + - / „ )• Hinh 79 - + — + - = (pt doan chan) ^ 2) Goi r la ban kinh hinh c^u npi tiep tii dien O A B C K h i r chinh la khoang each tiir I den cac mat phang (OAB), (OBC), ( O A C ) va (ABC) Goi M la trung diem ciia IJ thl M: 1 ,3 1 ° 1 "3 Vay toa d6 ciia I(r; r; r) Khoang each tir O de'n mp(ABC): «;n 151 Dang chinh tac: « = , , 13 1 , -.V,yJ(-;-;-) _ X _ ^ ^ -1 ""iT'"! b) mp Oxy c6 pt: z = c6: cos^ = 128 l ) D s : 15x- l l y - z + = 2) Viet PT mp (P) qua A va vu6ng goc vdri (d) Mp (P) nhan vecto chi phirong u = (3; 4; 1) cua (d) lam vecto phap tuye'n, nen PT mp (P): 3(x - 1) + 4(y - 2) -f- z - 12 = 130 I ) c ( d ) f 2x — v — 11 = ^ ^ y = -ll+2x x-y-z+5=0 Datx = t nen CO PT (d): Toa giao diem B cua (d) va (P) la nghiem ciia he: £=>Lzl.z + 3 3x + 4y + z - = B 33 35 67 26 26 V8987 17 93 AB = (—; )=> AB = —V7' +17' +93' = 26 26 ^26 26 26' 129 a) Pt mp (KHI) di qua ba diem: 2x + y - z - l PT giao tuye'n cua hai mp: X + z= X Dat z = t, ta CO PT tham so: 2) Ta c6: Duofng thang (d) Dudng thing ( A ) Di q u a d i e m M , ( ; - l l ; 16) Di qua diem M2(5; 2; 6) = (1;2;-1) M,M, Bie'u thiic [M, , = -t y = - +—t 2 z= t = (2;1;3) = (5;13;-10) ] M,M^ f PTmpla: = => (d) va A cung thu6c mot mp -1 -1 1 3 2 1J = (7;-5;-3) x - ( y + 1 ) - (z-16) = 7x-5y-3z-7 =0 3) Mp (P) chiia (d) va chie'u (d) theo phuong ( A ) nen no la PT mp of cau2 Vay PT tdng quat ctia hinh chieu la: r7x-5y-3z-7 152 ^ Mat phang c6 vecto phap tuye'n: =0 2x + 2y - 9z - = ^ - * u = ( l , 2,-1) [y = - l l + 2t = fl4x - lOy - 6z - 14 = 3x-2y-2z-l = j+Sx - lOy - lOz - = 4) X - 4z +9 = => X = -9 + 4z Duomg thing A phai tim song song v6i (?) nen song song v6i (d,), nghia x = - + 4t la CO vecto chi phuong u^ PT ciia ( A ) la: y = -14 + 5t Dat z = t ta CO PT tham s6': x = + 43t z=t PT chinh tac: x+9 y + l4 y = - + 30t z z = - + 23t 132 Goi hai mp da cho theo thii tu I la (P) va (Q) V i (P), (Q) cung • vu6ng g6c vdi mp (R) ntn giao : tuye'n ( A ) ciia chung phai vu6ng goc vdi (R) Do do, 131 Trudc het, viet PT mp (Q) chijra ducmg thSng (d) da cho va di qua A Dudng thang (d) c6: X = PT: I + 3t ^ y = - + 2t z = + 2t Hinh 104 No di qua B(2; -4; 1) va vecW chi phuong u = (3; 2; 2) va AB = (nen c6 PT: ( x - ) ^ 2 _^ + ( y + 4) _^ c -1 3 ^ + (z - 1) ^ ^ = 1 -1 -12 3 133 ) ^ = (2;0;2) = 2(1;0;1) Dudng thang AB c6 vecto chi phuong M = (1; 0; 1) di qua diem A(0; 0; - ) , nfen AB c6 phuong trinh: 12x - 8y - 12z - 28 = [x=0 + t 248 43 -30x + 43y + 248 = => x = — + — v 30 30- vay (d,) CO vector chi phucmg w, = (43; 30; 23) 2x + 3y + z = G '-42x + Sly + 12z + 276 = Dat y = t, ta CO PT (d,): -12 No cung la vecto phap tuyen ciia mp (R) nen PT mp (R) la: f-14x + 17y + 4z + 92 = Dudng thang (d,) c6 PT: BC thi mat CDD'C la mat ben Idn Theo tmh chat hinh hop thi BC mp CDD'C nen CD' la hinh chieu cua BD' tren mp Vay goc tao bai BD' va mat ben 1dm Hinh 106 CDD'C la goc BD'C = /? Tam giac vu6ng BCD' cho: BC = BD' sin /? = d sin p, CD' = BD'cosy5 =dcosy9 |Tam giac vuong BDD' cho: D'D = BD'sin« = d s i n a Tam giac vu6ng CDD' cho: CD = Vc'i)^ -DD^ De tim SH, ta dira vao tarn giac vuong SHK va phai ti'nh H K De tim H K , ta phai dua vao cac tarn giac vuong K H A va C H I : = d.^cos^ p - sin^ a V = AB.BC D D ' = d^sina sin p Vcos^y^-sin'a (*) HA = Bie'n doi: l + cos2y9 cos p - s m a = l-cos2a _ HC = Thay vao (*) ducfc dpcm » a = HK 135 Cho A A B C c B = l v v a A = a , A C = a.Theod£ubM:(SAQ ( A B Q chop thi SH e(SAC) SI BC thi H I la hinh chie'u HI cos a cos a + sin a a sin 2a sin a cos a 2V2sin(45°+a)' Tarn giac vuong SHK cho :SH = H K t g Ke cua SI tren day ma SI Thay vao (*) duac dpcm Chii y: V i d^u bai kh6ng noi ro BC Chan ducmg cao thu6c mien day nen H I BC (theo dinh l i ba ducmg vu6ng goc) VSy sin a BC = H K ( ^ ^ + - ^ • ) v i ( H K = H I ) sma cos a = -(cos2/? + c o s a ) = c o s ( a + > ) c o s ( > - a ) nen ke ducmg cao SH ciia hinh HK Hinh chop hay khong, nen H c6 SIH the la giao diem cua A C keo dai la goc phang ciia nhi didn BC, theo d^u bai SIH = /? Xac va tia phan giac ngoai cua goc Hinh 107 B (hinh: 108) NeuAB>BC dinh tuong t u ta co S K H = /3 De tha'y A SHI = A SHK => H I = H K , nghia la H each diu hai canh goc vuong => H thuoc dudng phan giac cua B : HA = HC = H B I = H B K = 45? => B I H K la hinh vu6ng C V = ^dt(AABC).SH Dt A A B C = - A B A C s i n a sin a HI cos a HK = A B = AC.cos a = acos a = - a ^ sin a cos a = -a^sin2a va V = < 45") thi: HK a = HK(—!^ sin a De tim dt ( A ABC) ta t i m A B tuf tarn giac vu6ng ABC: (a ~ ) cos or a sin 2a.tgP 2V2sin(45°-a) a\s,\n^ laigp 24.V2sin(45°-a) = ^^^"^"'^^^ 2V2sin(45°+a) - Neu AB < AC ( « >45") (hinh 109) thi: 12 Khi th^ tich lom nha't la • - HI HC = cos a HK HA = sin a a = HK ( V= •'' 1 a.smla.tgp >HK = ^ a' • 12 ~ 48 A B / / C D => AB//(SCD)ma 2V2sin(a-45'') (P) chiia AB, M N = (P) n (SCD) =>AB// M N , de dang chiing minh A N = BM, ndn ABMN la hinh thang can 24.V2sin(a-45°) 136 1) Ta CO aS 137.1) Goi O la tam ciia day hinh chop, P, Q theo thii tu la trung diem canh day AB, CD, R = M N n SQ •) sin a cos a 'y BH ± AC (vi H la true tarn cua A ABC d^u) /' ' \\ /'' ' \\ A/M[ \ Ta CO SP AB, OP AB nfin BH M A (vi A M (ABC)) Suy BH (MAC) => MC MC , * MC 1 OS? la goc phang cua nhi ' dien tao bod mat ben va day BH BK (K la true tarn A BMC), suy MC ± (BHK) (BHK) => HK MC cung chiing minh tucfng tu ta c6: HK >HK (BMC) a Dodo M N = ^;PR = MB 2) Goi D la trung diem ciia BC thi K e M D Da dang chiing minh BC (MAD) Trong mp (MAD) ke LO AD (O e AD), de dang chiing minh dugfc: ^KABC = ~ => OPS = 60°, de dang chiing minh tam giac SPQ la tam giac d^u Ta CO PR la ducmg cao, M N la dudng trung binh cua A SCD H i n h 1 vay dien tich thie't dien ABMN, la: a _ AB + MN '^ABMN ^ t'K - a + 2- aS 2) Theo c6ng thiic tmh M tich khoi chop: ABC- OK Vay VKABC lo'n nha't va chi OK Idn nha't Do A H K D vudng K (HK (BMC) => K nam tr6n du5ng tron ducmg kfnh HD ndn O K \6n nha't ^ V=i.Bh B = dt hinh vu6ng ABCD Hinh no B = a^ H = SO cung la chieu cao cua tam giac d^u SPQ nen: _^S.a^ ~ Dieu kien du: Co he thiic be + 2a' = a( b + c) thi A ABC vuong a A h= so = aS That vay, ta c6: AB' + AC' = 2a' + b' + c' - a(b +e) = 2a' + b' + e' - (be + 2a') vay: = b' + c'-tK: = B C Theo dinh l i dao Pitago thi A ABC vuong A VAy3CD= ^ - Ta lai c6: ~ V SCO = > _ 1 SD ' _ i V A S B M = — Vs.ABCD' V A S M N = ' S.ABCD 3) Ke ducmg cao A H Ke A I ± OB thi H I la hinh chie'u eiia A I tren day ntn H I ± OB theo dinh l i ba du5ng vu6ng goe Tuorng tu: ke A K I O C eung CO HK ± OC Cac tarn giac vu6ng OAI va KAO bang (OA Chung, mot goe 60") =>AI = A K => H I = HK nghia la H each deu cac canh A c Hinh 112 ciia goe yOz = 60" n6n OH la tia 138 1) Da tha'y OABC la tii dien deu vi cac tarn giac OAB, OBC, OCA la nhirng tarn giac deu bang Khoang each tCr A den mp Oyz la ducmg phan giac goe HOI = HOK = 30°, OA = a => = - a va A I = cao tii dien, va tinh dugc bang V = - d t (AOBC).AH = tarn giac vu6ng HOI cho HI = O I t g " = A ^ = 12 2) Ap dung dinh if ham so cosin vao tarn giac OAB, OBC, OCA ta c6: ^ Ap dung dinh l i Pitago vao tarn giac vu6ng H A I cho: A B ' = OA^ + OB' - 2.OA.OBcos60" = a' + b' - ab BC' = OB' + OC' - OB.OC.COS60" = b' + c' - be AH' = A I ' - H I ' = 2.a' C A ' = OC + O A ' - OC.OA.COS60" = c' + a' - ac Dieu kien can: Cho BAC = Iv thi be + 2a' = a(b+c) That vay, ap dung dinh If Pitago vao tarn giac vuong ABC, c6: BC' = A B ' + AC' « b' + c' - be = 2a' + b ' + c' - a(b + c) =;>dpem V = dt (A BOC).AH =1.1 Theo cau 2, ta c6 be + a' = a(b +c) QK OC.sin60".AH = b,c la nghiem cua phuong trinh: - dx + ad - 2a'= (*) Co A = d' - 4(ad - 2a') = (d - 2a)' + 4a' >0 Nen phuong trinh (*) chic chin c6 hai nghiem khac Vi b, c > nen a(d - 2a) > => d > 2a Day la di^u kien d^ tinh duoc b, c Goi f(x) = x' - dx + ad - 2a' Co f(a) = - a' < 0, f(2a) = a (2a - d) < (do d > 2a ) z:>a va 2a d khoang hai nghidm ciia (*) => dpcm 139 a) Ke OK d thi OK = a va SK i d theo dinh If ba dudng vuong goc => d mp(KOS) ii>mp(S,d) ± mp (KOS) Tam giac vuong KAO cho: OA = KO sina a 3sina sma Tam giac vuong SAO cho: SA' = SO' + OA' o 25a' 64a' Ila tam mat cau ngoai tiep tir dien SOAB 10 cat SM tai G Ta phai chiing minh G la tam A SAB That vay Do A //SO (vi chiing cung vuong goc v6i (P) nen theo Talet, ta c6: ^ = ^SO = l2 ^ G S = 2GM(dpcm) ' y , 140 Bai toan khong sai neu ta cho hinh cau va hinh non long vao cho chan ducmg cao hinh non triing \6'\p diem cua hinh cau va (P), liJC tam hinh cau thuoc ducmg cao hinh non Hinh 114 la thiet dien qua true hinh non, cat hinh non theo mot thiet dien la tam giac can, cat hinh cau theo mot thiet dien la hinh tron Idn (qua tam hinh cau), c6 ban kinh bang R Do (Q) // (P) => A'l //AB nen theo Talet ta c6: SI DC SH HA R (h -x) Tam giac vuong HA'J cho: h - x IKi IK = — h h R A'l' = IJ.IH = (2R -x)x S = ;r (IK' + lA") = n: R' (h-xf +(2R-x)x (1) Khi h < X < 2R (hinh 114), thurc hien phep tinh nhu tren, ta van dugrc bieuthirc(l)vi (x - h)^ = (h - x ) ' b Biend6i.(l): ,R S= n: i^-\)x^ voi < X < < X +2R0 R O a) )x + R Sn,in X 2R Do thi la parabol c6 dinh cue tieu I o Rh 141 a V s A B c - IsA.dt (AABC) ma SA = a, A ABC la tarn giac deu canh a n^n 2:i.R'.h R +h' R +h R > h thi parabol c6 be 16m quay Ifin tren nSn c6 la tung dp dinh: 2nR'h S^i^ = R +h dt(AABC)=^ (hinh 116) Vay V s A B c = - « - - — R - Neu —- - < R < h thi Parabol c6 be 16m quay xu6ng dudi ntn Hinh 118 ^V3 12 CO cue dai S^ax la tung dinh: Sn^^^ = R +h (hinh 117) Goi I la trung diem eiia BC - Nfi^u R = h thi S = ;r R^, thi la doan thing song song \6i true hoanh (hinh 117c) A SBC can dinh S (SB = SC) ^ SI I B C , nen dt (A SBC) = - a SI ma SI = -JSA'+Al' ^dt(ASBC)=ia.^ = 2 V Neu • T I I I O D ^ V s A B c = - h d t ( A S B C ) = Ih ^ H ^ ^ goi h la khoang each tCr A den = mat phang (SBC) thi = 12 h = Hinh 116 167 b) Ghon he toa d6 nhu hinh ve 118 ta c6 A(0; 0; 0) la g6'c toa d6, S(0; 0; a), C(0; a; 0) Vi SIC = SAC = 90" suy tii dien S AIC noi tie'p mat c§LU c6 ducmg kinh la SC = Neu goi K la trung di^m cua AC, thi K = K R= 2'2) la tam mat cdu 142 a) AB = (2; 3; 4) Ducmg thang d di qua A va B nhan vecta AB lam vecto chi phuofng va di qua A(0; -2; 0) ndn c6 phuong trinh tham s6' la: = 2t y = - + 3t z = 4t b)M ed => M (2t;-2 + 3t; 4t) 2t-2 + 3t-4t + d(M,(a))= — X t+3 V5 d(M,(«)) = 2.V3 o - ^ = 2.V3 t+3 = V29 Ta c6: d(I, ( a ) ) = V3 7V3 V29 (S) CO tam la trung diem I ciia AB „ ——- < —— = R nen mat phang ( a ) cat mat c^u (S) 143 a) OA = (4;0;0) => OA =4 Tuong tu ta tinh duoc OA = OC = AB = BC = => ABCO la hinh thoi Co OC = (0;4;0) =^OA.OC = ^ OA IOC => ABCO la hinh vu6ng Mat khac ta lai c6 SO = SC = SA = SB = VlT => SABCO la hinh chop d^u b) Dien tich hinh vu6ng ABCO la : OA^ = 16 Mat phang (ABCO) c6 hai vecta chi phuofng OA = (4; 0; 0), OC = (0; 4; 0) nen no chinh la mat phang Oxy, nen phuong trinh mat phang (ABCO) la: z = 0; d(S, (ABCO)) = I Vay VsABco=^-6.16 = 32(dvtt) t=3 t = -9 Vdi t = ^ M, (6; 7; 12) V6i t = -9 M2 (-18; -29; -36) c) Mat ciu (S) CO ducmg kinh AB va CO ban kinh R = AB Mat c^u (S) CO phuofng trinh: (x -1)^ + (y + )2 + (z _2)2 = ^9 ngoai tiep tii dien SAIC Vay phuong trinh mat c^u ngoai tiep tii dien SAIC la: x^ + ( y - f ) ^ + ( z - 2^ ) ^ = ^ 2) 1(1; c) Goi M la trung diem ctia SO => M = (l; 1;3) Mat phang (P) la mat phang trung true ciia SO di qua M c6 vecta phap tuyeh Ik OM = (1; 1; 3) Vay phuong trinh ciia (P) la: ( x - l ) + ( y - l ) + 3(z-3) = / iW i\\ /' // /' \ //'' \\ / > \ ^.1 \ \ (0:4.i)) U ' ' ' / -'X B (4:4:(n Hinh 119 ^ =>x + y + z - l l = T De dang c6 toa 1(2; 2; 0) 7S = (0; 0; 6) Vay phuong trinh tham so cua ducmg thang X=2 IS: ] y = z = 6t Tarn I cua mat c^u la giao diem cua mat phang (P) va ducmg thing IS, thay gia tri x = 2; y = 2; z = 6t vao phuomg trinh mat phang (P) ta c6: + + t - l l = = > t = — J ( ; 2; - ) 18 ,, ,^ =i>R = OJ = 121 _ n ~ Do phuong trinh mat ciu ngoai tiep hinh chop S OABC la: AH ±b ^ (^t - y ) + l(t + 16) + ( t - ) = - r = - y «t = - s u y r a H ( - ; - l l ; - ) Dia'm A' la dilm doi xirng ciia diem A qua A2 nen H la trung diem ciia AA' suy AA' = 2AH mkAH= (-10; -13; -8) ^ A'= (-17;-24;-11) b) Goi (Q) la mat phing di qua A, va song song A2 phuong trinh mp (Q) CO dang: a ( x - y + 3z-5)+ /?(x + y - z ) = vdia^+y9'^0 /? = Vay phuong trinh mat phAng (Q): 4x + 3y + z-5 = c) Ta CO d (A,, A2) = d( A2 (Q)) = d(B.(Q)) Goi H la hinh chieu vuong goc ciia A len Aj 4.(-y) + 3(-14) + 1.0-5 H(-Y + ^ t ; - - t ; t ) Taco AH = (1- t - —17; - t - ; t - ) 2 ncnng lb rigL = ^(2a + J3)-l(-a+2/3)+1(3a-J3) (x-2)^ + (y-2)^ + ( z - ^ ) ^ = 1211 ^ 11 27 145 a) A,: 69 ^/4'+3'+l' x+y+z-4=0 Dat z = t ta c6: 2x - y + 5x - = X = A,: - 2t = (3; 5: 6) y =2+t Tim mot diem M e A z =0+t M (1; - ; 1) Vay phuofng trinh chinh tac ciia ducmg thang A la: => A, di qua M , (2; 2; 0) va c6 vectachi phucmg u^ = (-2; 1; 1) A2 di qua diem M2 (1; 0; 2) va c6 vecto chi phuong x-l _y + = (-2;3;1) l^2-1 ~ c) Goi H la hinh chieu vuong goc cua A tren mat phang (P), H la [u[, ].M^M-^ =6^0 phuong trinh d u ^ g thing fx - 9y + 5z + = hirffng I K H O l D A D I E N V A THE TfCH CLIA C H U N G [x - 2y - 5z + = hirang I I M A T N O N , M A T T R U , M A T C A U A , : •< M6t vector chi phuomg ciia A la M J phuofng v6i M , va = (55; 10; 7), M J kh6ng ciing => A cit A , va A J 148 a Ta da bie't neu hai mat phing vu6ng goc \6i thi hai vecta phap cua hai mat phang vu6ng goc vdi va ngugc lai ^ = (1;6;2) J^' ^ = 2.1 - 1.6 + 2.2 = o ^ o ( a ) (/?) b) Mat phang (P) chiia giao tuyen ciia hai mat phang {a)\kifi) phuong trinh dang: (P): A ( x - y + z - l ) + ;/(x + y + 2z + 5) = o ( A + / / ) x + (->l+6//)y + (2/l+2//)z + / / - A = Mat phang (P) di qua gd'c toa d6 O nfin D = => // - A = cho A = => fi =>{V): l l x + y + 12z = =1 ntn c6 19 I Mat non, hinh non, khdi n6n 19 I I Mat tru, hinh tru, khoi tru 23 I I I Mat c^u 27 On tap chuong I I 32 hUtfng I I I P H U O N G P H A P T O A D O T R O N G K H O N G Tac6;C=(2;-l;2) GIAN 32 I He toa d6 khdng gian 35 I I Mat phang 48 I I I Dudng thang 58 On tap chirang I I I 70 NTAPCUOINAM 74 [...]... phuong ciia (D2): =-5 XB B(-5; / 1;0) U2 z,=0 = \ Vi du 11 -1 -1 2 -1 M,M^ 3 \ + 2z -2 3 2 [M1,W2]= =0 1 1 2 2 2 - J = 5(1;-1;-1) =(l;-3;-3) Viet phucfng trinh hinh chieu ciia ducmg thang: J5x-4y-2z-5 5 3 -1 3 1 I -1 -1 1 2 II = (l;4;-3) -1 =0 [M,,M2].M,M2 = 1.1 +4.(-3) + (-3).(-3) = - .2 trSn m p 2 x - y + z - l = 0 U„U2 Giai: Ducmg thang da cho thupc chiim mp: m(5x - 4y - 2z - 5) + n(x + 2z - 2) = 0 «(5m... - 4 z + 12 = 0 nhan di qua diem (-5; 7; 0) va vecto chi phuong z =t x = l + 3t x = l + 3t y =3+t y =3+ t z =-4-2t • z = -4-2t 3x + y - 2 z = 0 3(l + 3t) + 3 + t - 2 ( - 4 - 2 t ) = 0 x = -2 u = (2; - 2 ; 1) Viet PT mp (a) qua P(4;l;6) va vu6ng goc v6i (d) nSn y =2 nhan u ciia (d) lam vecta phap tuydn z = -2 V a y P T m p ( a ) l a : 2 ( x - 4 ) - 2 ( y - l ) + z - 6 = 0 o 2 x - 2 y + z - 1 2 = 0 Tim... = T2 AA, AH i 1- cos a 2 oc cos — AK • ^ a sin« 2 a cos — 2 \o Jcos^ — 11 Ke A ' H 1 (ABCD), PiK l ' A B ( K e AB) 2 ^ M p ~ ^MEHK • A A i cos 2 a fSi 2 cos a '^ 1 • cos^a = 2a si a 2 \ cos 2 a H e (ABCD) va ke H M 1 _ AD (M e A ' M H = 60°, A ' K H = 45° A'K = xa, 8^. 82 S2 — — x y s i n a a = ya 1 S, 2 a a sin a a sin a 2a cos 2 a Theo dinh ly 3 ducmg vuong goc ta c6 /s.D 1 A ' M , A B 1 82 xy.sin«... 9 = ( 32; - 12; -16) = 4(8;-3;-4) M p (P) C O cap vecto chi phuong u, va u^ nen c6 vecto phap tuyen: Up = [u, ]= v -4 3 3 2 2 -4 -3 -4 2 8 8 -3 = (25 ; 32; 26 ) 15 0 '^^'^^ S^'^ m(x - 3y + 7z + 36) + n(2x + y - z -15) = 0 ( m H n^ ^ 0) (m + 2n)x + (-3m + n)y + (7m - n)z + 36m - 15n = 0 Theo eong thiic khoang each, ta c6: 36m-\5n ^|(m + InY 25 x + 32y + 26 z + 55 = 0 M p (Q) chiia (d^): x-3 y+\ -2 3 vecto... D X 120 Cho phuong trinh tham so ciia ducmg thang (A) CAU HOI T R A C NGHlfiM phuong trinh chinh tSc cua dudng thang (A) la: 113 Trong khong gian Oxyz, toa do cua vecto a = 3 i + 2 k la A.a= (3; 0; 2) B a = (3; 2; 0) C a = ( 3 ; l ; 2 ) D a = ( l ; 3 ; 2 ) A x -2 _y z + \ B x -2_ y _ z+1 ~4~~ f 6 I 114 Cho vecto a {2; 3; -1), b(0; 1; 4), c ( l ; 0; -3) Xac dinh toa d6 cua vecto 2 a -b - 2 c A. (2; 5;l),... toa d6 cua vecto 2 a -b - 2 c A. (2; 5;l), B (2; 5; 0), C.(l;4;l), x +2 C 4 A 115 Cho 4 diem M ( - l ; 5; -10), N(5; -7; 8), P (2; 2; -7), Q(8; -10; 11) A Hinh binh hanh C i IC B Hinh vuong ffinhthoi -2 _z + l ~~5~ B x-3_y-6_z+l -2 ~ -4 ~ 5 y-6 2 z +1 5 D x-3 4 x-Z_y-6 x-3 4 + + 4x - 2y - 20 = 0 A.(1; -2; 0);R = 5, B ( -2; 0; 1); R = 5, C.( -2; 1;0);R = 5, D ( -2; 1; 0); R = 4 1 A Cat nhau; > 117 Viet phuong... ciia (d) vao PT ( a ): 2( -5 + 2t) - 2( 7 - 2t) + 1 - 12 = 0 t = 4 Thay lai vao PT tham s6' ciia (d) dugc toa d6 A(3; - 1 ; 4) Theo tinh chat doi xiing thi A la trung die'm PQ, de dang c6: t = -\ I( -2; 2; -2) Theo tinh chat doi xiing thi I la trung di^m AB nfin de dang tinh dugc toa do diem I -2 =|(1+^.) 3 = ^(4 + ^ , ) y, =\^z,) 2 = ^ ( 3 + ;^,) - 2 = l ( - 4 + z,) ^A = T ( 2 / ' + Z O ) 4 = ^(6... 2) = 0 «(5m + n)x - 4my + 2( -m + n)z - 5m - 2n = 0 (P) = V l ' + 4 ' + ( - 3 ) ' =:,V26 f A p dung c6ng thiic tinh khoang each ciia hai dufimg thang cheo n hau vao bJli toan ta c6: V i P 1 Q (mp Q: 2x - y + z - 1 = 0) ndn phai c6: Hp ± Hp HQ [u,, d(D,A) = =0 HQ U2 M,M2 2 ^1 /26 _U,,U2_ 2( 5m + n) + 4m + 2( -m + n) = 0 ce> 3m = - n T Chon m = - 1 , n = 3, thay vao (P) duoc mp: -2x + 4y + 8z - 1 = 0 BAITAP... +2= 0 x+l =0 3 '~ -2 _z -2 ~ 2 ^ • ^ (Dai hoc Bach khoa Ha Noi, nam 1997) a) Chiing minh rangi ducmg thang (d) va dudng thang A B cung nam trong mot mp | b) Tim diem I e (d) sao cho A I + BI nho nha't 1 02 Cho mat phang (P): 2x + y + z - 1 = 0, va du6ng thang (d): x + l _y -2 _ z -2 thang (d) CO PT: 3 — 2 2 Goi N la diem doi xiing cua M qua ducmg thang (d) Tinh do dai cioan thing M N 2) Cho A ( l ; 2; ... de thi vao dai hoc khoi A - 20 04) 111 Trong khong gian Oxyz cho 2 du5ng t h i n g : d, ' -2 + y^ + z + 2 -1 1 'x = -l + 2t vad.,: \ = \ t z = 3 1) Chung minh d , va 02 cheo nhau 2) Viet phucmg trinh dudng thang d vu6ng goc vdi mat phang (P): 7x + y - 4z = 0 va cat hai ducmg thang d , va d2 (Trich de thi vao dai hoc khoi A - 20 07) C 3x - 3y + 2z + 1 = 0 + 3y + 2z + 1 = 0 2 + 4; y = -6/ the tlii z =