Thiết kế bài giảng hình học 12 (tập 2) phần 2

Phan cAc BAI SOAN §1 He toa khong gian (tiet 1, 2, 3, 4) MUC TIEU Kien thurc HS n i m duoc: Khai niem tao vecta khdng gian, tpa diem va dp dai vecto Bieu thiic tpa dp ciia cac phep toan : cdng, trii vecto; nhan vecta vdi mdt so thirc Bieu thiic tpa dp ciia tfch vd hudng ciia hai vecta Riuong trinh mat cau KT nang • Thuc hien thao cac phep toan ve vecto, tfnh dp dai vecto, • Vie't dupe phuang trinh mat ciu Thai • Lien he dupe vdi nhieu van de thue te' khdng gian • Cd nhieu sang tao hinh hpc • Humg thii hpc tap, tich cue phat huy tfnh dpc lap hpc tap II CHUAN DI CUA GV VA H& 1, C h u a n bi cua G V : • Hinh ve 3.1 den 3.3 • Thudc ke, phan mau, Chuan bj cua HS : Dpc bai trudc d nha, cd the lien he cac phep bie'n hinh da hpc d Idp dudi III PHAN PHOI THOI LUONG Bai duoc chia tie't: Tie't Tii dau de'n het phan I Tie't Tie'p theo den he't phan II Tiet Tie'p theo de'n het phan III Tie't Phan IV va hudng dan bai tap IV TIEN TDINH DAY HOC a DRT VA'N Di Cau hdi Nhic lai khai niem hinh hop, hinh chdp Cau hdi Cho hinh lap phuong ABCDA'B'CD' a) Chiing minh cac canh ciia hinh lap phuong xuit phat tii mdt dinh vudng gdc vdi b) Cho canh ciia hinh lap phuong la a, tfnh dp dai dudng cheo ciia hinh lap phuong n isni MOI HOATDONCl I TOA DO CUA DIEM VA CUA VECTO He toa GV mo ta he true toa khong gian va neu cau hoi : HI Hai vecto i, j cd vudng gdc vdi hay khdng? H2 Vecto k cd vudng gdc vdi tat ca cac vecta thudc mat phing (Oxy) khdng? 42 • GV sir dung hinh 3.1 SGK va dat van de: H3 Hay dpc ten cac mat phing tpa dp H4 Hay ke' ten cac vecto dan vi H5 Cd the ed them mdt gdc tpa dp niia khac O hay khdng? H6 Hay neu cac tfnh chit ciia mat phing tpa dp, vecto don vi? -2 — -2 — -2 H7 Tfnh i = i.i, j = j j , k = k.k H8 Tfnh i.j,j.k, k.i • Thue hien ^ phiit Su dung hinh ve 3.2 GV cho HS len bang ve lai hinh va hudng din HS thuc hien z N Hoat ddng cua GV Cdu hoi Bieu dien OM theo OE va ON Cdu hdi Bieu dien OE theo OH va OK Hoat ddng cua HS Gffi y trd loi cdu hoi OM = OE + ON Ggi y trd loi cdu hdi OE=OH+OK 43 Cdu hoi Ggi y trd Idi cdu hoi Tim cac mdi quan he giiia cac OK = x.i OH = y.j,ON == z.k vecto ON OH va OK Cdu hdi Ggi y trd Idi cdu hdi Bieu dien OM theo i, j va k OM = x.i + y.j + z.k Toa dp cua diem GV sir dung hinh 3.2 va dat cau hdi: H9 Cho ba sd thuc x, y va z Cd bao nhieu diem M thda man OM = x.i + y.j + z.k HIO Cho OM = x.i + y.j + z.k Cd bao nhieu bd sd sd thuc x, y va z thda man he thiie tren • GV tra Idi va neu dinh nghia : Bd ba sd thuc (x; y z) thda mdn OM = x.i + y.j + z.k ggi la tga diem M vd ki hieu M (x ; y ; z) hoac M = (x ; y ; z) HI Cho M (0 ; ; 0) Hay chi M HI2 Cho M(0 ; ; 2) Hdi M thudc true nao ? HI3 Cho M(l ; ; 2) Hdi M thudc true nao ? H14 Cho M(l ; ; 0) Hdi M thudc true nao ? Toa dp vecto • GV neu dinh nghia : Trong khdng gian cho vecta a Bg ba sd (x ; y ; z) thda mdn a = x.i + y.j + z.k ggi la tga cua vecta a Ki hieu a(x;y;z) hoac a = (x;y;z) HI5 Vecta OM va diem M cd ciing tpa dp khdng? 44 • GV neu nhan xet SGK: Tga cua OM chinh Id tga cua M • Thue hien A2 phiit Su dung hinh ve 3.2 GV cho HS len bang ve lai hinh va hudng din HS thuc hien Hoat ddng cua GV Cdu hoi Tim tpa dp cua AB Cdu hoi Tim tpa dp ciia AC Cdu hoi Tim tpa dp ciia A C Cdu hdi Hoat ddng cua HS Ggi y trd Idi cdu hoi AB(a;0;0) Ggi y trd Idi cdu hoi AB(a;b;0) Gffi y trd Idi cdu hdi AB(a;b;c) Ggi y trd Idi cdu hdi Tim tpa dp ciia AM AM(-;b;c) HOATDQNC II BIEU THtrC TOA D O C U A CAC PHEP TOAN V E C T O • GV neu dinh If: Trong khdng gian Oxyz cho ba vecta a(ai;a2;a2) vab(bi ;b2;b3) fa cd : a + b = (aj +bi;a2 + b2;a3+b3) a - b = (a, - b p a - b ; a - b ) ka = (kai;ka2;ka3), dd k la mgt sdthuc • GV hudng din HS chiing minh dinh If tren H16 Hay so sanh cac tpa dp ciia a va b a = b • GV neu he qua 1: Hai vecta bdng thi cdc tga tuang img bdng HI7 Hay viet cac bieu thiic tpa dp cua he qua • GV neu he qua 2: Vecta cd cdc tga bdng HI8 Hay viet cac bieu thiic tpa dp ciia he qua • GV neu he qua 3: Hai vec ta cdng phuang thi mdi tga cua vec ta ndy bdng k Idn tga tuang Umg cda vec ta HI9 Hay vie't cac bieu thiic tpa dp cua he qua • GV neu he qua 4: Khi bie't tga ciia AvdB ta cd tga ciia AB bdng cdch lay mdi tga tuang img cda B trit di tga tuang img ciia A H20 Hay viet cac bieu thiic tpa dp cua he qua 46 • Trong sach GK khdng cd vf du nhung GV nen lay vf du minh hpa cho dinh If va he qua Vi day la kien thiic rat quan trpng Vfdu.ChoA(l ;1 ; 1),B(-1 ; 2; 3) va C (0 ; ;-2) a) Hay tim cac tpa dp ciia AB va AC b) Tim tpa dp ciia vec to 3AB c) Tfnh AC + 3AB • GV gpi mdt HS giai cau a Hoat ddng ciia GV Hoat ddng ciia HS Cdu hoi Ggi y trd Idi cdu hoi HS tu giai Tim AB Ggi y trd Idi cdu hoi Cdu hoi HS tu giai Tim AC • GV gpi tie'p HS thii tra Idi cau b va sau dd gpi HS thii ba lam cau c HOATDONC III TICH VO HUONG Bieu thurc toa cua tich vd hudng • GV ndu dinh If Trong khdng gian Oxyz cho ba vecta aia^;32;a2) vab(b, ;h2;h-^) ta cd: a.b = ajb2 + a2b2 + a3b3 • GV hudng din HS chiing minh dinh If tren tTng dung a) Do ddi cua vectff • GV neu cau hdi sau : H21 Trong boat ddng 2, hay tfnh dp dai AC 47 GV neu dinh nghia : Cho a(a| ;a2 ;a3) do ddi ciia vecta aki hieu va: 3.1 ~r 3,T ~r 3.-5 b) Khodng cdch gida hai diem H22 Cho A (XA ; yA ;ZA) va B (Xg ; y^ ;z^) Xac dinh AB Tfnh AB GV neu ket qua 2: Khodng cdch gida hai diem AB Id AB = AB = 7(XB-XA)^+(yB-yA)^+(ZB-ZA)^ c) Gdc giita hai vectff • GV neu cdng thiic tfnh gdc giiia hai vecto : Cho cdc vecta iij = (x,; yj; Zj), ta cd M2 = (-^2 ! }'2 > ^2) ^^ ^^ ^ ^^y >*' ^1-^2+yi>'2+^1^2 COS(M] , M2) = ^x\+y1 + z\ yjxj+yj+zl H23 Khi nao hai vecto vudng gdc vdi ? • GV neu he qua : M] J U2 o iii.ii2 = d Khi dd tdm mat cdu la diem I(-a ; -b ; -c )vd bdn kinh mat cdu la r = yja^ +b' +c^ -d H28 d phai thoa man deu kien gi de x^ + y" + z^ + 2ax + 2by + 2cz + d = la phuang trinh ciia mat ciu ? • GV cho HS thuc hien vi du SGK HOATDONCL TOM T ^ Bfit HOC Cho cac vecto u^ = ( x , ; y i ; z , ) , ^2 = (A^ ; y2 : '-2) va sd k y, ta cd : 1) »i = fh xi = X2, yi = y2 zi = ^2 2) », - U2 = (AI + x , : y, + y , ; ?i + -2) 3) ;ii - i?2 = (-^'i - ^'2; >'i - >''2; ^i - ^2) 4) kn^ = (A-A'i; ky^ ; fe,) 5) N|.U2 = Vj.vo + 3'iy2 + z,Z2 6) K | = V"i" = v-^T +>'i" +^\ 7) cos((7, iin) = "•"'"- —'•'V-T +.^T + - r ''^ vdi ii, ?t 0; U2 ^ \-^'2 +-^': + - : 8) M, -L ih i[...]... Tfnh diO, ia)) diO, ia)) |2. 0 - 2 0 - 0 + 31 ^2' +{-2f+{-lf Cdu hoi 2 TmhdiM,ia)) Ggi y trd Idi cdu hdi 2 diO, ia)) _ | 2 1 - 2 ( - 2 ) - 1 3 + 3| 4 ^22 +(_2f+(-1 )2 3 71 • 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...3 Cho hai diem Aix^ ; y^ ; z^) va Bixg; VB ' ^fi)1) AB = ixg-X/^;yg-y^;zB-z^) i 1 9 1 2) A 5 - ^ ( A : g - x ^ ) +{yB-yA) +{^B-^A) 4 Mat ciu tam I(a ; b ; c), ban kfnh r cd phuang trinh (x-fl )2+ (y-fc)^+(z-c )2= r2 2 2 2 •) - Phuong trinh x + y + z + 2ax + 2by + 2cz + d = 0 la phuofng trinh cua mat cau 2 2 2 •' khi va chi khi a + b + c > d Khi dd tam mat cau la diem I ( - a ; - b ; - c) va ban kfnh... I(-l ; 2 ; 3) (c) Ban kfnh ciia hinh cau la 2 (d) Ban kfnh ciia hinh cau la yf2 Trd Idi a b c d D S S D 52 U U D D Chon khang djnh diing trong cac cau sau: Cdu 5 Trong cac cap vecto sau, cap vecto ddi nhau la (a)a = ( l ; 2 ; - l ) , b = ( - l ; - 2 ; l ) ; (b)a = ( l ; 2 ; - l ) , b = ^ ( l ; 2 ; - l ) ; (c)a = ( - l ; - 2 ; l ) , b = ( - l ; - 2 ; l ) ; (d)a = ( l ; 2 ; - l ) , b = ( - l ; - 2 ; 0... l;z-l) AB = (-1 ;2; 1) va DC = AB Ta cd C (2; 0; 2) Tfnh toa dd A' bang each : AA' = DD' ta ed A'(3 ; 5 ; 6) Tuong tu ta cd B' (4 ; 6 ; -5), D'(3 ; 4 ; -6) 56 Bai 4 Hudng ddn Dua vao tfnh chat ciia tfch vo hudng hai vecto U\.U2 =x\X2+yiy2+h ^2 a) a.b = 6 b) c.d = - 2 1 Bai 5 Hirdng ddn Dua vao phuong trinh mat cau a) Phuang trinh mat cau dupe vie't dudi dang : ( x - 4 f + ( y - l ) 2 + z2=16 Tir dd ta... vecto Caua.Tacd 4a = {8; -20 ; 12) ; — b 3 ^ ' 0;- 2 I ; 3c-(3 ;21 ;6) 3' 3 Tii dd ta cd ke't qua caub Tacd ; 4b = (0;8;-4); -2c = ( - 2 ; - 1 4 ; - 4 ) Tilr dd ta cd ket qua Bai 2 Hirdng ddn Dua vao tfnh chat chit XQ - - ( x ^ + Xg + x^ ); y G = ^ ( y A + y B + y c ) ; ZG= gl^A+ZB+zc) Bai 3 Hudng ddn Dua vao tfnh chat cua phep toan toa dp Hai vecto bing nhau A(1 ;0;1) B(0 ; 1; 2) D(1 ; -1 ; 1 C'(4 ; 5... (P) Cdu hoi 2 Xac dinh m de (a) = (P) 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 69 cau c Hoat ddng cua HS Hoat ddng cua GV Cdu hdi 1 Neu dieu kien de (a) cit (P) Cdu hdi 2 Xac dinh m de (a) cit (P) 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... ( 2 ; l ; - 2 ) Gffi y trd Idi cdu hdi 2 AC = ( - 12; 6;0) Gffi y trd Idi cdu hdi 3 n = (l ;2 ;2) phing (ABC) 61 HOATDONC 2 II PHUONG TRINH TONG QUAT CUA MAT PHANG • GV neu va cho HS giai bai toan 1 (Sii dung hinh 3.5) M(x ;y ; x) Hoat ddng cua HS Hoat ddng cua GV Gffi y trd Idi cdu hdi 1 Cdu hdi I Tim toa dp vecto M Q M HS tu tfnh Cdu hdi 2 Gffi y trd Idi cdu hdi 2 MoM.n=0 M Q M va n cd quan he vdi nhau... C(z-Zo) = 0 • Tliuc hien ^ 2 trong 5 phut GV gpi ba HS len bang dien vecta phap tuye'n vao d trdng sau: HSl: (a) 4x - 2y - 6z + 7 = 0 4x + 2 y - 6 z + 7 = 0 4x + 2y -+6z + 7 = 0 4x + 2 y - 6 z - 7 = 0 VTPT:n HS2: (a) VTPTin 63 HS3: 4x - 2y - 6z - 7 = 0 (a) 4x +2y + 6z + 7 = 0 VTPT:n • Thuc hien ^ 3 trong 5 phiit Hoat ddng cua HS Hoat ddng ciia GV Cdu hdi 1 Xac dinh MN Cdu hdi 2 Xac dinh MP Cdu hdi 3 Xac... : 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 I -m -2 4 m +2 m -4 m -2 Caub Hoat ddng cua HS Hoat... - 4 ; 5 ) • (c)2b-a cdtoaddIa(5;-4;5) [] (d) Ca ba khing dinh tren deu sai [j Trd Idi a b c d D D s s Cdu5 Cho a = ( l ; 2 ; 3 ) , b = ( - 2 ; 3 ; - l ) Khi dd a + b c d t o a d d i a (a) a.b = 1 • (b)a.b=-l • (c)2b.a =2 [] (d) Ca ba khing dinh tren deu sai Q Trd Idi a b e d D S D S Cdu 4 Cho hinh cau cd phuong trinh : (x -1)^ + (y + 2f + (z + 3)^ = 2 (a) Tam ciia hinh cau la 1(1 ; -2 ; -3) (b) Tam ... M2 = (- ^2 ! } '2 > ^2) ^^ ^^ ^ ^^y >*' ^1- ^2+ yi> '2+ ^1 ^2 COS(M] , M2) = ^x+y1 + z yjxj+yj+zl H23 Khi nao hai vecto vudng gdc vdi ? • GV neu he qua : M] J U2 o iii.ii2 = 'i - >' '2; ^i - ^2) 4) kn^ = (A-A'i; ky^ ; fe,) 5) N|.U2 = Vj.vo + 3'iy2 + z,Z2 6) K | =... Viet phuong trinh mat ciu H.hoc 12/ 2 Ggi y trd Idi cdu hoi I a = 1, b = -2 vac = Ggi y trd loi cdu hdi r = Ggi y trd Idi cdu hdi (x-l )2+ (y + 2) 2 +(z-3)^ =25 49 H27 Hay neu mdt dang khac ciia phuong