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6 Cho hinh lang try d u n g ABC.A' B ' C c6 day ABC la tam giac can AB = A C = a va •6At: = 120° Biet goc tao boi A ' B va mat p h i n g ( B C C ' B ' ) bSng ^ cho sin(t) = — 2v3 Gpi M la trung diem ciia A A ' Tinh the tich khoi chop B ' M C C va khoang each t u trung diem N ciia BB' den mat phang ( B ' M C ) PHlTOfNG PHAP TOA DO TRONG MAT PHANG Phumtgphdp toa dp mat phang: - Xdc dinh toa ciia diem, vecta - Duattg tron, ba ducrtig conic - Viet phuang trinh duang thang - Tinh goc; tinh khoang each tkdiem den duang thang * Tat ca cac vi du va bai tap cua phan deu xet mat phang toa Oxy BAI T O A N X U N G Q U A N H T A M G I A C V"e vecta Nam dugc khai niem hai vecto bang nhau: Toa tuong l i n g bang Cach viet toa ciia mot vecto: Toa dp ngpn trir cho tpa dp goc Hai vecto song song nhau: ti so hai tpa dp bang Ve dien tich Thuong dung: S = —ah^ = —bh = —ch^ hoac S = —ab.sinC = — bc.sin A = —ca.sinB T i n h chat tam Trong tam ciia tam giac chia cac d u o n g trung tuyen ba phan, each dinh hai phan va each day mot phan Cac khai niem True tam - Giao ba d u o n g cao Trpng tam - Giao ba duong trung tuyen Tam duong tron ngoai tiep - Giao ba duong trung true Tam duong tron npi tiep - Giao ba duong phan giac T i n h chat duang phan giac Chia canh d o i dien hai doan thang ti le voi hai canh ke hai doan t h i n g ay Cong thijfc tinh khoang each tir mot diem M ( x Q ; y p ) toi duang thang d:ax + by + e = = > d ( M ; d ) = axg+byo+c 186 Cong thuc t i n h cosin ciia goc giua hai d u o n g thang d j , d t h e o cac V T P T : 8^82 + bjb2 " i = ( a i ' t ' i ) ; " = ( a ; b ) : ^ c o s ( d i , d ) = cos(nj;n2 BAI T O A N X U N G Q U A N H H I N H CHLf N H A T Hai canh ke vuong goc v6i Hai dirong cheo cat tai trung diem moi duong Hai each doi dien bang Theo tinh chat hai d u o n g thang song song bi cat boi mot cat tuyen t h i - Cac goc so le bang nhau; - Cac goc cung phia bang BAI T O A N X U N G Q U A N H H I N H V U O N G Xet hinh vuong ABGD c6 tarn I Cac canh doi mot vuong goc va b3ng Hai duong cheo bang va vuong goc voi Bon tarn giac vuong can: AIB, BIC,CID va A I D bang Neu canh hinh vuong bang a thi hai d u o n g cheo c6 dai la aV2 Hinh vuong noi tiep d u o n g tron c6 tarn I ban kinh R = BAI T O A N X U N G Q U A N H H I N H T H O I Co hai cap canh doi song song va bang Tat ca cac canh bSng Hai duong cheo vuong goc tai trung diem moi duong Hai duong cheo la hai true doi xung cua hinh thoi Moi duong cheo chia hinh thoi cac tarn giac can bang Dien tich hinh thoi bSng niia tich hai d u o n g cheo BAI T O A N X U N G Q U A N H DU'ONG T R O N Duong tron c6 tam I(a;b) c6 ban kinh R c6 phuong trinh: (x - a)^ + ( y - b)^ = R^ Phucng trinh: + y^ + 2ax + 2by + d = la phuong trinh cua duong tron tam I( - a; - b) ban kinh la R = Va^ + b^ - d voi dieu kien a"^ + b^ - d > Vi t r i tuong doi giua mot diem M(xQ;yQ) va d u o n g tron (C): Neu: PM/(C) = ('^O ~ + (YO ~ ' ' ) ^ ~ > 0, thi M nam ngoai d u o n g tron (C) 187 Neu: PM/(C) ={'^0 '^{YO "^f Neu: P[^y((3j = ( X Q ~a) + ( y g - b ) - R = 0, thi M nSm tren duong tron (C) < 0, thi M nam ben duong tron (C) Vj tri tuong doi giiia duong thang d : Ax + By + C = va (C) c6 tarn I(a;b) Neu: d(l;(C)) = Neu: d(l;(C)) = aA + bB + C aA + bB + C > R, thi d khong cat (C) = R , thi d tiep xuc voi (C), d gpi la tiep tuyen ciia N/A^+B^ duong tron (C) Neu: d(l;(C)) = aA + bB + C < R , thi d cat (C) tai hai diem phan biet M, N Vi tri tuong doi giiia hai duong tron: ( q ) : ( x - a , f + ( y - b , f =R? va{C,):{x-a,)\{Y-b,f=Rl Hai duong tron khong c6 diem chung + R < Ijl2 , => (Cj) ngoai ( C j ) Ngugc lai R - R < ^I^T = ^ (^2)'^Vng (Cj) hoac (Cj) d u n g ( C ) Hai duong tron c6 mot diem chung: - Neu Rj + Rj = l]l2 ^ ( ^ ) *^'^P - Neu R2 - R = I1I2 ngo^i voi ( C ) (C^) tiep xiic voi ( C j ) Hai duong tron c6 hai diem chung R^ + Rj > I1I2 =^ ( ^ i ) (^2) diem phan biet M, N va duong thSng di qua hai diem M, N goi la true d§ng phuong cua-hai duong tron Tinh chat cua true dang phuong: Moi diem nSm tren c6 phuong tich doi voi hai duong tron bang Tij' mot diem M bat ki, ke qua M duong thMng d cat (C) tai hai diem phan biet, goi d la khoang each tu M den I, ta eo: MA.MB = d^ - R^ la hang so (khong phu thupe vao vi tri cua diem M) Dac biet, neu MT la mot tiep tuyen cua (C) thi MA.MB = MT^ Duong thSng qua tam duong tron va diem giu-a cua mot day cung thi duong th3ng vuong goc voi day cung Ngugc lai, mot duong thang qua tam duong tron va vuong goc voi mot day cung thi duong thang di qua diem giiia cua day cung Tu mot diem M ke hai tiep tuyen toi (C) Gpi I la tam duong tron va A , B la hai tiep diem ciia hai tiep tuyen thi ta c6 mot so ket qua sau: - M A = MB va lA M A ; IB MB - Tii giac MAIB npi tiep mot duong tron c6 duong kinh la MI - MI la phan giac cua hai goc ^ K T A ^ va X l ^ 188 Tiep tuyen cua d u a n g tron (C) tai diem MQ(xQ;yQ)c6 phuang trinh la: (x-a)(xo-a) + ( y - b ) ( y o - b ) = R ^ 10 Duong thang d : Ax + By + C = la tiep tuyen cua (C) khi: (aA + b s f = R ^ ( A ^ + B ^ ) o d ( l ; ( C ) ) = R 11 Trong mot duong tron, neu cung AB Ion hon cung CD thi cang gan tam thi goc d tam chan cung cang Ion) 12 Cach xac djnh tam vi tu va tam vi tu ngoai (day cung Gia su tiep tuyen chung ngoai cat duong thang noi hai tam I^l2 tai mot diem M, do: - Neu diem M nam ngoai doan I^Ij thi M la tam vj tu ngoai; - Neu diem M nam doan I^Ij thi M la tam vi tu 13 Neu OM = kON thi ta c6 phep vi tu tam O, ti so vj tu la k da bien diem N diem M Vi vay, neu N chay tren mot duong (G) thi M chay tren duong (G') la anh cua duong (G) qua phep vi tu tam O c6 ti so vi tu la k BAI TO AN XUNG QUANH ELIP Elip (E) la tap hop cac diem M cho MFj H-MFj = 2a (voi F^Fjla hai diem co djnh) Neu FjFj =2c thi Fj;F2 la hai tieu diem cua (E) va 2c la tieu cu ctia (E) va 2 phuong trinh chinh tSc cua (E) la: ^ + ^ = (a > b) (b^ - - c ^ ] a b ^ ' Elip (E) noi tiep mot hinh chu' nhat co so c6 ki'ch thudc la 2a va 2b Tieu diem cua elip (E) bao gio cung n^m tren true Ion: - Neu 2a > 2b thi tieu diem nam tren true hoanh; - Neu 2a < 2b thi tieu diem nam tren true tung Phuong trinh chinh tac ciia elip (E) la chin doi voi x va y cho nen (E) co thi nhan hai true toa dp lam true doi xiing va goc toa dp la tam doi xung Ti so: - = e goi la tam sai ciia (E) (e < 1) Duong thang x = — va x = - - gpi la hai a e e duong chuan cua (E) cho nen voi mot diem M bat ki tren (E) thi ta luon co MH = MF (trong do, H la hinh chieu cua M tren duong chuan tuong ung, F la tieu diem ung voi duong chuan tuong ung) c c Voi M bat ki tren (E) ta co: MF^ = a + - x va MFj = a — x (gpi la ban kinh qua tieu) 189 BAI TO AN XUNG QUANH Dl/ONG THANG Cac vi duj V i d u 1 Cho diem C ( ; - ) va d u o n g tiiang A : x ~ y + = T i m tren duong thang Ahai diem A va B doi xung qua I 2;—J cho dien tich tam giac ABC bang 15 Cho tam giac ABC c6 BC = 2AB Phuong trinh duong trung tuyen xuat phat t u dinh B la x + y - = Biet XSt: = 120° va A ( ; ) Tim toa cac dinh cua tam giac ABC Lcri gidi Thay toa I vao p h u o n g trinh A ta dugc 3.2 - 4.^ + = (luon diing) nen I G A Vi A e A nen gia su A (4a; 3a +1) ma B d o i x i i n g v o i A qua I nen I la trung diem A B B(4-4a;4-3a) T u C d u n g C H A B tai H thi C H = d ( C , A B ) = 3.2-4(-5) + / = Theo gia thiet S^,^ = 15 o ^ C H A B = 15 o ^ ^ ( - a ) ' + ( - a ) ' = 15 A/25 ( - a ) ' = < » a - l = l o 'a=l=> a=0^ A(4;4),B(0;1) A(0;1),B(4;4) Vay hai diem can t i m la (4; 4) va (0; 1) Theo d i n h l i cosin A A B C , ta c6: A C ' = A B ' + B C ' - A B B C c o s : ^ = A B ' ,2 AB'+BC' Theo cong thiic d u o n g t r u n g tuyen, ta c6 B M ' = • T u day ta tinh duoc cos^I^BK^ = "^^ + B M AM AC' 3AB' ^^ 2AB.BM Suy AB B M Do p h u o n g trinh d u o n g th3ng AB d i qua A va vuong goc voi B M la ( x - ) - ( y - l ) = o x - y - = Toa diem B la nghiem cua he x+y-2 = B(2;0) x-y-2=0 Goi C ( a ; b ) v o i a;^2;3 va b ^ O ; l , A C = ( a - ; b - l ) ; BC = ( a - ; b ) ; AB = ( - l ; - l ) V i A C = V A B va BC = A B nen ta c6 he 2b'+4b-4 = (a-2)'+b'=8 a = -b ( a - ) ' + ( b - l ' ) = 14 190 T u d o s u y r a C = (a;b) = ( l - > / ; - l + Vs) hoac C = (^1+S;-l-S) V i du Cho tam giac ABC c6 B(1;2), dudng phan giac A K c6 phuong trinh: 2x + y - l = va khoang each t u C den duong thang A K bang Ian khoang each t u B den duong thSng A K Tim tea cac dinh A va C biet C thuoc tmc tung Z Cho hinh thang A B C D vuong tai A va D c6 phuong trinh duong thing A D : 2x + y + = M(2;5) la trung diem BC va DC = BcV2 = 2AB T i m toa cae d i n h h i n h thang biet A CO tung duong L&i gidi l G g i C(0;c); d(B; A K ) = ^ ; d ( C , A K ) = c-1 V5 B va C nam khac phia bo la A K va khoang each t u C den A K bang fan khoang each tuB den A K nen c - = -6 c = -5 => C ( ; - ) Goi B' la diem do! xi i ng v d i B qua duo ng thSng A K =j> B' e A C B B ' l A K = > B B ' : x - y + = Toa dp I la nghiem he phuong trinh: diem BB', suy toa diem B' •CB' = 29 •5' X = 2x + y - l = x-2y+3=0 ^=5 -1 -1 I ( y ; - ) I la trung "5'5 = ^ ( - ; ) ^ D u o n g thSng A C c6 VTPT n ^ ^ = (29;7) ==> AC : 29x + 7y + 35 = Toa diem A la nghiem he phuong trinh: 2x + y - l = 29x + 7y + 35 = 14 x = - < 14 33 Vay A " 5' 33 r Gia su A B = a ^ CD = 2a;BC = aV2 Ke BE CD Goi N la trung M N : x - y + = diem AD => M N A D Toa N la nghiem he p h u o n g trinh: N M = (6; 3) M N = 3V5 = y 2x + y + = x-2y+8=0 14 33^ 5' ;C(0;-5) CE = ED = a Phuong x = -4 y = ^ trinh BE = a = A D duong thing N(-4;2) => a = 2N/5 191 A, D thupc d u o n g tron (T) tarn N ban ki'nh R = | = >/5 => ( T ) : (x + 4)^ + (y - 2)^ = Toa A , D la nghiem he phuong trinh: 2x + y + = ^2 , (x + ) ^ + ( y - r = ^ [y = - x - ^2 5(x + ) ^ = « x = -3;y = x = -5;y = Vi A CO tung d u o n g nen A(-5;4); D ( - ; ) AB = - N M = (4; 2) => B ( - l ; ) M la trung diem BC nen C(5;4) Vay A ( - ; ) ; B ( - l ; ) ; C ( ; 4); D ( - ; ) V i d u Cho hinh vuong ABCD c6 M la trung diem cua canh BC, phuong trinh duong thSng D M : x - y - = va C ( ; - ) Xac d j n h toa cac d i n h A , B , D biet diem A thuoc d u o n g t h i n g d : x + y - = Cho h i n h chij' nhat ABCD c6 phuong trinh d u o n g thang AB: x - 2y + = 0, phuong trinh d u o n g thang BD: x - 7y + 14 = 0, d u o n g thang A C d i qua M(2; 1) T i m toa dp cac d i n h cua hinh c h i i nhat ABCD LOT gidi Cacft 1; Gia su A ( a ; - a ) e d va M ( b ; b - ) E D M V i M la trung diem BC nen toa diem B ( b - ; b - l ) V i A B = DC nen ta suy duoc toa dp diem D(a - 2b + 6; -3a - b ) V i diem D thuoc d u o n g thang D M nen: ( a - b + ) - ( - a - b ) - = 0c5.4a + = o a = - l (1) V i A B C D la h i n h vuong nen: BA.BC = » ( a - b + ) ( - b ) + ( - a - b + ) ( - b - ) = (2) Thay (1) vao (2) dupe: (2 - 2b)(6 - 2b) + (6 - b ) ( - b - 2) = o b = hoac b = Truong hap V o i a = - l , b = ta c6: A ( - ; ) , B ( ; ) , C ( ; - ) , D ( - ; - ) Kiem tra thay dai cac canh AB, BC khong bang Ta loai t r u o n g hop Tntmghapl V d i a = - l , b = ta c6: A(-1;5),B(-3;-1),C(3;-3),D(5;3) Kiem tra thay dp dai cac canh A B , BC,CD, D A bang nen ta nhan truong hop Vay cac d i n h can t i m la: A ( - ; ) , B ( - ; - ) , D ( ; ) Cdch2:G\asu A ( a ; - a ) e d va M(b;b-2)eDM Vi M la t r u n g diem BC nen tpa dp diem B ( b - ; b - l ) 192 Tarn giac ABC vuong can tai B nen ta c6: BA.BC = f(a-2b + 3)(6-2b) + (-3a-2b + ) ( - b - ) = ( l ) BA2=BC2 < [ ( a - b + 3) +(-3a-2b + 3) = ( - b ) + ( - b - ) (2) fa-2b + = k { - a - b + 3) Tu (1) ta SUV tBn tai so thuc k de: < , , [-2b-2 = -k(6-2b) Thay ket qua (2) ta thu duoc phuong trinh: k^ (-3a - 2b + 3^ + (-3a - 2b + 3)^ = (6 - 2bf + k^ (6 - 2b)^ o k2+lj(-3a-2b + f =(l+k^)(6-2bf o ( - a - b + f =(6-2b)^ » a = - l hoac a = - ^ ^ Trubng hap Voi a = - thay vao he (1),(2) ta suy b = Truang hap Voi a = ^^'^'^ thay vao he (1),(2) ta suy duoc he v6 nghiem Tom lai, cac dinh can tim la: A ( - ; ) , B ( - ; - ) , D ( ; ) Cflc/i3;Giasii Aj,A2 lanluytcohf sogoc k^,k2 va (ApA2) = a thi tana = l + kjkj Duong thang D M : x - y - = c6 he so goc bang Gia su phuong trinh duong thang D C : y = k ( x - ) - c6 he so goc k Vi ABCD la hinh vuong nen tanlClDt: = MC _ DC " 1-k 1+k = — k = - hoac k = 3 • Truang hap Voi k = - ta c6 phuong trinh DC : y = - (x - 3) 3 Khi toa dp diem D la nghiem cua he y = i(x-3)-3 x-y-2=0 x = -3 y = -5^ D(-3;-5)., Giasu A ( a ; - a ) e d , vi DA DC nen DA.DC = c= 6(a + 3) + 2(7 -3a) = o 32 = (v6 li) Truang hap Voi k = ta c6 phuoiig trinh DC : y = 3(x - 3) - Khi toa dp diem D la nghiem cua he y = 3(x-3)-3 X =D x-y-2=0 y = 3^ D(5;3) Giasu A ( a ; - a ) e d , VI DA DC nen DA.DC = ci> -2(a - ) - ( - a - l ) = cs a = - => A ( - ; ) 193 V i A B = D C n e n ta d e d a n g xac d i n h d u o c toa d o d i e m Cdch 4: G g i C B(-3;-1) la g i a o d i e m cua D M va A C t h i G c h i n h la t r o n g t a r n t a m giac BCD D o d o C A = C G , h a y la d i e m A c h i n h la a n h cua d i e m G q u a p h e p v i t u D u a v a o t i n h chat cua p h e p v i t u ta c6: D M : x - y - = - V(C,3) •A : x - y V(C,3) + = V i d u o n g t h a n g D M d i q u a d i e m G n e n d u o n g t h a n g A d i q u a d i e m A S u y ra: Toa d o d i e m A la n g h i e m ciia he: Gpi ^'^ ^ ^ ^ ^ x-y +6 = y = ^ =>A(-];5) ^ I l a t r u n g d i e m A C thi l ( l ; l ) D u o n g thang BD d i qua d i e m l ( l ; l ) v a v u o n g goc v o i A C n e n c6 p h u o n g t r i n h : X - 2y +1 = D u o n g tron ngoai tiep ( c ) CO p h u o n g t r i n h : ( x - l j h i n h v u o n g c6 t a m +(y ~ l j va b a n k f n h R = l A = 2V5 nen = 20 x - y + l =0 x = 5,y = l o a d o ciia cac d i e m B, D la n g h i e m ciia he: ( x - l f + ( y - l f =20 x = -3,y = - Ket h o p v o i d f e u k i e n d i e m D t h u o c d u o n g t h a n g D M : x - y - = s u y toa d o cac d i n h can t i m la: A ( - ; ) , B ( - ; - l ) , D ( ; ) nhan De Cdchl: thay B la g i a o cua x-2y +l = BD voi AB cho nen toa d o B la n g h i e m cua he: ^fTAU x - y + 14 = l^5'5 D u o n g t h S n g (BC) q u a B(7;3) va v u o n g goc v o i ( A B ) cho n e n c6 V T C P x = u=(l;~2)==>n;3C=(2;l)=^(BC): M a t khac: C O S T B D ; B C ) = 21,t 13 , y = - - t 2-7 N/5.5V2 2a + b = '10 Khi do: N/TO 2a+ b = COSTAQBC) = , ,2 o ( a + b ) = a ^ + b ^ « a + a b + b2 n^c = ^ ( - b ; b ) / / n ^ = ( - l ; l ) = > ( A C ) : - ( x - ) a = -b =0 b = -7a +y - l = o - x " A C = (a;~7a)//n^ = ( l ; - ) = ^ ( A C ) : ( x - ) - ( y - l ) +y + l=0 = o x - y +5 = 194 21 ^ X =—+t (AC) cat ( B C ) tai C:>t = - = > C ( ; ) ; C -x + y + = 21 ^ X = —+t hoac ( A C ) cat ( B C ) tai C y =—-2t « t = -=>C ^ 5 x-7y+5=0 '24.6^ - X + y +1 = [x = ( A C ) cat ( A B ) tai A \| xo- 2j y^ + l =< ^0 A ( 1[;y0 )=; iioac cijng c6: x + 7y + = x - y + l =0 x — O A < 11 ' ' Tim tpa cua I : A C cat BD, sau t i m D doi xiing v o i B qua I Cdch 2: Theo tinh chat hinh chii nhat: = "66^; gpi VTPT cua (AC) la: n = ( a ; b ) ; vecto phap tuyen ciia AB la = ( l ; - ) va vecto phap tuyen cua (BD) la n2(l;-7) thi ta c6: — \\b + 14 cos n;n2 =cos n , ; ! ! a-2b 15 = = = = - = - = o ( a - b f = ( a + b ) o a + a b + b2 = 0 = = = =— — = a = -b b = -7a T u ta lap dupe A C qua M(2;l) va c6 n = ( - b ; b ) / / r V = ( - l ; l ) = > ( A C ) : - ( x - ) + y - l = 0 A(1;1;1) loai v i A = I Vai =2=>B(3;5;5) x = 3+ t Phuang trinh A qua M va B la: y = + t : Truong hop loai v i A khong cat A, z = 5-2t Voi t , = - = > B ( - ; - ; - ) x = - l + 5t f;2;0 Phuong trinh d u o n g A : y = - + 10t=>A cat A, tai A z = - + 6t Goi O la tarn mat cau theo gia thiet O thuoc (P) ( b ; 2b - 2; O) Ta c6: V i (S) tiep xuc v a i A, va Aj Ian luot la A va B Nen O A = OB ^ b = -11 80 40 80 Cdch khdc: De thay A,; A^; A dong phSng Do vay ta phai c6: B la trung diem A M Goi A (a; a - : - a ) => B ^ - ^ ; a + ; - a • V ' [ T u d o ta t i m A ( ; - ; ) ; B(2;3;3) Ta goi tarn l ( m ; n ; ) thi dua vao d ( l ; A , ) = d ( l ; A , ) , phan lai danh cho ban doc u^ = ( l ; l ; - ) la mot VTCP cua A H , H G A H = > H = (2 + a;3 + a ; - a ) • C H = (a - ; + a ; - a ) H e BC ^ CH.u = a = => H ( ; ; ) x = 3-t => BC CO phuong trinh: y = 2+t (te#) z =3 U j = ( l ; - ; l ) lampt VTCPcuaBD,B = B C n B D ^ B ( l ; ; ) I e BD zi 1(1 + b; - 2b; + b); I H BD : ^ I H i ^ = =^ I 2 H , la diem doi xiing ciia H qua BD => I la trung diem cua H H , => H , ( l ; 3; 4) ^ B H , = (O; - ; l ) , H , e AB => AB c6 phuong trinh: y = - s (s z = 3+s A = A B n A H => A = (l;2;5) => A B = A C = BC = 2V2 Cv^Bc = A B + A C + BC = 672 329 Ta c6: d x-3_y-2_z-l x-2_y-5_z+3 -; d , :4 ' -3 > d , va d j cheo M N = ( - ; - ; - l ) => D u o n g thang chiia M N c6 p h u a n g trinh: u = ( l ; - ; ) la mot VTCP cua d , ; M N u = x+1 y-1 z+1 M N d, M N c= ( ? ) = M N ± A B => M N ( Q ) ; ( Q ) la mat phSng chua d , va A B :=> ( Q ) : 5x + 3y + z - 2 = 0, ( Q ) n ( d , ) = B B ( ; ; - ) => M B = 76 A e d , =^ A = (3 +1; - 3t; + 4t) => M A = J(t - ) ' + (2 + t ) ' + ( l + t ) ' M G (P) => M A = M B A CO toa dp: ( ; ; l ) hoac 30,53,-23 J3'l3' 13 Gpi E la trung diem ciia A B ; (P) la mat phang trung true cua AB (P) qua E va nhan AB la mot VTPT Vdi A = ( ; ; l ) = i > E = ; AB = ( ; ; - ) = > ( P ) c6 phucmg t r i n h : y - z - = J Voi A = 13 13'13' 23 30 53 E = ^69 46 26 13 49' 26 ; AB = ^9 -14 ^13 13 -3^ 13^ => (P) CO phuong trinh: 9x - 14y - 3z + 20 = Vay p h u a n g trin h cua mat phang (P) la y - 3z - = hoac 9x - 14y - 3z + 20 = Bai tap : Gpi d la d u o n g thang bat k i d i qua B ( - ; ; ) va cat d u o n g th§ng A Giasir d cat A tai M ( l + t ; ; - t ) Khi d c o V T C P B M = (2 + t ; - ; - t ) Taco B A = ( ; - l ; - l ) , BM,BA BM,BA = (2-t;2-2t;4-t) ^(2-t)^(2-2t)^(4-tf d(A,d) = t ^ - t + 12 V t^+2t + BM f (t) = ^ ^ ^ ^ = o t = ±2 (t2^2t + f Xet ham so f ( t ) = t^+2t + Vay khoang each t u A t o i d Ion nhat bSng ^ / l l k h i t = - l i n g v a i M ( - ; ; ) x=- l D u o n g thang d can t i m c6 p h u o n g trinh: y = 2-t z=t 330 Ox CO VTCP i = (1;0;0) qua 0(0; 0; 0) va A B CO VTCP AB = ( - l ; l ; - ) va i.AB = - ^ => Ox va A B khong vuong goc Taco [i,AB].OA = + + = nen A B va O x c h e o n h a u x=t Phuang trinh tham so cua Ox : < y = 0, M e Ox => M ( t ; 0; 0) z=0 S = M A + M B = ,/(t ~3f+0 + 4+s]{t - i f + ^ =s]{t -3f+4+^J{t - i f + Ta phai t i m t cho S dat gia tri nho nhat Trong mat phang toa Oxy, xet cac diem M( (t; O) e Ox va hai diem A , ( ; ) , B ( ( ; l ) thi S = M , A , + M , B t Ta thay A^, Bj nSm cung phia v a i Ox nen ta lay A / ( ; - ) d o i x i i n g v d i Atqua Ox Phuong trinh d u o n g thSng A ( ' B j : x + y - = S= A j + M , B ( nho nhat M la giao diem ciia Ox va A^' Bj => 3t - = hay t = ^ • Vay M ( - ; ; ) la diem can t i m Cdch khdc: Ta cd the tim diem M bang phucmg phdp khdo sat ham so' T u bieu thuc S = ^(t - ) ^ + + ^/(t - ) ^ + T a x e t h a m s o f ( t ) = J(t - ) ^ + + ^/(t -if+1 t-3 t-2 Co dao ham f ' ( t ) = f'(t) = t-3)%4 t-3 ^/(T^ ^ { t - l f + t-2 :+ , (te#) = J(t-2f+1 ^ =0 (t-2) -(t-3) O ^ ' J(t-3)%4 J(t-2f+1 vai dieu kien < t < ta c6: (*) o (t - ^ [(t - i f +1] = (t - i f [(t - 3)^ + 4] «(t-3f=4(t-2f t - = 2(t-2) t - = -2(t-2) t = lg[2;3] Lap bang bien thien cua ham so f ( t ) Suy m i n f ( t ) = f /38+VlO 3, Vay M A + M B dat gia tri nho nhat bang ^ ' ^ ^ ^ ^ , dat duoc tai t = - , tiic la M ( - ; 0; ) 3 331 D u o n g thang d c6 p h u o n g trinh tham so x = l + 2t y = + 2t z =l+ t d qua diem N ( ; ; ) , C6 V T C P u = (2;2;l) va AB = ( ; ; - ) Ta CO u C D = + - = ^ =:>d khong vuong goc v6i A B va [ u , A B l N A = - = => d va A B cheonhau Chu v i tam giac M A B la 2p = ( M A + MB + A B ) , A B khong doi nen 2p dat gia tri nho nhat k h i M A + MB dat gia t r i nho nhat Xet diem M e d => M ( l + 2t; 2+2t; +1), ta phai t i m t de M A + M B dat gia trj nho nhat Xet f ( t ) = M A + MB =^|{2^ + 2f +{2t + lf +1^ + yjiltf +{2t-2f+{t + \f => f (t)=V9t^ + 12t + + V9t^ - 6t + = ^/(3t + 2)^ +1 + ^/(3t -1)^ + Co dao ham f'(t) = f'(t) = O o 3t + \/(3t + ) + 3t + 3t-l 7(3t + ) + o - 3t + j = 3t-l J(3t-l)2+4 =0 J(3t-l)2+4 -(3t-l) , = - = ^ = = = vol = V(3t + ) + l ^/(3t-l)2+4 — ^ t< - 3 cp^j^ t = - - , phuong t r i n h d u o n g thang can t i m la: x+l _ y _ z+ l ~4^~ a) d n d ' = M = ^ M ( - l + t ; t ; - t ) , t G i J ? ' , V T C P c u a d : U j = A M ( t - l ; t + l ; - t ) AB,Ud AB(2;2;-1); AB;Ud = ( l - t ; l ; - t ) = > d(B,d) = Xet ham so f(t) = I j ^ +24t + 54 2t^+4t + 12t^-18t + 18 V 6t^-2t + rnaxf(t) = f(0) = 18; minf(t) = f(2) = — 11 • ^ ^ d ( B , d ) < Vl8 x = 3t min(d(B,d)) = t = Phuong trinh d u o n g thang can t i m la: y - - l + 3t z = 2-2t x = -t max(d(B,d)) = v l t = Phuong trinh d u o n g thang can t i m la: y = - l+ t z = 2-t b) d n d ' = M = ^ M ( ^ l + t ; t ; - t ) , t e # , V T C P c u a d : U j = A M ( t - l ; t + l ; - t ) T u p h u o n g t r i n h A =o u^==(2;-2;l) va N = ( ; ; ) e A = (t l;4t-l;6t) AN(5;l;-2); AN - = 3., d(A,d) = (2 + t)^ (53t''"10t + = 3.Vfa) Xet ham so f(t) = — ^ ^ ^ ^ r:> maxf(t) = f ( — ) = - - , max(d(A,d)) = V26 53t^-10t+2 37 x = 29t Phuong t r i n h d u o n g thang can t i m la: y = -l-41t z = + 4t Goi VTCP ciia d u o n g thSng d la: u(a; b; c), a^ + + >0 d//(P)c=>Ud.nQ = c j c = a - b ; u ( l ; - ; ) Goi goc giua hai mat phSng la cp (0 < cp < — ) , 334 5a-4b cos((p) = V a ' - a b + 2b2 (5a-4br ^ \ a - a b + 2b^ Truong hap 1: Neu b = 0, coscp = -.^5 Truong hap 2: Neu b # , dat t = - , coscp = b Xet ham so f(t) = (5t-4)^ -, < cos(p < j-^^L_i)_ ^-.Mt) 3\5t2_4t + 5^3 5t^-4t + So sanh trudng hgp va t r u o n g hgp thi ^ cos(p < y min(cos{p) = => cp^^^ = ^O*^ J = ^ • b x-1 Phuong trinh d u o n g thang can t i m la: — ^ = max(cos{p) = a 5/3 y + ^ = z-2 Phuong trinh d u o n g thang can t i m la: X y + z-2 = ^—^ = — ^ Goi VTCP cua d u o n g thang d la: u(a;b;c),a^ + b"^ + c^ > dc(P)ouj.np =0 b = 0, chgn a = 1, c = X Phuong trinh d u o n g thang can t i m la: =l +t y =0 z =t max(d(B,d)) = \ / l o a = - b c h g n b = - l , a = l , c = - l 335 P h u a n g t r i n h d u o n g thSng can t i m la: x = ~ l - t y =t z =t 10 a) P h u a n g t r i n h m a t p h ^ n g (P) c h u a d c6 V T P T la n ( A ; B ; C ) , A ^ + B + C ^ > c6 d a n g A ( x - l ) + B ( y + 2) + C z = Taco: d c (a) « Uj".!!^ =0 o C = A + 2B G o i goc g i i i a h a i m a t p h S n g la (p, (0 < (p < ^ ) , coscp = A + 2B \/2A^ + A B + B ( A + 2B)^ Trudmg hap 1: N e u B = 0, coscp = 2 A + A B + 5B^ (1) A , cos(q)) = —(t + 2)^ Truong hap 2: N e u B ?t , d a t t = — B V2t2+4t + X e t h a m so f ( t ) = —l^t^) 2t2+4t + V a y m a x coscp = /30 , m a x f ( t ) = - tai t =1 h a y — = - • B (2) So sanh t r u o n g h o p v a t r u o n g h o p t h i (p^^^ coscp = — ^ v o i — = — B P h u o n g t r i n h m a t p h a n g c a n t i m l a : x + 2y + 5z + = b) P h u a n g t r i n h m a t p h S n g (P) c h u a d c6 V T P T : n ( A ; B ; C ) , A ^ + B^ > c6 d a n g : A ( x - l ) + B ( y + 2) + C z = Ta c6: d c ( a ) n^ = o C = A + 2B G o i goc g i i i a m a t p h a n g (P) v a d u o n g t h a n g d ' la T , ( ^ T < —) sin^ = A + 3B (4A + V A + A B + 5B2 3Br ' V A + A B + 5B Truatig hap 1: N e u B = 0, sin4^ = 2V2 N e u B *0,Aai t = —, s i n T = - j - ^ ^ ^ ' • B' 3V2t2+4t + Truanghapl: * 336 Xet ham so f(t) = 5V3 (4t + 3)^ 25 —, maxf(t) = — tai t =-7 hay 2t^+4t + sin \\i = So sanh truong hap va truong hap suy ra: ^ ^ ^ ^ sinT = voi = • Phuong trinh mat phang can tim la: 7x - y + 5z - = c) Phuong trinh mat phSng (P) chua d c6 VTPT n(A;B;C), > c6 dang A ( x - l ) + By + C(z-2) = Taco: d c (a) o u^.n^ = o B = - A - C A+C ' (A+C)^ = V A + A B + 5C2 " V A + A B + 5C2 Tru&ttghop 1: Neu C = 0, d(A,(a)) = d(A,(a)) = Trumg hap 2: Neu C # 0, dat t = —, d(A,(a)) = Xet ham so f(t) = (t+ir = 9^/f(t)• V5t^+8t + (t+ir va f'(t) = Oc=>t = ± l ; f ( - l ) = 0;f(l) = - , lim f(t) = - 5t^+8t + t^.±co Lap bang bien thien suy maxf(t) = - tai t =1 Vay maxd(A,(a)) = 3V2 ^ = So sanh truong hop va truong hop suy ra: A = C va B = -4C Phuong trinh mat phSng can tim la: x - 4y + z - = Bai tap 3: Lay tren duong thang d da cho hai diem M ( l ; 0; - l ) , N (O; 2; O) Mat phSng (p) can tim c6 phuong trinh dang: Ax + By + Cz + D = (A^ + B" + ^ ' [MG(P) [A-C +D = fC = [Ne(P) [2B + D = [ D= -2B > o) A-2B =>(p):Ax + By + ( A - B ) z - B = Hai mat phSng (P) va ( a ) : x + 2y - 2z - = c6 VTPT: n^ = ( A ; B ; A - B ) ; n ^ = ( l ; ; - ) Goi X la goc hop boi hai mat phang (p) va ( a ) : x + 2y - 2z - = thi cosx = - A + 6B —» np n 3V2A'-4AB + 5B' , (A-6B)' • cos X =9 ( A ' - A B + B ' ) ' 337 Xet B-O=>cosx = —- A -12 Xet B^O=>cos^x = 'A^ -4 Xet ham so f(t) = ^A^ vBy 36 +5 t ' - t + 36 ^ ^ A ( t ' - t + 5)' B t' -12t + 36 -; t e 2t' - t + -V 53 „, , 20t'-134t + 84 ^ t= l a co: f (t) = ——; ~ ; f (t) = 16 10 • \10, ~ ' ( t ' - t + 5y [ t = 6=:^f(6) = De thay goc x nho nhat cos x Idn nhat o f(t) Ion nhat A ( 7\ = — => — = — ; A = Lap bang bien thien cua ham so f(t) => max f(t) = f ,10, 10 B 10 B = 10 Phuong trinh mat phSng can tim: (P): 7x +10y - 13z - 20 = C M A T C A U Bai tap : Goi I la tam duong tron (C) cho nen I la hinh chieu cua A tren mat ph§ng (P) Goi A la duong thang qua A va vuong goc voi (P) nen A c6 phuong trinh tham so la x =l +t A:I la giao diem cua A va (P) va I e A i:> l ( l + 2t;2+ 2t;3 + 2t) z = + 2t Ie(P)=:>l + t + 2(2 + 2t) + 2(3 + t ) - = 0=>t = - l i z > l ( ; ; l ) Duong thSng d tiep xiic vdi duong tron (c) d(l;d) = R Duong th§ng d c6 VTCP la a = ( l ; - l ; - ) va di qua O(0;0;0) Taco IO = ( ; ; - l ) : IO;a = ( - l ; - l ; ) = ^ R = d ( l ; d ) = Ta laico l A = (l;2;2) ^ lA = O = AC n BD r = VlA'+R- = iO;a 28 (s): (x - l)^ +(y -2)^ +(z -3)^ = 28 Vay mat cau can tim la Goi Vi S A B C D la chop deu nen lO Suy I e ( S A C ) Ma B D ( S A C ) nen (a) chinh la ( S A C ) Nhu vay, yeu cau bai toan thuc chat la viet phuong trinh mat phang di qua ba diem: S, A , C 338 T a c : n ( ) = SA,SC = [ ( , , l ) , ( l , , - ) ] = (0,10,0) Vay ( a ) : y = Nhan thay: A, O thuoc mat cau Goi M la trung diem cua A O , suy M ( ; ; ) GoiB(a,b,c) Ta c6: O A ( ; ; ) ; M B ( a - ; b - ; c ) , tam giac A B O can tai B nen O A ± MB =^ a + b = (1) Taco: 4>/3 = S^„„ = ^ A O B M = | V B M =^ B M = ^6 = ^ ( a - ) ' + ( b - ) ' + c ' = (2) Do B e ( l ; R ) nen ( a - ) ' + ( b - ) ' + ( c - ) ' =12 (3) T u ( l ) , (2), ( ) s u y ra: B ; B 2- V Bai tap 2: Mot VTCP cua d u o n g thSng d la u ( ; l ; - l ) Mot VTPT cua mat phSng (P) la n ( l ; ; l ) Goi la goc giiia d u o n g thang d va mat phang (P) + 2-1 Ta c6: sin cp = cos u , n = - = > ( p = t M X = 30° Goi R la ban kinh mat cau (S) => l A = R Tam giac I M A vuong tai A Ta c6: T M X = 30° ^ A M = R>/3.S„,^ = 3V3 o |lA.A = M3V3 o R = N/6 Gia sir l ( l + t ; l + t ; ~ t ) , t < - - 3t-3 T u gia thiet ta CO khoang each d ( l , ( P ) ) = R c ^ — p - = V o t = - l =>l(-l;0;l) Phuong trinh mat cau (S): (x +1)" + y" + (z -1)^ = Bai tap 3: I , ( ^ l ; ; - ) ; R, = 2^2 Goi r la ban kinh d u o n g tron giao tuyen I , (d) o r = AR] - d ( l , ; ( P ) ) = I , ( t - ; t;2t - ) , d ( l , , { P ) ) = ^ R ^ - r ' = 7i '3V2I VTi ^ 4~2 t = 0:^(S'):(x + l ) ' + y ' + ( z + l ) ' =4 t = -2=i>(x + ) ' + ( y + ) ' + ( z + ) ' = 339 Bai tap 4: Goi I la trung diem cua AC, I 2 V\D la hinh chop deu nen ABCD la hinh vuong Ma AC = 3>/2 BI = ; B thuoc (Q) => B(a; b; - a - 2b) BI.AC = b= 24-5a ABCD la hinh vuong: BI = 3- — [2 a = 3=i.b = l 57 a=— 107 -a + J f5 — -h [2 (a e .(a.2b-5)^4 J 57 (loai) ==>B(3,1,-2) vol a = - 107 > ( A B C ) : x + 2y + z - = (cung chinh la phuong trinh ( A B C D ) ) x = - + 2t d y = - + 2t ^ z = -2 +1 Gpi d la duong thang qua I va vuong goc vdi ( A B C D ) Vi S A B C la hinh chop deu nen S thuoc d => S - + t ; - + 2t;-2 + t 2 Trong mat ph3ng (SAl) dung trung true cua SA cat SI tai O Vi O thuoc d la true va O thuoc trung true cua SA nen SO = OA = OB = OC = OD = R Vay O la tam mat cau ngoai tiep hinh chop S.ABC ban kinh R = SO Ma ^ R X ; I X cung nhin OA duoi mot goc vuong nen OHAI npi tiep duong tron duong kinh OA Vi S nSm ngoai duong tron • SH.SA = SI.SO SH.SA SO = R = SI SA' 2SI 2h Ma theo eia thiet — = - h = — ^ h h = SI = - S I = (-2; - t ; - t ) o S l ' = — =^t = - = > S 2-3-^ 2,3,^ ^ ' 16 hoac t = — => S l ; ; - ^ 4 Bai tap 5: A e d = > A(4 + 3t;4 + 2t;4 + t ) 340 Goi I la tam mat cau (S) =i> I = (l;2;3) => lA = (3 +3t;2 + 2t;l +1) =^ AI = |l +1| Vl4 (1) Dat AB = a; AI n(BCD) = H; (s) c6 ban kinh R = 14 AH±(BCD)=>AH = ^ ; =^AI = AH + IH = ^ IH = VR^-BH^ =J^^T^ + ^ | ^ ^ (2) Lai c6: Al' = R ' +a= r^ AI = VR' +a' (3) TLr(2)va(3) ^ V R ^ ^ = ^ + « a^ = 2R^ =^ A I = RVS = ^/l4 (4) Diem A C O cac toa do: (4;4;4); ( - ; ; ) Tir (1) va (4) + t = Bai tap 6: Ta CO tam mat cau (S) la l(l;l; -2), ban kinh la: R,s, = 4, d(l;(P)) = - Vi d(l;(P))