Ebook bồi dưỡng học sinh giỏi toán hình học 12 phần 2 ths lê hoành phò

C h u o n g I I I : P H U O N G P H A P T O A T R O N G K H O N G D O G I A N §1 H E T O A D O T R O N G K H O N G G I A N A KIEN THUC C O B A N Toa khong gian Ba vecto don v i i , j , k tren true Ox, Oy, Oz : i = ( ; ; ) , ] = ( ; 1;0), k = (0; 0; 1) M M(x, y, z) hay M = (x, y, z): >T>~ OM = x i + y j + z k a(x, y, z) hay a = (x, y, z): a = x i + y j + z k • Hai vecta: u = (x, y, z) va v = (x', y', z') thi: u ± v = (x ± x' ; y ± y'; z ± z') ; k u = (kx; ky, kz) u v = xx' + yy' + zz' ; | u Vx + y + z 2 x.x + y.y + z.z' cos(u, v) >/x + y z Vx' +y' z' 2 + 2 + Hai diem A ( x i , y i , z\) va B(x , y , z ) thi: 2 AB = (x - x i ; y - y i , z - z\) AB = yj( 2 - x ^ + ( y - y i ) +(z - z r 2 x M chia AB theo ti so k * 1: M x i - k x _y - k y l-k l-k kz., l - k Tich co hirotig: cua a = (x, y, z) va b = (x', y', z') la vecto: n = [a , b ] = y z z y' z' z x X y x' J X ' y' • Ket qua: - Vecto [ a , b ] vuong goc vdi a , b - Do dai cua vecto [a , b ] : | [a , b ] | = | a | | b | sin(a , b ) - vecto a , b cung phuong: - vecto a , b , c dong phang: [a, b ] = [a, b ] c =0 - vecto a', b , c khong dong phang: 154 [a , b ] c * Dien tich va the tich Dien tich tam giac ABC: S=-|[AB,AC]| The tich tii dien ABCD: V = — | [ AB AC AD | The tich hinh hop ABCD.A'B'C'D': V = | [ A B , AD ] AA~' | The tich hinh lang try A B C A ' B ' C : V = | | [ A B , ADJ.AAT' | Phirong trinh mat cau: Mat cau (S) tam I(a, b, c) ban kinh R: (x - a) + (y - b) + (z - c) = R hay: x + y + z + 2Ax + 2By + 2Cz + D = 0, A + B + C - D > 2 2 2 2 2 co tam I ( - A , - B , - C ) va ban kinh R = \ / A + B + C - D Chu y: - Tam duong tron I ngoai tiep tam giac ABC khong gian: f l A = IB = IC 2 [ I e (ABC) Tam K mat cau ngoai tiep tii dien ABCD: KA = KB = KC = KD D A N G 1: TOA D O DIEM, VECTO Ba vecto don v i tren true Ox, Oy, Oz : i = ( ; ; ) , j = ( ; 1;0), k = (0; 0; 1) Hai diem A ( x i , y i , Z\) va B(x , y , z ) thi: 2 AB = (x - x i ; y - y i , z - Zi) 2 AB = f x - x ) 2 +(y - y ^ 2 + (z - z ^ 2 ^ yi - y i k - Diem M chia A B theo ti k * 1: M , z k z 2 l-k l-k l-k Toa trung diem M cua doan A B , tam G cua tam giac ABC tam E cua t i i dien ABCD voi toa dp A(x y zO, B(x , y , z ), ys, Z ) D(X4 y4 , z ): +x + x +x +x i + x + x X = l X = + y + y yi + y E y = y i + y + y + y.4 G v M y = y + z +z + z +z f z Zl + z zz= i z= Z— 1; 1( 2 x x X l 2 2 3 Y l Z l 3 z 4 Phep toan ciia hai vecto: u = (x, y, z) va v = (x , y' z') thi: u + v = (x + x' ; y + y'; z + z') k u = (kx; ky, kz), u - v = (x - x' ; y - y'; z - z') u v = xx' + yy' + zz' ; 155 [u , v ] = f y Vy ' z z' j Z X z x' x x' y s y' > - Quan he cac vecta Vecta [ a, b ] vuong goc v o i a, b Vecto a vuong goc voi b » a b = Hai vecto a, b cung phuong b = k a Hai vecto a, b cung phuong [ a, b ] = Ba vecta a, b , c dong phang < > [ a , b ] c = = B6n diem A , B , C, D dong phang [ A B , A C ] A D = Ba vecto a, b , c khong dong phang [ a , b ] c # Bon diem A , B, C, D khong dong phang [ A B , A C ] A D * Chu y: Ung dung toa dp khong gian de giai cac bai toan hinh khong gian co dien, quan he song song, vuong goc, dp dai, goc, khoang each, vi tri tuong doi, V i du 1: Cho ba vecto a = (2; - ; 3), b = (0; 2; - ) , c = (1; 7; 2) a) T i m toa dp ciia vecto e = a - 4b - c b) Tim toa ciia vecta f = a - — b + 3c Giai a) e = a - 4b - 2c = (2 - - 2; -5 - - 14; + b) f = a - - b +3c = (8 + + ; - - - + 21; 12 + 3 V i du 2: Tim toa dp cua vecto m cho biet: a ) a + m = a v a a = ( ; - ; 1) 4) = (0; -27; 3) 1 K K - +6) = ( 1 ; - : — ) 3 V b) a + m = b va a = ( ; ; - l ) , b = ( ; - ; 3) Giai a) a + m = a => m = a = (0; -6; 3) b) a + 2m = b = > m = a + — b = —; ;2 2 { 2 ) V i du 3: Cho hai bp ba didm: A ( l ; 3; 1) B(0; 1; 2), C(0; 0; 1) va A ' ( l ; 1; 1;), B'(-4; 3; 1), C'(-9; 5; 1) Hoi bp ba diem nao thang hang? Giai Taco CA = ( ; 3; 0), CB = ( ; 1; 1) V i cac toa dp khong tuong xiing ti le nen khong co s6 k nao de CA = kCB , suy A , B, C khong thang hang Ta co t T T = (10; - ; 0), G B = (5; - ; 0) => C^A' = C^"' Do A', B', C thing hang 156 V i du 4: Tinh tich vo huong ciia hai vecto moi truong hop sau: a) a = (3; 0; -8), b = (2; - ; 0) b) a = (1; - ; 2), b = (4; 3; -5) c) a = (0; a) N/2 ; S ) , b = (4; S ; - ) Giai a.b = 3.2 + 0.(-7) + (-8).0 = b) a b = 1.4 + (-5).3 + 2(-5) = - c) a b = 0.4 + N/2 73 + V3(-%/2) = V i du 5: Cho ba vecto: a = ( ; - ; 1), b = ( ; ; - l ) , c = (3; ; - ) Tinh: a) (a b ) c , a (b c ) b) a b + b c + c a , a c + b - c Giai 2 : 2 a) Taco: a.b = 1.4 + ( - l ) + l ( - l ) = D o d : ( a b ) c = c = (9; 6;-3) Ta co b c = 4.3 + 0.2 + (-1)(-1) = 13 Do a ( b c ) = 13a = ( ; - ; 13) b) Taco i = 3, b = 17, c = 14 nen a b + b c + c a = 3b + c + a = (77; 20; - ) va a c = 2 2 2 b4a.c + b -5c 2 = -53 Vi du 6: Cho I u 1=2 | v I = , goc giua hai vecto u va v bang — Tim k de vecto p = k u + 17 v vuong goc voi vecto q = u - v Giai l-i ^ ,_ | - _ 2n Ta co | u | = 2, I v | = 5, cos(u v ) - cos— — - Do p lq p.q = 0(ku +17v)(3u - v) = o k u - 17 v 2 + (51 - k)u v - < > 3k.4 - 17.25 + (51 - k)2.5— = = < > 17k - 680 = < > k - 40 = = Vi du 7: Cho ba diem A(2; 0; 4) B(4; S ; 5) va C(sin5t; cos3t; sin3t) Tim t de A B vuong goc voi OC Giai Ta co AB = (2; 73 ; 1), OC = (sin5t; cos3t; sin3t) Hai dudng thang A B va OC vuong goc vdi va chi khi: A B OC = < > 2sin5t + 73 cos3t + sin3t = = < > sin5t + — cos3t + — sin3t = < > sin5t + sin(3t + — ) = = = 157 Jl o sin5t = sin(-3t ) Vay t = -— + -k*.t=— + kn, k e Z 24 V i du 8: Xet su dong phang cua ba vecto a) a = (-3; 1; - ) b = (1; 1; 1), c' = (-2; 2; 1) b) a = (4; 3; 4), b = (2; - ; 2), c = (1; 2; 1) Giai a) T a c [ a , b ] = ^ ] J ~ j = (3; 1;-4) Do [a, b ] c = -6 + - = -8 * Vay vecto khong dong phang b) Taco [ a , b ] = (10; 0;-10) => [ a , b ] c = Vay vecto dong phang V i du 9: Cho a = ( : - ; 1), b = ( ; 1;2), c = ( ; ; ) v a d = (2; 7; 7) " a) Chung minh cac vecto a , b , c khong dong phang b) Hay bieu thi vecto d theo cac vecto a , b , c Giai a) Taco [a", b ] = ( - ; - ; 1)=> [a", b ] c = - * Vay vecto khong dong phang b) Gia sir d = m a + nb + pc m + 4p = f m = -2 < - m + n + 2p = [AB A C ] = (-3;-3; 3) Va AD = ( ; - ; - ) = > [ A B , A C ] AD = 0: dpcm V i du 11: Chung minh cac tinh chat sau day cua tich vo huong a)[a,a]=6 c) [ k a , b ] = k [ a , b ] = [ a , k b ] 158 b) [a , b ] = - [ b , a ] Giai Yi y a) [ a , a ] i Zj z b) [a, b] z i z x i x' x y Xj y : l Zj Xj yi i x = (0; 0; 0) = : yi y ) = (yi - Y2 i ; zix - z xi ; xiy - x y ) = - ( y z i - yiz ; z Xi - Z!X ; x y - x y ) z z x x 2 z 2 2 y yi x A x 2 z 2 '"2 |z i' k z 2 x i x x i : x y yi = - [ b , a] Ket qua [ a , a ] = - [ a , a ] => [ a , a ] c) k [ a , b ] = yi Ik'" y i z ky y z i z ,k z kz x z kz x z i x x , k x x 2 yi y ) kxj k y j 2 kx x i x x x y = [ka, b] Tucmg tu: k [ a , b ] = [ a , k b ] V i du 12: Chung minh cac tinh chat sau day cua tich vo huang a) [ c , a + b ] = [ c , a ] + [ c , £ ] b) a [b , c ] = [a , b ] c c) I [a b ] | = | a b | -(a.b) Giai 2 y yi + y a) [c, a + b] y yi z z 3i z l Z y z - [a.b]c z 2 j : z x z l x z = [c, a] y z z z y i Jy i |y y yi b) a [ b c ] = Xj z i x + x i l z z x -1 + [c, b] x y x +yi Z, 3 y l l l +y +z x X, y Z 2 x X X Z 3 Z 2 i |z '|z Z x x 3 x ; yf + y y j , Jy y yi x Z x X c) VP = |a | | b | - ( a b ) = | a | | b | - | a P I b | W a = Ia | | b (1 - cos a) = | a | | b sin a 2 2 2 = |[a.b]| 2 2 2 = VT V i du 13: Trong khong gian cho ba vecto a , b , c tirng doi khong cimg phuong Chung minh rang dieu kien can va dii de vecto tong: a + b + c = la [ a , b ] = [ b , c ] = [ c , a ] Giai Tira + b + c = - ^ a = - ( b + c ) = > [ a , - b - c ] = D o d [ a , - b - c ] = [ a - b ] - [ a c ] = =>[c, a ] = [a b J Tuong tu ta cung c o [ b , c ] = [ a , b ] Vay: [ a , b ] = [ b , c ] = [ c , a ] Nguoc lai, tir [ a , b ] = [ b , c ] => [ b , a + c ] = Mat khac, [ b , b ] = = > [ b a + b + c ] = = > b cimg phuong voi vecto a + b + c Chung minh tuong tu ta cung co vecto a ciing phuong voi vecto a + b + c Nhung a va b khong cimg phuong, vay a + b + c = V i du 14: Cho diem M(a; b; c) a) Tim toa hinh chieu ciia M tren cac mat phang toa va tren cac true toa b) Tim khoang each tir diem M den cac mat phang toa do, den cac true toa Giai a) Goi M i ( x ; y; 0) la hinh chieu ciia diem M(a; b; c) tren mp(Oxy) thi: M M j = (x - a; y - b; -c) V i MM I = va M M ^ T = nen x - a = l y - b = VayM,(a;b;0) Tuong tu, hinh chieu ciia M tren mp(Oyz) la Mi(0; b; c) hinh chieu ciia M tren mp(Oxz) la M (a: 0; c) Gia sir M ( x ; 0; 0) la hinh chieu ciia M(a; b; c) tren true Ox thi x MM X = (x - a; - b ; - c ) V i M M i = nen x = a, do M (a; 0: 0) V x Tuong tu, hinh chieu ciia M(a; b; c) tren true Oy la M ( ; b; 0), hinh chieu ciia M(a; b; c) tren true Oz la M (0; 0; c) y z b) d(M; (Oxy)) = M M , = V ( a - a ) + ( b - b ) + ( c - ) 2 = |c | Tuong tu d(M; (Oyz)) = | a | , d(M; (Ozx)) = | b | Ta co d(M; Ox) = M M , = x/(a - a) + (b - 0) + (c - 0) =x/b +c Tuong t u d(M; Oy) = Va + c 160 2 d(M; Oz) = Va + b 2 2 V i du 15: Cho hai diem A ( x i ; y i ; z\) va B(x ; y ; z ) T i m toa diem M chia doan thang A B theo ti s6 k * Giai Voi diem M(x; y; z) ta co: MA = (xi - x; y, - y; z\ - z), 2 MB = (x - x; y - y; z - z) Diem M chia doan A B theo ti so k * va chi 2 x = —l-k yi - ^ MA =kMB Yi ~ Y = k ( y - y) o l-k Zj - z = k(z - z) Z =— l-k V i du 16: Trong khong gian Oxyz cho ba diem A ( l ; 2; 4), B(2; - ; 0), C(-2;3;-l) a) Gpi (x; y; z) la cac toa dp cua diem M nam tren mat phang (ABC) Tim su lien he giua x, y, z b) Tim toa dp cua diem D biet rang hinh ABCD la hinh binh hanh Giai a) AB = (1; - ; - ) , AC (-3; 1; -5), A M = (x - 1; y - 2; z - 4) Xj - x = k(x - x) k y = 2 Ta co M nam uen mat phang (ABC) [ A B , AC] A M = < > 19(x - 1) + 17(y - 2) - 8(z - 4) = < > 19x + 17y - 8z - 21 = = = b) V i ABCD la hinh binh hanh nen AB = DC : = -2 - x D -3 = - y » y =6 D D -4 = - - z Vay D(-3; 6; 3) n V i du 17: Cho A(2; l ; 3), B(4; 0; 1), C(-10; 5; 3) a) Chung minh rang: A, B, C la ba dinh ciia mpt tam giac b) Tim chan duong phan giac ngoai ciia goc B ciia tam giac ABC Giai r a) BA = ( - ; - l ; X B C = ( - ; 5; 2) Ta co BA va BC khong ciing phuong nen A B, C la ba dinh ciia mpt tam giac b) Gpi BE la ducmg phan giac ngoai ciia goc B, do: BA EA EA BC " EC ^ EC ~ 15 " V i vecto E A , EC ciing huong nen E chia doan AC theo t i k = — E(5;~;3) 161 V j du 18: Cho hinh hop ABCD.A'B'C'D* biet A ( x y ; z ), C(x ; y ; z ), B'(x' ; y' ; z' ), D'(x' ; y' ; z' ) Tim toa cua cac dinh lai Giai Goi Q = AC n BD, Q' = A ' C n B'D' thi Q, Q' la trung diem cua AC, B'D' nen: f i + yi+y ,z + z l ' "* i; 2 x 4 x x 3 x n 3 V /x' +x I 2 _ y + y , z +z ' ' 4 Tu AA*' = BB ' = CC"' = DTI' = QQ?, suy ( x + x ' + x ' " yi + y ' + y - y 2 J I f x ' + x - - HX _ y' + y - y i + y z ' + z ' - z + z ) ) 2 J I z +z +z — z l 2~ yi + y + y '2-y'4 J 2 { A C B x 2 X + 4 X + X X l 3 X X : 3 ( i+ -x' -f x y -y2+y4 l 3~ 2 I V i du 19: Cho hinh hop ABCD.A'B'C'D' co cac diem A ( l ; 0; 1), B(2; 1; 2), D ( l ; - ; 1) va C'(4; 5; -5) Tim cac diem lai Giai Ta co ABCD la hinh binh hanh nen: x - 2=0 x =2 y - = - » < y = Do do: C(2; 0; 2) BC = AD < > = x x D z + z z + z c c c c z - 2=0 c L z =2 c Va A A ' = BB' = DD' = CC' = (2; 5; -7) Nen A'(3; 5; -6), B'(4; 6; -5), D'(3; 4; -6) V i du20: a) Tim toa diem M thuoc true Ox cho M each deu hai diem A(l;2;3)vaB(-3;-3;2)' b) Tren mat phang (Oxz) thi diem M each deu ba diem A ( l ; 1; 1), B ( - l ; 1; 0), C(3; 1; -1) Giai a) M thuoc Ox nen M(x; 0; 0) Ta co MA = MB o MA = M B o (1 - x) + + = (-3 - x) + (-3) + o x - - Vay M ( - l ; 0; 0) b) M thuoc (Oxz) tren M(x; 0; z) Ta co: M A = M B = M C 2 (AM = B M 2 A M = CM 162 2 ((x-l) + + (Z-1) 2 2 = ( X + 1) +l + z (x - ) +1 + (z - ) = (x - 3) + + (z + l ) 2 2 » { 4x + 2z = 4x-4z = x= VayNlf-jO;-6 ! o - J v Vi du 21: Cho tam giac ABC co A(-2; 1; 0), B(0; 3; - ) , C ( - l ; 0; 2) a) Chung minh tam giac ABC co goc B nhon b) T i m toa diem H la hinh chieu ciia A tren canh BC Giai a) Taco BA = ( - ; - ; 1), BC = ( - l ; - ; 3) BA.BC 11 Nen cos B = > => goc B nhon BA BC 3V19 b) H(x; y; z) thuoc BC nen B H = tBC Do x = - t , y - = -3t, z + = 3t => x = - t , y = - 3t, z = - + 3t Ta co A H BC nen A H BC = (-t + ) ( - l ) + (-3t + 2)(-3) + (-1 + 3t).3 = Vay hinh chieu H f - — ; - —; — 3 Vi du 22: Cho b6n diem A(-3; 5; 15), B(0; 0; 7), C(2; - ; 4), D(4; - ; 0) Hoi hai duong thang A B va CD co cat hay khong Neu chiing cat nhau, hay tim toa giao diem Giai Ta co: AB = (3; - ; - ) , AC = (5; - ; -11) v AD - (7; - ; -15), CD = (2; - ; -4) Do [ A B A C ] = (7; - ; 7) => [ A B A C ] AD = nen A B CD dong phang, hon nua A B , CD khong ciing phuong do duong thang A B va CD cat Goi M ( X M ; >M, ZM) la giao diem ciia A B va CD Dat MA = k M B , MC = k'MD Taco: k x A kx - 4k = l-k 1-k' l-k' i-k - l + 3k y - y _y y = i-k i-k' l-k' l-k 15-7k A ~ B_ y ~ = l-k l-k' l-k l-k' [—3 Giai duoc k' = — nen M — ; —; 11 I 11 12 X T B k A Z k B c k z D k c 163 Bai 53: Lap phirong trinh chinh tac qua E (3, 4, 1) va song song voi dudng x = + 2t thang d: