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"Laboratoire 12 : Droite s-1" }}}{EXCHG {PARA 261 "" 0 "" {TEXT -1 52 "Par Claude St-Hilaire, c laude.sthilaire@videotron.ca" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "restart:with(linalg):with(student):" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 316 51 "Principales commandes utilis\351es dans ce laboratoire" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "vector, matrix, evalm, equate d e la biblioth\350que student, genmatrix, norm, gausselim, backsub, dot prod, angle, " }}{PARA 0 "" 0 "" {TEXT -1 61 "plots, disk de la biblio th\350que plottools, seq, display, subs." }}{PARA 0 "" 0 "" {TEXT -1 69 "Note : S\351lectionner un mot et utiliser l'aide pour plus d'infor mation" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 323 49 "Repr\351sentation gr aphique de la droite vectorielle" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "with(plots):with(plottools):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "Soit A = (1,2,3) et B = (0,3,1), 2 points d\351finissant une dr oite D." }}{PARA 0 "" 0 "" {TEXT -1 72 "Illustrons que le vecteur OP = OA + k*AB a son extr\351mit\351 sur la droite D" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "OA:=vector([1,2,3]):OB:=vector([0,3,1]):AB:=e valm(OB-OA):OP:=evalm(OA+k*AB):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "Repr\351sentons OP(rouge) = OA(brun) + k*AB(bleu), le point P en n oir et la droite en magenta" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 668 "k:=-2.0:\nvecteuruOA:=arrow([0,0,0],OA,0.2,0.6,0.2,color=brown): \nvecteurOP:=arrow([1,2,3],evalm(k*AB),0.2,0.6,0.2,color=blue):\nfl \350cheOP:=arrow([0,0,0],[OP[1],OP[2],OP[3]],0.2,0.6,0.2,color=red):\n pointP:=sphere([OP[1],OP[2],OP[3]],0.1):pointA:=sphere([OA[1],OA[2],OA [3]],0.1):pointB:=sphere([OB[1],OB[2],OB[3]],0.1):\nvecteurAB:=arrow([ OA[1],OA[2],OA[3]],AB,0.2,0.6,0.2,color=green):\ndroite:=line([1,2,3]- (abs(k)+1)*[-1,1,-2],[1,2,3]+(abs(k)+1)*[-1,1,-2],color=magenta):\nlet tres:=textplot3d([[1,2,3,`A`],[0,3,1,`B`],[OP[1],OP[2],OP[3],`P`]],col or=red):\ndisplay(vecteurAB,vecteuruOA,vecteurOP,fl\350cheOP,pointP,po intA,pointB,lettres,droite,axes=normal,orientation=[35,110]);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "OP(rouge) = OA(brun) + (k*AB)(bleu ) et AB en vert." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 322 22 "Changez la v aleur de k" }{TEXT -1 101 " (entre -3 et 3 pour un bon graphique) et f aites des retours pour obtenir un autre point P v\351rifiant " }}} {EXCHG {PARA 0 "" 0 "" {TEXT 324 11 "Animation :" }{TEXT -1 92 " Les \+ points P (en noir) v\351rifiant OP = OA + k*AB, sont sur une droite. ( k varie de -2 \340 2) " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 795 " k:='k':droite2:=line([1,2,3]-2.5*[-1,1,-2],[1,2,3]+2.5*[-1,1,-2],color =magenta):\nw1:=seq(sphere(convert(subs(k=t/5,op(OP)),list),0.15),t=-1 0..10):\nsphere2:=display(w1,insequence=true):\nanimvecteurOP:=display (seq(subs(k=t/5,fl\350cheOP),t=-10..10),insequence=true):\nanimOp:=dis play(seq(arrow([0,0,0],subs(k=t/5,op(OP)),0.2,0.6,0.2,color=red),t=-10 ..10),insequence=true):\nlettres2:=textplot3d([[1,2,3,`A`],[0,3,1,`B`] ],color=red):\nZ:=subs(k=t/5,op(OP)):\nanimlettreP:=display(seq(displa y(textplot3d([[Z[1],Z[2],Z[3],`P`]],color=red)),t=-10.1..10),insequenc e=true):\nv:=evalm(k*AB):\nanimAP:=display(seq(arrow([1,2,3],subs(k=t/ 5,op(v)),0.2,0.6,0.2,color=blue),t=-10.1..10),insequence=true):\ndispl ay(droite2,vecteuruOA,lettres2,sphere2,animOp,animAP,animlettreP,vecte urAB,axes=normal,orientation=[35,80]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 325 19 "Animez le graphique" }{TEXT -1 86 " illustrant que les p oints P (en noir) v\351rifiant OP = OA + k*AB, sont sur une droite. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 259 " " 0 "" {TEXT 270 65 "\311quations vectorielles, param\351triques et sy m\351triques d'une droite" }{TEXT -1 1 " " }{TEXT 271 6 "de R^3" } {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT 301 7 "Rappels" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "Un " }{TEXT 306 7 "vecteur" }{TEXT -1 49 " est repr\351sent\351 par vector([a1,a2,...,an]) et un " } {TEXT 307 5 "point" }{TEXT -1 30 " par une liste [a1,a2,...,an] " }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 232 "Un vecteur peut aussi \352tre rep r\351sent\351 par une liste [a1,a2,..,an]. Cependant certaines command es Maple n\351cessitent vector (norm, basis, GramSchmidt, etc) alors q ue d'autres sont compatibles (angle, dotprod, crossprod dans R^3, etc) ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 152 "Remarque: le vecteur OA et \+ le point A sont le m\352me n-uple dans R^n, o\371 O est le point [0,0, ...,0]. On convient que le vecteur OA repr\351sente le point A. " }} {PARA 0 "" 0 "" {TEXT -1 166 "Dans Maple, evalm(A) transforme un point A en vecteur : evalm(A) repr\351sente le vecteur OA. De m\352me, si A et B sont des points alors evalm(B-A) repr\351sente le vecteur " } {TEXT 326 12 "AB = OB - OA" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "Les deux situations suivantes d\351finissent une droite" }}{PARA 0 "" 0 "" {TEXT -1 21 "a) Deux points A et B" }}{PARA 0 "" 0 " " {TEXT -1 91 "b) Un point A et un vecteur v parall\350le \340 la droi te, appel\351 vecteur directeur de la droite." }}{PARA 0 "" 0 "" {TEXT -1 48 "Le cas a) se ram\350ne \340 b) en prenant v = OB - OA " } }{PARA 0 "" 0 "" {TEXT -1 154 "Une droite peut aussi \352tre d\351fini e par l'intersection de 2 surfaces, en particulier par l'intersection \+ de 2 plans (cas vu au laboratoire 15 Plans-droites)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "Soit A(a1,a2,a3) un point d'une droite D et v = (a,b,c) un vecteur parall\350le \340 cette droite" }}{PARA 0 "" 0 "" {TEXT -1 60 "Soit P(x,y,z) un point quelconque de D et O = (0,0,0) alo rs " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 302 48 "L'\351quation vectorielle de D est : OP = OA + t*v " }{TEXT -1 24 "o\371 t parcourt les r\351el s." }}{PARA 0 "" 0 "" {TEXT -1 7 "On a : " }{TEXT 305 31 "(x,y,z) = (a 1,a2,a3) +t*(a,b,c)" }{TEXT -1 4 ".(1)" }}{PARA 0 "" 0 "" {TEXT -1 53 "Chaque vecteur OP(x,y,z) repr\351sente le point P(x,y,z)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 " On tire de (1) : " }{TEXT 303 80 "Les \+ \351quations param\351triques de D sont : x = a1+t*a, y = a2 + t*b et \+ z = a3 + t*c" }{TEXT -1 4 " (2)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "Remarque 1 : Si le param\350tre t varie de t1 \340 t2, on obtient un segment de droite au lieu d'une droite" }}{PARA 0 "" 0 "" {TEXT -1 178 "Remarque 2 : Les \351quations x = f(t), y = g(t) et z = h(t) d \351finissent une courbe dans l'espace. Si les fonctions f(t), g(t) et h(t) sont lin\351aires alors la courbe est une droite." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "Si a<>0, b<>0 et c<>0 alors en isolant t \+ dans (2) :" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 304 66 "Les \351quations s ym\351triques de D sont (x-a1)/a = (y-a2)/b = (z-a3)/c" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "Remarque : Si la droite D est d\351finie par 2 points A et B alors l'\351quation vectorielle de D est OP = OA+ t*(OB-OA) " }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 277 7 "Exemple" }{TEXT -1 1 " " }{TEXT 276 2 "1)" }{TEXT -1 54 " : Soit D, la droite passant \+ A(3,-7,12) et B(12,-5,7) " }}{PARA 0 "" 0 "" {TEXT 274 73 "a) Trouver \+ les \351quations vectorielles, param\351triques et sym\351triques de D ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "OA:=vector([3,-7,12]); OB:=vector([12,-5,7]);v:=evalm(OB-OA);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "eqvectD:=vector([x,y,z])=evalm(OA)+k*evalm(v);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 285 33 "Les \351quations param\351triques de D " }{TEXT -1 5 "sont " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "equate(lhs(eqvectD),rhs(eqvectD));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "En isolant k, on obtient " }{TEXT 286 4 "les " }{TEXT 287 21 "\351quations sym\351triques" }{TEXT -1 38 " de D : (x-3)/9 = ( y+7)/2 = (z-12)/-5." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 273 58 "b) le poi nt C(-18.6,-11.8,24) appartient-il \340 la droite D?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "k:='k':with(student):equate([-18.6,-11.8, 24],[3+9*k, -7+2*k, 12-5*k]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve(%,k);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Le point C a ppartient \340 la droite D." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "Un e autre solution consisterait \340 v\351rifier si AC //AB, c'est-\340- dire AC = r*v" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 272 49 "c) le point D(5 ,7,-10) appartient-il \340 la droite?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "equate([5,7,-10],[3+9*k, -7+2*k, 12-5*k]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve(%,k);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Pas de solution, donc le point D n'appart ient pas \340 la droite. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "L' \351quation vectorielle du segment de droite passant par les points A \+ et B peut s'\351crire : " }}{PARA 0 "" 0 "" {TEXT -1 71 "(x,y,z) = A + k*(B-A) o\371 0 <= k <= 1 ou encore (x,y,z) = (1-k)*A + k*B" }}} {EXCHG {PARA 256 "" 0 "" {TEXT 275 10 "Exemple 2)" }{TEXT -1 88 " Trou ver les \351quations param\351triques du segment de droite AB o\371 A( 2,-3,5) et B( 9,-9,-5)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 " L'\351 quation vectorielle est (x,y,z) = OA + k*(OB-OA)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "OA:=[2,-3,5];OB:=[9,-9,-5];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "eqvectD:=evalm([x,y,z])=evalm(OA)+k*evalm (OB-OA);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "Les \351quations para m\351triques de D : x = 2 + 7k, y = -3 - 6k, z = 5 - 10k" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 298 38 "Droites parall\350les, perpendiculaires, " }{TEXT 299 36 "droi tes s\351cantes et droites gauches " }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Deux droites de R^3 peuvent \352tre, soit : " }}{PARA 0 "" 0 " " {TEXT -1 2 "a)" }{TEXT 289 11 " parall\350les" }{TEXT -1 70 ", si le urs vecteurs directeurs v1 et v2 sont parall\350les i.e. v1 = k*v2" }} {PARA 0 "" 0 "" {TEXT -1 3 "b) " }{TEXT 290 9 "s\351cantes " }{TEXT -1 44 "(concourantes), si elles ont un point commun" }}{PARA 0 "" 0 " " {TEXT -1 3 "c) " }{TEXT 291 7 "gauches" }{TEXT -1 45 ", si elles ne \+ sont ni parall\350les, ni s\351cantes" }}{PARA 0 "" 0 "" {TEXT -1 3 "d ) " }{TEXT 292 10 "confondues" }{TEXT -1 55 " (la m\352me droite), si \+ elles sont parall\350les et s\351cantes" }}}{EXCHG {PARA 258 "" 0 "" {TEXT 262 39 "A) droites parall\350les, perpendiculaires" }{TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 256 10 "Rappels : " }{TEXT 320 84 "droites parall\350les, perpendiculaires, angle entre 2 droites, produ it scalaire, norme" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "Soit D1 et D 2, 2 droites de vecteurs directeurs v1 et v2 respectivement." }}{PARA 0 "" 0 "" {TEXT 268 52 "D1 et D2 sont parall\350les si v1 // v2 i.e. v 1 = k v2 " }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 269 44 "D1 et D2 son t perpendiculaires si v1ov2 = 0 " }}{PARA 0 "" 0 "" {TEXT 257 27 "L'an gle a entre 2 vecteurs " }{TEXT -1 10 "v1 et v2 :" }}{PARA 0 "" 0 "" {TEXT -1 49 "a) v\351rifie cos(a) = (v1ov2) / (||v1|| ||v2||) o\371" }{TEXT 293 13 " 0<= a <= Pi " }{TEXT -1 9 "radians. " }}{PARA 0 "" 0 " " {TEXT -1 61 "b) est donn\351 en radians, par la commande Maple : ang le(v1,v2)" }}{PARA 0 "" 0 "" {TEXT 258 29 "L'angle t entre les 2 droit es" }{TEXT -1 25 " D1 et D2 est donn\351 par :" }}{PARA 0 "" 0 "" {TEXT -1 10 "cos(t) = (" }{TEXT 259 14 "valeur absolue" }{TEXT -1 31 " (v1ov2)) / (||v1|| ||v2||) o\371 " }{TEXT 294 15 "0<= t <= Pi/2 " } {TEXT -1 8 "radians." }}{PARA 0 "" 0 "" {TEXT -1 56 "Remarque 1: L'an gle t entre D1 et D2 = angle(v1,v2) si " }{TEXT 295 15 "0<= t <= Pi/2 " }{TEXT -1 50 "et l'angle t entre D1 et D2 = Pi -angle(v1,v2) si " } {TEXT 296 14 "Pi/2< t <= Pi" }}{PARA 0 "" 0 "" {TEXT -1 98 "Remarque \+ 2: L'angle entre deux droites est d\351fini m\352me si les 2 droites n e sont pas concourantes. " }}{PARA 0 "" 0 "" {TEXT -1 41 "Dans Maple : la norme d'un vecteur v est " }{TEXT 260 10 "norm(v,2)," }{TEXT -1 35 " le produit scalaire de u et v est " }{TEXT 261 12 "dotprod(u,v)" }{TEXT -1 38 " et les angles sont donn\351s en radians." }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 278 10 "Exemple 3)" }{TEXT -1 55 " Soit D1 passa nt par le point A1(4,6,-2) et B1(-4,12,8)" }}{PARA 0 "" 0 "" {TEXT -1 71 "D2 passant par le point A2(2,7,0) et parall\350le au vecteur v2 = \+ (-4,3,5)" }}{PARA 0 "" 0 "" {TEXT -1 48 "D3 passe par le point A3(6,-7 ,11) et B3(1,-2,-8)" }}{PARA 0 "" 0 "" {TEXT 282 2 "a)" }{TEXT -1 41 " Trouver la position relative de D1 et D2" }}{PARA 0 "" 0 "" {TEXT 283 2 "b)" }{TEXT -1 31 " Trouver l'angle entre D1 et D3" }}{PARA 0 " " 0 "" {TEXT 288 2 "c)" }{TEXT -1 65 " Trouver l'angle entre v1 et v3 \+ o\371 v1 = OB1-OA1 et v3 = OB3 - OA3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:with(linalg):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "a) Trouver la position relative de D1 et D2" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "OA1:=vector([4,6,-2]);OB1:=vector([-4,12,8]); v1:=evalm(OB1-OA1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "D1 // D2 \+ car v1 = 2 v2. D1 et D2 sont-elles confondues ? V\351rifions si le poi nt A2(2,7,0) de D2 appartient \340 D1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "eqD1:=vector([x,y,z])=evalm(OA1+k*v1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "with(student,equate):equate([2,7,0] ,[4-8*k, 6+6*k, -2+10*k]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve(%,k);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "Pas de solutio n donc A2 n'appartient pas \340 D1. D1 et D2 sont des parall\350les no n confondues" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "b) Trouver l'ang le entre D1 et D3" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "OA3:=vector([6 ,-7,11]);OB3:=vector([1,-2,-8]);v3:=evalm(OB3-OA3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "L'angle entre les droites D1 et D3, de vecteurs directeurs v1,v3, cos(t) = (" }{TEXT 279 14 "valeur absolue" }{TEXT -1 28 "(v1ov3)) / (||v1|| ||v3||) " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "evalf(arccos(abs(dotprod(v1,v3))/(norm(v1,2)*norm(v3, 2))));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "L'angle entre v1 et v3 \+ est d\351fini par : cos(t) = (v1ov3) / (||v1|| ||v3||) " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "evalf(angle(v1,v3));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 121 "Dans ce cas-ci, l'angle entre les droite s D1 et D3 est diff\351rent de l'angle entre les vecteurs directeurs d e ces droites." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 280 10 "Exemple 4)" }{TEXT -1 51 " Soit A (2,-5,6), B(5,-7,6) et C(7,-4,6), 3 points " }}}{EXCHG {PARA 0 "" 0 " " {TEXT 281 1 " " }{TEXT -1 43 "Le triangle ABC est-il isoc\350le ? re ctangle?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "OA:=vector([2,- 5,6]);OB:=vector([5,-7,6]);OC:=vector([7, -4, 6]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "AB:=evalm(OB-OA):AC:=evalm(OC-OA);BC:=evalm (OC-OB):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "norm(AB,2);norm (BC,2);norm(AC,2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Le triangle est isoc\350le, AB = BC. Est-il rectangle? " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "dotprod(AB,BC);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "ABC est un triangle isoc\350le et rectangle." }}}{EXCHG {PARA 257 "" 0 "" {TEXT 267 54 "B) Droites s\351cantes (concourantes) \+ et droites gauches " }}}{EXCHG {PARA 256 "" 0 "" {TEXT 263 64 "Deux dr oites D1 et D2, non parall\350les, sont s\351cantes ou gauches." }}} {EXCHG {PARA 256 "" 0 "" {TEXT -1 84 "Soit A un point quelconque (mobi le) sur D1 et B, un point quelconque (mobile) sur D2" }}{PARA 256 "" 0 "" {TEXT -1 63 " D1 et D2 ont un point d'intersection si A = B a une solution. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:with (linalg):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 284 10 "Exemple 5)" }{TEXT -1 59 " Trouver, s'il existe, le point d'intersection de D1 et D2 " }} {PARA 0 "" 0 "" {TEXT -1 33 "D1 : (x-2)/3 = (y-3)/2 = (z-1)/3 " }} {PARA 0 "" 0 "" {TEXT -1 32 "D2 : x-2 = (y-1/3)/-2 = (z-3)/3 " }}} {EXCHG {PARA 0 "" 0 "" {TEXT 300 25 "Solution comme \340 la main " } {TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 64 "1) On trouve les \351qu ations param\351triques de D1 et D2 en posant :" }}{PARA 0 "" 0 "" {TEXT -1 28 " (x-2)/3 = (y-3)/2 = (z-1)/3" }{TEXT 317 4 " = k" }{TEXT -1 31 " et x-2 = (y-1/3)/-2 = (z-3)/3 " }{TEXT 318 3 "= r" }}{PARA 0 " " 0 "" {TEXT -1 104 "b) On r\351soud les 3 \351quations obtenues en \+ \351galisant les abscisses, les ordonn\351es et les cotes de D1 et D2 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "a) D1: x = 2 + 3k, y = 3 + 2k , z = 1 + 3k ; D2 : x = 2 + r, y = 1/3 -2r, z = 3 + 3r" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "Il y a un point d\"intersection lorsque l es abscisses, les ordonn\351es et les cotes sont \351gales" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 "b)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "e q1:=2+3*k=2+r;eq2:=3+2*k=1/3-2*r;eq3:=1+3*k=3+3*r;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "M:=genmatrix([eq1,eq2,eq3],[r,k],flag);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "gausselim(M);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "backsub(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "r = -1, k = -1/3. Le point d'intersection est :" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "subs(r=-1,\{x = 2 + r, y = 1 /3 -2*r, z = 3 + 3*r\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 321 16 "Autr e solution :" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "Soit A un point q uelconque sur D1 et B un point quelconque sur D2 :" }{TEXT 264 1 " " } }{PARA 0 "" 0 "" {TEXT -1 55 "A = [2+3*k, 3+2*k, 1+3*k] et B:=[2+r, 1 /3-2*r, 3+3*r];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "A:=[2+3*k,3+2*k,1+3*k];B:=[2 +r,1/3-2*r,3+3*r];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "Le point d' intersection est obtenu, s'il existe, lorsque A = B" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "with(student):eq:=equate(A,B);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "rk:=solve(eq,\{k,r\});" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "V\351rification :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "pointA:=subs(rk,A);pointB:=subs(rk, B);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "S'il n'y a pas de point d' intersection, on obtient un syst\350me de 3 \351quations incompatibles " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 315 7 "Exemple" }{TEXT -1 1 " " } {TEXT 314 3 "6) " }{TEXT -1 57 "Trouver, s'il existe, le point d'inter section de D1 et D2" }}{PARA 0 "" 0 "" {TEXT -1 36 " D1 : (x-2)/3 = (y -3)/2 =(z-1)/3 = k" }}{PARA 0 "" 0 "" {TEXT -1 33 "D2 : x-2 = (y-4)/-2 = (z-3)/3 = r" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "Soit A sur D1 e t B sur D2 :" }{TEXT 265 1 " " }{TEXT -1 49 "A = [2+3*k,3+2*k,1+3*k] \+ et B:=[2+r,4-2*r,3+3*r];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "A :=[2+3*k,3+2*k,1+3*k];B:=[2+r,4-2*r,3+3*r];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "with(student,equate):" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 16 "eq:=equate(A,B);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "rk:=solve(eq,\{k,r\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "Les deux droites n'ont pas de point d'intersection alors \+ elles sont, soit gauches, soit parall\350les" }}}{EXCHG {PARA 0 "" 0 " " {TEXT 266 14 "V\351rification :" }{TEXT -1 164 " On peut v\351rifier si on a v1 //v2 ou avec : Les droites sont gauches <=> det([AB,v1,v2] ) <> 0, o\371 A est un point de D1 et B est un point de D2, v1 // D1 e t v2 // D2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Prenons A(2,3,1) et B(2,4,3)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "v1:=vector([3, 2,3]);v2:=vector([1,-2,3]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "v1 n'est pas parall\350le \340 v2 donc les droite sont gauches" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "OA:=vector([2,3,1]);OB:=vect or([2,4,3]);AB:=evalm(OB-OA);v1:=vector([3,2,3]);v2:=vector([1,-2,3]); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "det(matrix([AB,v1,v2])) ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 297 10 "Exemple 7)" }{TEXT -1 51 " \+ Soit A(2,-5,6), B(5,-7,6) et C(7,-4,6), 3 points " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 181 "Trouver le point d'intersection des m\351dianes \+ du triangle ABC sachant que les m\351dianes dans un triangle se rencon trent en un point situ\351 au 2/3 de chaque m\351diane \340 partir du \+ sommet. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "Le point P est l'inte rsection des m\351dianes si AP = 2/3 AM o\371 M est le point milieu de BC donc " }{TEXT 319 17 "OP = OA + 2/3 AM " }}{PARA 0 "" 0 "" {TEXT -1 75 "Remarque : le point milieu de BC est M = (1/2)*(B+C) ou OM = ( 1/2)*(OB+OC)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "OA:=vector( [2,-5,6]);OB:=vector([5,-7,6]);OC:=vector([7,-4,6]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "OM:=evalm((1/2)*(OB+OC));# M est le point milieu de BC" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "AM:=evalm( OM-OA);# la m\351diane AM. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "OP:=evalm(OA+(2/3)*AM);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "V \351rifions pour les 2 autres m\351dianes" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 54 "ON:=evalm((1/2)*(OA+OC));# N est le point milieu de AC" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "BN:=evalm(ON-OB);# l a m\351diane BN" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "OP2:=eva lm(OB+(2/3)*BN);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "P = P2 car OP = OP2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "OQ:=evalm((1/2)*( OA+OB));# Q est le point milieu de AB" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "CQ:=evalm(OQ-OC);# la m\351diane CQ" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "OP3:=evalm(OC+(2/3)*CQ);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 130 "P = P2 = P3. Les m\351dianes se coupent \+ en un m\352me point qui est, dans chacun des cas, situ\351 au 2/3 de l a m\351diane \340 partir du sommet." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "V\351rifions avec la formule OP = (1/3)(OA + OB + OC) o\371 O e st l'origine (0,0,0)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "OP: =evalm((1/3)*(OA+OB+OC));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 328 9 "Exercices" }{TEXT 329 3 " \+ 1 " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:with(linalg): " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 332 5 "No 1)" }{TEXT 333 9 " Trouver " }{TEXT -1 123 " les \351quations vectorielles, param\351triques et \+ sym\351triques de la droite passant par les points A( 12,-33,7) et B(7 , 22,19) " }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 345 32 " Espace de travai l de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 330 5 "No 2)" } {TEXT 331 1 " " }{TEXT -1 24 "Voici 2 droites D1 et D2" }}{PARA 0 "" 0 "" {TEXT -1 58 "D1 passe par le point (2,16,17) et est // \340 v1 = [1,-4,2]" }}{PARA 0 "" 0 "" {TEXT -1 52 "D2 passe par le point (3,-1 2,-23) et est // \340 v2 = " }{XPPEDIT 18 0 "vector([-2, 24, 38])" "6# -%'vectorG6#7%,$\"\"#!\"\"\"#C\"#Q" }}{PARA 0 "" 0 "" {TEXT 336 2 "a) " }{TEXT -1 31 " Trouver l'angle entre v1 et v2" }}{PARA 0 "" 0 "" {TEXT 337 2 "b)" }{TEXT -1 49 " Trouver l'angle entre D1 et D2. Compar er avec a)" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 346 32 " Espace de travai l de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 334 5 "No 3)" } {TEXT 335 1 " " }}{PARA 0 "" 0 "" {TEXT -1 36 "D1 : (x.y.z) = (2,3,-5) + k(-2,-5,a)" }}{PARA 0 "" 0 "" {TEXT -1 38 "D2 : x = 2r, y = 8 + 6r, z = -12 + 7r." }}{PARA 0 "" 0 "" {TEXT -1 61 "a) Trouver la valeur de a pour que D1 et D2 soient parall\350les" }}{PARA 0 "" 0 "" {TEXT -1 67 "b) Trouver la valeur de a pour que D1 et D2 soient perpendiculaire s" }}{PARA 0 "" 0 "" {TEXT -1 108 "c) Trouver la valeur de a pour que \+ D1 et D2 aient un point d'intersection et donner le point d'intersecti on." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 347 32 " Espace de travail de l' \351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 340 5 "No 4)" } {TEXT -1 49 " D1 passe par les points (4,-7,1 ) et (12,9,-43)." }} {PARA 0 "" 0 "" {TEXT 341 3 "a) " }{TEXT -1 120 "Trouver 2 droites D2 \+ et D3 perpendiculaires \340 D1 et passant par le point M(-2,5,9). De p lus D2 est perpendiculaire \340 D3 " }}{PARA 0 "" 0 "" {TEXT 342 2 "b) " }{TEXT -1 126 " Trouver 2 droites D2 et D3 perpendiculaires \340 D1 \+ et passant par le point M(-2,5,9). De plus D2 n'est pas perpendiculair e \340 D3" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 348 32 " Espace de travail de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 338 5 "No 5)" } {TEXT 339 50 " Soit 3 points A(2,6,13), B(-5,6,11) et C(7,3,-6)." }} {PARA 0 "" 0 "" {TEXT 343 3 "a) " }{TEXT -1 34 "Trouver les angles du \+ triangle ABC" }}{PARA 0 "" 0 "" {TEXT 344 3 "b) " }{TEXT -1 49 "V\351r ifier que la somme des angles donne Pi radians" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 349 32 " Espace de travail de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}} {SECT 1 {PARA 3 "" 0 "" {TEXT 310 26 "Trajectoires et collisions" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "restart:with(plots):" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 203 "Soit 2 mobiles qui se d\351placen t en fonction du temps t sur 2 droites (ou en g\351n\351ral, sur 2 cou rbes). \253\034Les trajectoires se coupent-elles? \273 et \253les mobi les se frappent-ils?\273 sont 2 probl\350mes diff\351rents. " }}} {EXCHG {PARA 0 "" 0 "" {TEXT 311 11 "Exemple 1 :" }{TEXT -1 146 " Deux mobiles M1 et M2 se d\351placent chacun en fonction du temps t selon \+ une trajectoire lin\351aire, M1 sur D1 et M2 sur D2. Y aura-t-il colli sion? " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "D1 : (x,y) = (-5,0) + \+ t(1,1)" }}{PARA 0 "" 0 "" {TEXT -1 31 "D2 : (x,y) = (6,-2) + t(-1,1.2) " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "Soit A sur D1 et B sur D2 :" }{TEXT 327 1 " " }{TEXT -1 34 "A = [-5+t,t] et B = [6-t,-2+1.2*t]" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "with(student):equate([-5+t,t ],[6-t,-2+1.2*t]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve (%,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "Maple ne r\351pond rien donc " }{TEXT 308 26 "il n'y a pas de collision." }{TEXT -1 59 " Tra \347ons le graphique avec l'animation des mobiles M1 et M2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 340 "droite1:=plot([-5+t,t,t=0..10],col or=blue):droite2:=plot([6-t,-2+1.2*t,t=0..10]):\nwith(plottools):\nM1: =seq(disk([-5+t,t],0.2,color=black),t=0..10):\nM2:=seq(disk([6-t,-2+1. 2*t],0.2,color=green),t=0..10):\ntrajet1:=display(M1,insequence=true): \ntrajet2:=display(M2,insequence=true):\ndisplay(trajet1,trajet2,droit e1,droite2,scaling=constrained);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "Cliquer sur l'animation et sur << pour ralentir l'animation" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "Trouvons le point d'intersection des trajectoires. Les 2 mobiles passent en ce point en des temps diff\351rents t1 et t2." }} {PARA 0 "" 0 "" {TEXT -1 91 "C'est pourquoi on va utiliser des param \350tres diff\351rents t1, t2, pour repr\351senter le temps. " }} {PARA 0 "" 0 "" {TEXT -1 45 "Red\351finissons D1 et D2, avec ces param \350tres. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "D1 : (x,y) = (-5,0) + t1(1,1)" }}{PARA 0 "" 0 "" {TEXT -1 32 "D2 : (x,y) = (6,-2) + t2(-1 ,1.2)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "with(student):equa te([-5+t1,t1],[6-t2,-2+(6/5)*t2]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "solve(%,\{t1,t2\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "subs(%,[-5+t1,t1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {XPPEDIT 18 0 "[1/11, 56/1 1];" "6#7$*&\"\"\"F%\"#6!\"\"*&\"#cF%F&F'" }{TEXT -1 87 " est le point d'intersection. M1 y passe au temps t1 = 56/11 et M2 au temps t2 = 6 5/11" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 309 71 "Si F(t) et G(t) repr\351 sentent 2 trajectoires en fonction du temps alors " }}{PARA 260 "" 0 " " {TEXT -1 53 "A) Pour trouver les collisions, on r\351soud F(t) = G(t )" }}{PARA 260 "" 0 "" {TEXT -1 67 "B) Pour trouver les points d'inter sections, on r\351soud F(t1) = G(t2)" }}{PARA 0 "" 0 "" {TEXT -1 40 "R emarque : Si t1 = t2 on a une collision" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 312 11 "Exemple 2 : " }{TEXT -1 74 " Trouver les points d'intersections et les collisions \+ les 2 trajectoires :" }}{PARA 0 "" 0 "" {TEXT -1 45 "F(t):=[-2+2*t,25- 6*t];G(t):=[-3+9/4*t,2-t/4];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "R \351solvons F(t1) = G(t2)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "F(t1):=[-2+2*t1,25-6*t1];G(t2):=[-3+9/4*t2,2-t2/4];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "with(student):equate(F(t1),G(t2)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "solve(%,\{t1,t2\});" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "Au temps t = 4, il y a collision. Trouvons le point d'intersection" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "subs(%,\{F(t1),G(t2)\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "La collision a lieu au point (6,1)" }}}}{SECT 1 {PARA 3 " " 0 "" {TEXT 350 11 "Exercices 2" }}{EXCHG {PARA 0 "" 0 "" {TEXT 351 5 "No 1)" }{TEXT -1 82 " Soit 3 mobiles dans l'espace se d\351pla\347a nt sur des droites en fonction du temps t " }}{PARA 0 "" 0 "" {TEXT -1 31 "M1 : F(t) = (-5 + 2t, t, -5+5t)" }}{PARA 0 "" 0 "" {TEXT -1 60 "M2 : G(t) = (-4 - t, 14 + 4t, 5)\nM3 : H(t) = (-2+t, t, 4+2t)" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "\300 midi, au temps t = 0 heure, d ire, " }}{PARA 0 "" 0 "" {TEXT 355 3 "a) " }{TEXT -1 103 "s'il y a eu, ou s'il y aura une collision entre les mobiles M1, M2 et si leurs tra jectoires se coupent." }}{PARA 0 "" 0 "" {TEXT 356 3 "b) " }{TEXT -1 36 "M\352me question qu'en a) pour M1 et M3" }}{PARA 0 "" 0 "" {TEXT 357 2 "c)" }{TEXT -1 37 " M\352me question qu'en a) pour M2 et M3" }}} {SECT 1 {PARA 3 "" 0 "" {TEXT 358 32 " Espace de travail de l'\351tudi ant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 352 5 "No 2)" }{TEXT -1 169 " Soit \+ D1 la droite passant par les points A1(-5,-12,3) et B1(16,72,45). Un m obile M1 part du point A1 dans la direction de B1 \340 la vitesse unif orme sqrt(21) unit\351s/sec. " }}{PARA 0 "" 0 "" {TEXT -1 128 "Soit D2 , la droite passant par les points A2(3,-12,-23) et B2(-3,60,91). Un m obile M2 part du point A2 dans la direction de B2. " }}{PARA 0 "" 0 " " {TEXT 353 2 "a)" }{TEXT -1 69 " \300 quelle vitesse uniforme doit al ler M2 pour qu'il y ait colision ? " }}{PARA 0 "" 0 "" {TEXT 354 2 "b) " }{TEXT -1 42 " Apr\350s combien de temps se frapperont-ils?" }}} {SECT 1 {PARA 3 "" 0 "" {TEXT 359 32 " Espace de travail de l'\351tudi ant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}}{MARK "2 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }