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" "6#-%$SumG6$*&&%\"uG6#%\"iG\"\"\"&%\"vG6#F*F+/F*;F+\"\"$" }}{PARA 0 "" 0 "" {TEXT -1 42 "D\351finition du produit scalaire dans R^n : " }} {PARA 0 "" 0 "" {TEXT -1 54 "Soit u = (u1,u2,....,un) et v = (v1,v2,.. ..,vn) alors " }{TEXT 274 31 "uov = u1v1 + u2v2 + ....+ unvn " }{TEXT -1 2 "= " }{XPPEDIT 18 0 "Sum(u[i]*v[i],i = 1 .. n)" "6#-%$SumG6$*&&% \"uG6#%\"iG\"\"\"&%\"vG6#F*F+/F*;F+%\"nG" }}{PARA 0 "" 0 "" {TEXT -1 12 "Avec Maple :" }{TEXT 275 14 " dotprod(u,v);" }}}}{SECT 1 {PARA 260 "" 0 "" {TEXT -1 8 "Exemples" }}{EXCHG {PARA 0 "" 0 "" {TEXT 279 11 "Exemple 1 :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "u:=vecto r([2,-3,5]);v:=vector([5,9,1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "dotprod(u,v);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 281 2 "a)" } {TEXT -1 12 " Calculons, " }{TEXT 278 16 "comme \340 la main," }{TEXT -1 25 " le produit scalaire uov " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "2*5+(-3)*9+5*1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 282 2 "b)" }{TEXT -1 6 " Avec " }{XPPEDIT 18 0 "Sum(u[i]*v[i]);" "6#-%$Sum G6#*&&%\"uG6#%\"iG\"\"\"&%\"vG6#F*F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "Sum(u[i]*v[i],i=1..3)=sum(u[i]*v[i],i=1..3);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 280 11 "Exemple 2 :" }{TEXT -1 10 " dans \+ R^6 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "u1:=vector([2,8,6,9 ,-2,8]);u2:=vector([-4,7,9,0,3,2]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "dotprod(u1,u2);2*(-4)+8*7+6*9+9*0+(-2)*3+8*2;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "v1:=vector([-4,7,6]);v2:=ve ctor([14, 29, -6]);v3:=vector([-4, 99, 6]);v4:=vector([372.6, 0, 248.4 ]);v5:=vector([a,34,76]);" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 283 32 "N orme d'un vecteur u : norm(u,2)" }{TEXT -1 3 " : " }}{PARA 257 "" 0 " " {TEXT -1 51 "La norme d'un vecteur u = (u1,u2,..un) est ||u|| = " } {TEXT 288 32 "racine carr\351e(u1^2+u2^2+..un^2) " }{TEXT -1 22 "= rac ine carr\351e(u o u)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "Remarque \+ : Dans Maple, norm(u,k) donne la racine k i\350me de (|u1|^k+|u2|^k+.. .+|un|^k) donc " }{TEXT 284 19 "|| u || = norm(u,2)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 285 11 "Exemple 3 :" }{TEXT -1 37 " Trouver la norme d e v2 = (14,29,-6) " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "# a) \+ comme \340 la main:\nsqrt(14^2+29^2+(-6)^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "# b) \nsqrt(dotprod(v2,v2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "# c)\nnorm(v2,2);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT 256 33 "Angle t entre 2 vecteurs u et v:" }{TEXT -1 1 " " } {TEXT 259 27 " u o v = ||u||*||v||*cos(t)" }{TEXT -1 6 " d'o\371 " } {TEXT 260 3 "t =" }{TEXT -1 2 " " }{TEXT 257 29 "arccos((u o v)/(||u| |*||v||))" }{TEXT -1 5 " ou " }{TEXT 258 23 "avec Maple: angle(u,v); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 286 11 "Exemple 4 :" }{TEXT -1 33 " \+ Quel est l'angle entre v2 et v3?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "arccos(dotprod(v2,v3)/(norm(v2,2)*norm(v3,2)));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "La r\351ponse est en radians" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "evalf(%);# pour obtenirla r \351ponse en d\351cimales" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "angle(v2,v3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "Si on veut \+ " }{TEXT 270 17 "l'angle en degr\351s" }{TEXT -1 30 ", on multiplie pa r 180/Pi ou " }{TEXT 261 19 "convert(%,degrees);" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 28 "convert(%,degrees);evalf(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 50 "Deux vecteurs u et v sont orthogonaux si u o v = 0" }{TEXT -1 41 ". Dans R^2 et R^3,on dit perpendiculaires" } }}{EXCHG {PARA 0 "" 0 "" {TEXT 287 11 "Exemple 5 :" }{TEXT -1 61 " Tro uver la valeur de a pour que v5 soit perpendiculaire \340 v2" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "dotprod(v5,v2)=0;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 13 "a=solve(%,a);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "Parmi v1, v2, et v3, trouver les vecteurs qui sont perpendiculaires \340 v4" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "dotprod(v1,v4);dotprod(v2,v4 );dotprod(v3,v4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "v4 est perpe ndiculaire \340 v1 et v3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 259 "" 0 "" {TEXT 268 103 "Projection orthogonale d'un vecteur u sur un vecteur v (sur la droite portant le vecteur v) \+ : proj(u/v)" }}{EXCHG {PARA 257 "" 0 "" {TEXT -1 58 "Notons le produi t scalaire des vecteurs u et v par u o v " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 264 35 " proj(u/v) = ((u o v) / (v o v)) v" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 11 "Exp lication" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "Par d\351finition, la \+ projection orthogonale d'un vecteur u sur un vecteur v, " }{TEXT 290 28 "proj(u/v), est le vecteur kv" }{TEXT -1 130 " o\371 k est un r\351 el, tel que u - kv est orthogonal v. On a alors (u - kv) o v = 0 et uov -kvov = 0. d'o\371 k = (uov)/(vov) et donc " }}{PARA 258 "" 0 "" {TEXT 289 4 "kv =" }{TEXT -1 1 " " }{TEXT 263 28 " ((uov)/(vov)) v = p roj(u/v)" }}{PARA 0 "" 0 "" {GLPLOT2D 282 247 247 {PLOTDATA 2 "60-%)PO LYGONSG6&7&7$$\"+quu\\\\!#7$\"+:y1rq!#87$$!+quu\\\\F*$!+:y1rqF-7$$!+vu \\\\!*!#5$\"+$*GH*H'!\"*7$$!+DD]]*)F6$\"+2rq+jF97%7$$!+**)*)z4\"F9$\"+ HdrriF97$$!\"\"\"\"!$\"\"(FH7$$!+755?qF6$\"+rUGGjF9-%&STYLEG6#%,PATCHN OGRIDG-%'COLOURG6&%$RGBG$\"*++++\"!\")$FHFHFen-%'CURVESG6#7*F'F.F3F@FE FKF:F'-F$6&7&7$$\"+6d@TVF*$!+#p%p![#F*7$$!+6d@TVF*$\"+#p%p![#F*7$$\"+% yecf$F9$\"+&p![-jF97$$\"+;7M/OF9$\"+0$>vH'F97%7$$\"+s8NEMF9$\"+)yF#*R' F97$$\"\"%FHFI7$$\"+G'[Ox$F9$\"+7Ax+iF9FP-FU6&FWFenFenFX-Fgn6#7*F]oFbo FgoFbpFgpFjpF\\pF]o-F$6&7&F]oFbo7$$\"+wk'z[#F9$\"+c\">SO%F97$$\"+3*[m \\#F9$\"+nx0fVF97%7$$\"+k!f'=BF9$\"+\\iwgWF97$$\"+p2BpFF9$\"+YQ:Y[F97$ $\"+?j&fm#F9$\"+u1JiUF9FP-FU6&FWFenFXFen-Fgn6#7*F]oFboFgqFbrFgrF\\sF\\ rF]o-F$6&7&7$$\"+k9rrFF9$\"+i]\\][F97$$\"+u+vmFF9$\"+IE\"=%[F97$$!*=wb D'F9$\"+pTF!y'F97$$!*Gif?'F9$\"+,m&*)y'F97%7$$!*6ZIA(F9$\"+dn'4h'F9FE7 $$!*N\"\\Q_F9$\"+8SEepF9FP-FU6&FWFHFHFH-Fgn6#7*FisF^tFctF^uFEFcuFhtFis -%%TEXTG6$7$$!#:FG$\"\"'FHQ\"u6\"-F^v6$7$FhpFcvQ\"vFfv-F^v6$7$$\"\"#FH F^wQ#kvFfv-F^v6$7$$\"#;FG$\"#kFGQ'u~-~kvFfv-%(SCALINGG6#%,CONSTRAINEDG -%%VIEWG6$;$!\"&FH$\"\"&FH;$!\"#FH$\"\")FH" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve \+ 4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" "Curve 9" "Curve 10" "Curve 11" "Curve 12" }}{TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 70 "Exemple graphique dans R^2: projection d'un vecteur u sur un vecte ur v" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "with(plottools):wit h(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "u:=vector([-1, 7]);v:=vector([4,7]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "pr oj(u/v):=(dotprod(u,v)/dotprod(v,v))*v;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Tra\347ons les vecteurs u, v, proj(u/v) et u - proj(u/v) " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 371 "flecheu:=arrow([0,0],u ,0.01,0.4,0.1,color = red):\nflechev:=arrow([0,0],v,0.01,0.4,0.1,color = blue):\nfleche(u/v):=arrow([0,0],evalm(proj(u/v)),0.01,0.4,0.1,colo r = green):\nw:=evalm(u-proj(u/v)): \nflechew:=arrow(evalm(proj(u/v)), w,0.01,0.4,0.1,color = black):\n\nnomsvecteurs:=textplot(\{[-1.5,6,`u` ],[4,6,`v`],[3.5,2,`proj(u/v)=((uov)/(vov))v)`],[1.6,6.4,`u - proj(u/v )`]\}):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "display(flecheu,flechev, fleche(u/v),flechew,nomsvecteurs,view=[-5..5,-2..8],scaling=constraine d);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 76 "Le vecteur (noir) u - proj(u/v) est perpendiculai re au vecteur v (bleu) et " }{TEXT 277 33 "proj(u/v) = ((u o v) / (v o v)) v" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "dotprod(v,u-proj( u/v));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 265 121 "Si on soustrait d'un \+ vecteur u, sa projection orthogonale sur un vecteur v, on obtient un v ecteur orthogonal au vecteur v" }{TEXT -1 2 ". " }}}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 63 "Trouver la projection orthogonale du vecteur u sur le vecteur v" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "u:=vector( [2,4,-3,7,8]);v:=vector([3,0,-4,7,0]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Proj(u/v):=(dotprod(u,v)/dotprod(v,v))*v;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "evalm(Proj(u/v));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 259 "" 0 "" {TEXT 266 8 "Exercice" }{TEXT 269 1 "s" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:with(linalg):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 297 5 "No 1 )" }{TEXT 298 1 " " }{TEXT -1 58 "Soit 3 points , A = (6,8,-7), B = (- 3,8,2) et C = (3,-3,8)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 " En uti lisant le produit scalaire, montrer que le triangle ABC est rectangle. " }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 267 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 -1 32 "#Pour les no s 2) 3) et 4), Soit:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 143 "v1 := vect or([10.12, -4.07, 7.82]); v2 := vector([5.77, -.61, 2.11]); v3 := vect or([11.535, -1.22, 4.11]); v4 := vector([-23.08, 2.44, -8.44]);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 299 5 "No 2)" }{TEXT -1 133 " Trouver l'a ngle en degr\351s, entre v1 et v2 de 2 fa\347ons diff\351rentes (en ut ilisant le produit scalaire et avec la commande Maple angle)" }}} {SECT 1 {PARA 3 "" 0 "" {TEXT 291 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 300 5 "No 3)" }{TEXT 301 1 " " } {TEXT -1 90 "Trouver la projection proj(v1/v2) b). Montrer que v1 - pr oj(v1/v2) est perpendiculaire v2" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 292 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 302 5 "No 4)" }{TEXT -1 126 " Parmi v1,v2 et v3, trouver quel ve cteur est paral\350le \340 v4 et dire si ces 2 vecteurs sont de le m \352me sens ou de sens contraire" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 293 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 304 5 "No 5)" }{TEXT -1 48 " Voici 2 vecteurs v1 = (3,-5,2), v2 \+ = (5,-7,9) " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 303 1 " " }{TEXT -1 96 " Trouver un vecteur v3 est perpendiculaire \340 v1 et qui est une comb inaison lin\351aire de v1 et v2 " }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 294 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 305 5 "No 6)" }{TEXT -1 318 " Soit A:=[2,5]; B:=[-3,55]; C:=[5,3 1]; 3 points. Trouver le point d'intersection des hauteurs du triangle ABC sachant que les 3 hauteurs d'un triangle se coupent en un m\352me point. \nSuggestion : Soit P = [x,y], le point cherch\351. On doit av oir AP perpendiculaire \340 BC, BP perpendiculaire \340 AC, CP perpend iculaire \340 AB, " }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 295 32 " Espace \+ de travail de l'\351tudiant" }{TEXT -1 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 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 306 5 "No 7)" }{TEXT -1 104 " V\351rifiez les propri\351t\351s s uivantes du produit scalaire avec les vecteurs v1, v2 v3 et le scalair e k = 10" }}{PARA 0 "" 0 "" {TEXT -1 66 "v1 = ( 6,-5 7), v2 = ( 9,12,- 4) v3 = ( -11,2,10)\np1: v1ov2 = v2ov1" }}{PARA 0 "" 0 "" {TEXT -1 30 "p2: v1o(v2+v3) = v1ov2 + v1ov3" }}{PARA 0 "" 0 "" {TEXT -1 46 "p3: k( v1,v2) = (kv1,v2) et k(v1,v2) = (v1,kv2)" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 296 32 " Espace de travail de l'\351tudiant" }{TEXT -1 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 }