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‫ﺗﻄﺒﻴﻘﺎت اﻟﺠﺪاء اﻟﺴﻠﻤﻲ ﻓﻲ اﻟﻔﻀﺎء‬
‫‪ -I‬اﻟﻤﺴﺘﻘﻴﻤﺎت و اﻟﻤﺴﺘﻮﻳﺎت ﻓﻲ اﻟﻔﻀﺎء‬
‫‪ -1‬ﺗﻌﺎﻣﺪ اﻟﻤﺴﺘﻘﻴﻤﺎت و اﻟﻤﺴﺘﻮﻳﺎت ﻓﻲ اﻟﻔﻀﺎء‬
‫أ‪ -‬ﺗﻌﺎﻣﺪ ﻣﺴﺘﻘﻴﻤﻴﻦ‬
‫ﻟﻴﻜﻦ )‪ (D1‬و )‪ (D2‬ﻣﺴﺘﻘﻴﻤــﻴﻦ ﻣﻦ اﻟﻔﻀﺎء ﻣﻮﺟﻬﻴﻦ ﺑﺎﻟﻤﺘﺠﻬﺘﻴﻦ ‪ u 1‬و ‪ u 2‬ﻋﻠﻰ اﻟﺘﻮاﻟﻲ‬
‫‪( D1 ) ⊥ ( D2 ) ⇔ u1 ⋅ u2 = 0‬‬
‫ب‪ -‬ﺗﻌﺎﻣﺪ ﻣﺴﺘﻘﻴﻢ و ﻣﺴﺘﻮى‬
‫ﺧﺎﺻﻴﺔ‬
‫ﻟﻴﻜﻦ )‪ (P‬ﻣﺴﺘﻮى ﻣﻮﺟﻪ ﺑﺎﻟﻤﺘﺠﻬﺘﻴﻦ ‪ u 1‬و ‪ u 2‬و )‪ (D‬ﻣﺴﺘﻘﻴﻢ ﻣﻮﺟﻪ ﺑﺎﻟﻤﺘﺠﻬﺔ ‪u 3‬‬
‫‪u2 ⊥ u3‬‬

‫و‬

‫‪u2 ⋅ u3 = 0‬‬

‫و‬

‫‪( D ) ⊥ ( P ) ⇔ u1 ⊥ u3‬‬
‫‪( D ) ⊥ ( P ) ⇔ u1 ⋅ u3 = 0‬‬

‫ج‪ -‬ﻣﻼﺣﻈﺎت واﺻﻄﻼﺣﺎت‬
‫* اﻟﻤﺘﺠﻬﺔ ‪ u‬اﻟﻤﻮﺟﻬﺔ ﻟﻤﺴﺘﻘﻴﻢ )‪ (D‬اﻟﻌﻤﻮدي ﻋﻠﻰ ﻣﺴﺘﻮى )‪ (P‬ﺗﺴﻤﻰ ﻣﺘﺠﻬﺔ ﻣﻨﻈﻤﻴﺔ ﻟﻠﻤﺴﺘﻮى )‪.( P‬‬
‫* اذا آﺎﻧﺖ ‪ u‬ﻣﻨﻈﻤﻴﺔ ﻟﻤﺴﺘﻮى )‪ (P‬ﻓﺎن آﻞ ﻣﺘﺠﻬﺔ ‪ v‬ﻣﺴﺘﻘﻴﻤﻴﺔ ﻣﻊ ‪ u‬ﺗﻜﻮن ﻣﻨﻈﻤﻴﺔ ﻟﻠﻤﺴﺘﻮى )‪(P‬‬
‫* اذا آﺎﻧﺖ ‪ u‬ﻣﻨﻈﻤﻴﺔ ﻟﻤﺴﺘﻮى )‪ (P‬و ‪ v‬ﻣﻨﻈﻤﻴﺔ ﻟﻤﺴــﺘﻮى )'‪ (P‬وآﺎﻧﺘﺎ ‪ u‬و ‪ v‬ﻣﺴﺘﻘﻴﻤﻴﺘﻴﻦ ﻓﺎن )‪ (P‬و)'‪( P‬‬
‫ﻣﺘﻮازﻳﺎن‬
‫* إذا آﺎن ) ‪ ( A; B ) ∈ ( P‬و ‪ u‬ﻣﻨﻈﻤﻴﺔ ﻟﻤﺴﺘﻮى )‪ (P‬ﻓﺎن‬
‫‪2‬‬

‫‪u ⊥ AB‬‬

‫ﺗﻤﺮﻳﻦ ﻓﻲ اﻟﻔﻀﺎء اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ‪.‬م‪(O ; i ; j ; k ) .‬‬
‫ﺣﺪد ﺗﻤﺜﻴﻞ ﺑﺎراﻣﺘﺮي ﻟﻠﻤﺴﺘﻘﻴﻢ )‪ (D‬اﻟﻤﺎر ﻣﻦ)‪ A(-1; 2 0‬و اﻟﻌﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻮى )‪ (P‬اﻟﻤﻮﺟﻪ ﺑـﺎﻟﻤﺘﺠﻬﺘﻴﻦ‬
‫)‪ u(1;−1;1‬و )‪v(2;1;1‬‬
‫ﺗﻤﺮﻳﻦ‬

‫ﻓﻲ اﻟﻔﻀﺎء اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ‪.‬م‪ (O ; i ; j ; k ) .‬ﻧﻌﺘﺒﺮ اﻟﻤﺴﺘﻮى)‪ (P‬اﻟﺬي ﻣﻌﺎدﻟﺘﻪ ‪ax-2y+z-2=0‬‬
‫و اﻟﻤﺴﺘﻘﻴﻢ )‪ (D‬ﺗﻤﺜﻴﻠﻪ ﺑﺎراﻣﺘﺮي‬

‫‪t ∈ IR‬‬

‫‪ x=2t‬‬
‫‪‬‬
‫‪ y =1+ 3 t‬‬
‫‪ z = − 2 + bt‬‬

‫‪ -1‬ﺣﺪد ﻣﺘﺠﻬﺘﻴﻦ ﻣﻮﺟﻬﺘﻴﻦ ﻟﻠﻤﺴﺘﻮى )‪(P‬‬
‫‪ -2‬ﺣﺪد‪ a‬و‪ b‬ﻟﻜﻲ ﻳﻜﻮن ) ‪(D )⊥(P‬‬
‫د‪ -‬ﺗﻌﺎﻣﺪ ﻣﺴﺘﻮﻳﻴﻦ‬
‫ﺗﺬآﻴﺮ ﻳﻜﻮن ﻣﺴﺘﻮﻳﺎن ﻣﺘﻌﺎﻣﺪﻳﻦ اذا و ﻓﻘﻂ اذا اﺷﺘﻤﻞ أﺣﺪهﻤﺎ ﻋﻠﻰ ﻣﺴﺘﻘﻴﻢ ﻋﻤﻮدي‬
‫ﻋﻠﻰ اﻟﻤﺴﺘﻮى اﻵﺧﺮ‪.‬‬
‫ﺧﺎﺻﻴﺔ‬
‫ﻟﻴﻜﻦ )‪ (P‬و )'‪ (P‬ﻣﺴﺘﻮﻳﻴﻦ ﻣﻦ اﻟﻔﻀﺎء و ‪ u‬و ‪ v‬ﻣﺘﺠﻬﺘــــــــــﻴﻦ ﻣﻨﻈﻤﻴﺘﻴﻦ ﻟﻬﻤﺎ ﻋﻠﻰ اﻟﺘﻮاﻟﻲ‬
‫) ‪ (P')⊥(P‬اذا وﻓﻘﻂ اذا آﺎن ‪u ⊥v‬‬

‫‪ -3‬ﻣﻌﺎدﻟﺔ ﻣﺴﺘﻮى ﻣﺤﺪد ﺑﻨﻘﻄﺔ و ﻣﺘﺠﻬﺔ ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ‬
‫‪ .a‬ﻣﺴﺘﻮى ﻣﺤﺪد ﺑﻨﻘﻄﺔ و ﻣﺘﺠﻬﺔ ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ‬
‫ﻣﺒﺮهﻨﺔ‬
‫ﻟﺘﻜﻦ ‪ u‬ﻣﺘﺠﻬﺔ ﻏﻴﺮ ﻣﻨﻌﺪﻣﺔ و ‪ A‬ﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻀﺎء‬
‫* اﻟﻤﺴﺘﻮى اﻟﻤﺎر ﻣﻦ‪ A‬و اﻟﻤﺘﺠﻬﺔ ‪ u‬ﻣﻨﻈﻤﻴﺔ ﻟﻪ هﻮ ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ‪ M‬ﻣﻦ اﻟﻔﻀﺎء ﺣﻴﺚ ‪AM ⋅ u = 0‬‬

‫‪Moustaouli Mohamed‬‬

‫‪1‬‬

‫‪http://arabmaths.ift.fr‬‬

‫* ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ‪ M‬ﻣﻦ اﻟﻔﻀﺎء ﺣﻴﺚ ‪ AM ⋅ u = 0‬اﻟﻤﺴﺘﻮى اﻟﻤﺎر ﻣﻦ‪ A‬و اﻟﻤﺘﺠﻬﺔ ‪ u‬ﻣﻨﻈﻤﻴﺔ ﻟﻪ‬
‫‪ .b‬ﻣﻌﺎدﻟﺔ ﻣﺴﺘﻮى ﻣﺤﺪد ﺑﻨﻘﻄﺔ و ﻣﺘﺠﻬﺔ ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ‬
‫ﺧﺎﺻﻴﺔ‬
‫* آﻞ ﻣﺴﺘﻮى )‪ (P‬ﻓﻲ اﻟﻔﻀﺎء و ) ‪ u(a;b;c‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ ﻳﻘﺒﻞ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻣﻦ‬
‫ﻧﻮع ‪ax + by + cz + d = 0‬‬
‫* آﻞ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻣﻦ ﻧﻮع ‪ ax + by + cz + d = 0‬ﺣﻴﺚ ) ‪ ( a; b ; c ) ≠ ( 0;0;0‬هﻲ ﻣﻌﺎدﻟﺔ ﻣﺴﺘﻮى‬

‫)‪ (P‬ﻓﻲ اﻟﻔﻀﺎء ﺑﺤﻴﺚ ) ‪ u(a;b;c‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ‬
‫*ﻓﻲ اﻟﻔﻀﺎء ﻣﻌﺎدﻟﺔ اﻟﻤﺴﺘﻮى ) ‪ ( P‬اﻟﻤﺎر ﻣﻦ ) ‪ A ( x0 ; y0 ; z0‬و اﻟﻤﺘﺠﻬﺔ ) ‪ u(a;b;c‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ هﻲ‪:‬‬
‫‪a ( x − x0 ) + b ( y − y0 ) + c ( z − z0 ) = 0‬‬

‫‪(4‬‬

‫ﻣﺒﺮهﻨﺔ‬
‫ﻟﺘﻜﻦ ‪ u‬ﻣﺘﺠﻬﺔ ﻏﻴﺮ ﻣﻨﻌﺪﻣﺔ و ‪ A‬ﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻀﺎء و ‪ k‬ﻋﺪدا ﺣﻘﻴﻘﻴﺎ‬
‫ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ‪ M‬ﻣﻦ اﻟﻔﻀﺎء ﺣﻴﺚ ‪ AM ⋅ u = k‬هﻲ اﻟﻤﺴﺘﻮى ) ‪ ( P‬اﻟﻌﻤﻮدي ﻋﻠﻰ ) ‪ D ( A; u‬ﻓﻲ اﻟﻨﻘﻄﺔ ‪H‬‬
‫ﺣﻴﺚ‬

‫‪k‬‬
‫‪AB‬‬

‫= ‪AH‬‬

‫; ‪u = AB‬‬

‫ﺗﻤﺮﻳﻦ‬

‫‪ x+y-2z+1=0‬‬
‫‪(P) : 2x-y+3z+1=0‬‬
‫ﻧﻌﺘﺒﺮ‬
‫‪(D): ‬‬
‫‪ x-y+z-2=0‬‬
‫‪ -1‬ﺣﺪد ﻣﺘﺠﻬﺔ ‪ u‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻰ )‪ (P‬وﻧﻘﻄﺔ ﻣﻨﻪ‪.‬‬
‫‪ -2‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى اﻟﻤﺎر ﻣﻦ)‪ A (2;0;3‬و )‪ n(1,2,1‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ‪.‬‬
‫‪ -3‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى اﻟﻤﺎر ﻣﻦ)‪ A' (2;0;3‬واﻟﻌﻤﻮدي ﻋﻠﻰ )‪(D‬‬
‫‪ -4‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى اﻟﻤﺎر ﻣﻦ)‪ A (2;0;3‬و اﻟﻤﻮازي ﻟـ )‪(P‬‬
‫‪ -6‬دراﺳﺔ اﻻوﺿﺎع اﻟﻨﺴﺒﻴﺔ ﻟﻠﻤﺴﺘﻘﻴﻤﺎت و اﻟﻤﺴﺘﻮﻳﺎت ﻓﻲ اﻟﻔﻀﺎء‬
‫أ‪ -‬اﻷوﺿﺎع اﻟﻨﺴﺒﻴﺔ ﻟﻤﺴﺘﻮﻳﻴﻦ ﻓﻲ اﻟﻔﻀﺎء‬
‫)‪ (P‬اﻟﻤﺴﺘﻮى اﻟﻤﺎر ﻣﻦ ‪ A‬و ‪ u‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ )'‪ (P‬اﻟﻤﺴﺘﻮى اﻟﻤﺎر ﻣﻦ '‪ A‬و ‪ v‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ‬
‫اﻟﺤﺎﻟﺔ ‪ u 1‬و ‪ v‬ﻣﺴﺘﻘﻴﻤﻴﺘﺎن‬
‫إذا آﺎن)'‪ A ∈ (P‬أو)‪ A' ∈ (P‬ﻓﺎن )‪(P') = (P‬‬
‫إذا آﺎن)'‪ A ∉ (P‬و)‪ A' ∉ (P‬ﻓﺎن )‪ (P‬و )'‪ (P‬ﻣﺘﻮازﻳﺎن ﻗﻄﻌﺎ‪.‬‬
‫اﻟﺤﺎﻟﺔ ‪ u 2‬و ‪ v‬ﻏﻴﺮ ﻣﺴﺘﻘﻴﻤﻴﺘﻴﻦ‬
‫ ‪(P')⊥(P ) ⇔ u ⊥ v‬‬‫ اذا آﺎن ‪ v‬و ‪ u‬ﻏﻴﺮ ﻣﺘﻌﺎﻣﺪﺗﻴﻦ ﻓﺎن )‪ (P‬و )'‪ (P‬ﻣﺘﻘﺎﻃﻌﺎن‪.‬‬‫ب‪ -‬اﻷوﺿﺎع اﻟﻨﺴﺒﻴﺔ ﻟﻤﺴﺘﻘﻴﻢ وﻣﺴﺘﻮى ﻓﻲ اﻟﻔﻀﺎء‬
‫)‪ (P‬اﻟﻤﺴﺘﻮى اﻟﻤﺎر ﻣﻦ ‪ A‬و ‪ u‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ )‪ (D‬اﻟﻤﺴﺘﻘﻴﻢ اﻟﻤﺎر ﻣﻦ '‪ A‬و ‪ v‬ﻣﻮﺟﻬﺔ ﻟﻪ‬
‫اﻟﺤﺎﻟﺔ ‪ 1‬إذا آﺎن ‪ u‬و ‪ v‬ﻣﺴﺘﻘﻴﻤﻴﺘﻴﻦ ﻓﺎن ) ‪(D )⊥(P‬‬
‫اﻟﺤﺎﻟﺔ ‪ u 2‬و ‪ v‬ﻏﻴﺮ ﻣﺴﺘﻘﻴﻤﻴﺘﻴﻦ‬
‫ ‪ (P) ⇔ u ⊥ v‬و )‪ (D‬ﻣﺘﻮازﻳﺎن‬‫ اذا آﺎن ‪ u ⊥ v‬ﻓﺎن‬‫‪ -7‬ﻣﺴﺎﻓﺔ ﻧﻘﻄﺔ ﻋﻦ ﻣﺴﺘﻮى‬
‫ﻧﺸﺎط‬
‫اﻟﻔﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ‬

‫)‪ (D‬ﻳﺨﺘﺮق )‪(P‬‬

‫)‬

‫(‬

‫‪ ( P ) . O; i ; j ; k‬اﻟﻤﺴﺘﻮى اﻟﻤﺎر ﻣﻦ ) ‪B ( xB ; yB ; z B‬‬

‫و ) ‪ n ( a; b; c‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ‪ .‬ﻟﺘﻜﻦ ) ‪ A ( x0 ; y0 ; z0‬ﻧﻘﻄﺔ ﻣﻦ اﻟﻤﺴﺘﻮى ‪ H ،‬اﻟﻤﺴﻘﻂ اﻟﻌﻤﻮدي ﻟﻠﻨﻘﻄﺔ ‪A‬‬
‫ﻋﻠﻰ ) ‪. ( P‬‬

‫أ‪ -‬أﺣﺴﺐ ‪ n ⋅ AB‬ﺑﺪﻻﻟﺔ ‪ HA‬و ‪n‬‬

‫‪Moustaouli Mohamed‬‬

‫‪2‬‬

‫‪http://arabmaths.ift.fr‬‬

‫ب‪ -‬أﺛﺒﺖ أن‬

‫‪n ⋅ AB‬‬
‫‪n‬‬

‫= ‪HA‬‬

‫د‪ -‬ﻟﻴﻜﻦ ‪ ( P ) : ax + by + cz + d = 0‬ﺣﻴﺚ‬
‫ﺑﻴﻦ أن‬

‫‪ax0 + by0 + cz0 + d‬‬
‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫‪a +b +c‬‬

‫) ‪( a; b; c ) ≠ ( 0;0;0‬‬

‫= ‪HA‬‬

‫‪ -1‬ﺗﻌﺮﻳﻒ و ﺧﺎﺻﻴﺔ‬
‫اﻟﻔﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ‪.‬م‪.‬م ) ‪(o;i ; j ;k‬‬
‫ﻣﺴﺎﻓﺔ ﻧﻘﻄﺔ ‪ A‬ﻋﻦ ﻣﺴﺘﻮى )‪ (P‬هﻲ اﻟﻤﺴﺎﻓﺔ ‪ AH‬ﺣﻴﺚ‪ H‬اﻟﻤﺴﻘﻂ اﻟﻌﻤﻮدي ﻟـ ‪ A‬ﻋﻠﻰ)‪ (P‬ﻧﻜﺘﺐ‬
‫‪A B •u‬‬
‫= ‪ d ( A ; ( P ) ) = A H‬ﺣﻴﺚ )‪ B ∈ (P‬و ‪ u‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻰ)‪(P‬‬
‫‪u‬‬
‫‪ -2‬ﺧﺎﺻﻴﺔ‬
‫ﻟﻴﻜﻦ )‪ (P‬ﻣﺴﺘﻮى ﻣﻌﺎدﻟﺘﻪ ‪ax + by + cz + d = 0‬‬

‫‪ax 0 + by 0 + cz 0 + d‬‬
‫‪a2 + b 2 + c 2‬‬
‫ﻣﺜﺎل‬
‫ﻟﻴﻜﻦ )‪ (P‬ﻣﺴﺘﻮى ﻣﺎر ﻣﻦ )‪ B ( 2;1;3‬و ‪u 1; −1; 2‬‬
‫ﺣﺪد‬

‫)) ‪d ( A ; ( P‬‬

‫)‬

‫(‬

‫و ) ‪ A ( x 0 ; y 0 ; z 0‬ﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻀﺎء‬

‫= )) ‪d ( A ; ( P‬‬
‫ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ ﻟﺘﻜﻦ )‪A (1;2;0‬‬

‫ﺗﻤﺮﻳﻦ‪1‬‬
‫ﻓﻲ ﻓﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ‪.‬‬
‫ﻧﻌﺘﺒﺮ )‪ A(1;-1;1‬و )‪ B(3;1;-1‬و )‪ (P‬اﻟﻤﺴﺘﻮى ذا اﻟﻤﻌﺎدﻟﺔ ‪ 2x-3y+2z=0‬و )‪ (D‬اﻟﻤﺴﺘﻘﻴﻢ اﻟﻤﻤﺜﻞ‬
‫‪ x = 3t‬‬
‫‪‬‬
‫ﺑﺎرا ﻣﺘﺮﻳﺎ ﺑـ‬
‫∈ ‪ x = − 2 − 3t t‬‬
‫‪ z = 2 + 4t‬‬
‫‪‬‬
‫‪ -1‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )‪ (Q‬اﻟﻤﺎر ﻣﻦ ‪ A‬واﻟﻌﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻘﻴﻢ )‪(D‬‬
‫ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )´‪ (Q‬اﻟﻤﺎر ﻣﻦ ‪ A‬و ‪ B‬واﻟﻌﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻮى )‪(P‬‬
‫‪ -2‬أﺣﺴﺐ ))‪ d(A;(P‬و ))‪d(A;(D‬‬
‫‪ -3‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )´´‪ (Q‬اﻟﻤﺎر ﻣﻦ ‪ B‬و اﻟﻤﻮازي ﻟﻠﻤﺴﺘﻮى )‪(P‬‬
‫ﺗﻤﺮﻳﻦ‪2‬‬
‫ﻓﻲ ﻓﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ‪.‬‬
‫ﻧﻌﺘﺒﺮ اﻟﻤﺴﺘﻮى)‪ (P‬ذا اﻟﻤﻌﺎدﻟﺔ ‪ 3x+2y-z-5=0‬و )‪ (D‬اﻟﻤﺴﺘﻘﻴﻢ اﻟﻤﻌﺮف ﺑـ‬

‫‪Moustaouli Mohamed‬‬

‫‪3‬‬

‫‪http://arabmaths.ift.fr‬‬

‫‪x − 2 y + z − 3 = 0‬‬
‫‪‬‬
‫‪ x− y−z+2=0‬‬
‫‪ -1‬ﺣﺪد ﺗﻤﺜﻴﻼ ﺑﺎرا ﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ )‪(D‬‬
‫ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )´‪ (P‬اﻟﺬي ﻳﺘﻀﻤﻦ )‪ (D‬و اﻟﻌﻤﻮدي ﻋﻠﻰ )‪(P‬‬
‫‪II‬־ اﻟﻔﻠﻜﺔ‬

‫اﻟﻔﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) ‪(O ; i ; j ; k‬‬

‫‪ -1‬ﻣﻌﺎدﻟﺔ ﻓﻠﻜﺔ ﻣﻌﺮﻓﺔ ﺑﻤﺮآﺰهﺎ وﺷﻌﺎﻋﻬﺎ‬
‫ﻟﺘﻜﻦ ) ‪ Ω ( a; b; c‬ﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻀﺎء )‪ (E‬و‬

‫‪*+‬‬

‫∈‪r‬‬

‫و ) ‪ S ( Ω; r‬اﻟﻔﻠﻜﺔ اﻟﺘﻲ ﻣﺮآﺰهﺎ ‪ Ω‬و ﺷﻌﺎﻋﻬﺎ ‪r‬‬
‫ﻟﻴﻜﻦ ) ‪ M ( x; y; z‬ﻣﻦ اﻟﻔﻀﺎء )‪(E‬‬
‫‪M ∈ S ( Ω; r ) ⇔ ΩM = r 2 ⇔ ( x − a ) + ( y − b ) + ( z − c ) = r 2‬‬
‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫ﻣﺒﺮهﻨﺔ‬
‫اﻟﻔﻀﺎء اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ‪. o;i ; j ;k‬‬
‫ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ ) ‪ S ( Ω; r‬اﻟﺘﻲ ﻣﺮآﺰهﺎ ) ‪Ω ( a; b; c‬‬

‫)‬

‫و ﺷﻌﺎﻋﻬﺎ ‪r‬‬

‫هﻲ‬

‫(‬

‫‪( x − a )2 + ( y − b )2 + ( z − c )2 = r 2‬‬

‫ﻣﻼﺣﻈﺎت و اﺻﻄﻼﺣﺎت‬
‫* إذا آﺎن ‪ A‬و ‪ B‬ﻧﻘﻄﺘﻴﻦ ﻣﻦ اﻟﻔﻠﻜﺔ )‪ S(Ω;r‬ﺣﻴﺚ ‪ Ω‬ﻣﻨﺘﺼﻒ ]‪ [A;B‬ﻓﺎن ]‪ [A;B‬ﻗﻄﺮا ﻟﻠﻔﻠﻜﺔ‬
‫* ﺗﻮﺟﺪ ﻓﻠﻜﺔ وﺣﻴﺪة أﺣﺪ أﻗﻄﺎرهﺎ ]‪ [A;B‬ﻣﺮآﺰهﺎ ‪ Ω‬ﻣﻨﺘﺼﻒ ]‪ [A;B‬و ﺷﻌﺎﻋﻬﺎ ‪r =½ AB‬‬
‫* ﻟﻠﻔﻠﻜﺔ ) ‪ S ( Ω; r‬ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻣﻦ ﺷﻜﻞ ‪ x 2 + y 2 + z 2 + ax + by + cz + d = 0‬ﺣﻴﺚ ‪ a‬و ‪ b‬و ‪ c‬و ‪d‬‬
‫أﻋﺪاد ﺣﻘﻴﻘﻴﺔ‪.‬‬
‫* اﻟﻔﻠﻜﺔ ) ‪ S ( O; r‬ﺣﻴﺚ ‪ O‬أﺻﻞ اﻟﻤﻌﻠﻢ ﻣﻌﺎدﻟﺘﻬﺎ ‪x + y + z = r‬‬
‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫ﻟﺘﻜﻦ )‪ S(Ω;r‬ﻓﻠﻜﺔ اﻟﺘﻲ ﻣﺮآﺰهﺎ )‪ Ω(a;b;c‬و ﺷﻌﺎﻋﻬﺎ ‪r‬‬
‫* اﻟﻜﺮة‬
‫اﻟﻜﺮة )‪ B(Ω;r‬اﻟﺘﻲ ﻣﺮآﺰهﺎ ) ‪ Ω ( a; b; c‬و ﺷﻌﺎﻋﻬﺎ ‪ r‬هﻲ ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ) ‪M ( x; y; z‬‬
‫ﺣﻴﺚ‬

‫‪( x − a )2 + ( y − b )2 + ( z − c )2 ≤ r 2‬‬

‫‪ -2‬ﻣﻌﺎدﻟﺔ ﻓﻠﻜﺔ ﻣﻌﺮﻓﺔ ﺑﺄﺣﺪ أﻗﻄﺎرهﺎ‬
‫‪ S‬ﻓﻠﻜﺔ أﺣﺪ اﻗﻄﺎرهﺎ ]‪[A;B‬‬
‫‪  AMB ⇔ M ∈ S‬زاوﻳﺔ ﻗﺎﺋﻤﺔ أو ‪ M=A‬أو ‪M=B‬‬
‫‪‬‬
‫‪‬‬

‫⇔ ‪MA ⋅ MB = 0‬‬
‫ﻣﺒﺮهﻨﺔ‬
‫‪ A‬و ‪ B‬ﻧﻘﻄﺘﺎن ﻣﺨﺘﻠﻔﺎن ﻓﻲ اﻟﻔﻀﺎء‬
‫ﻓﻲ اﻟﻔﻀﺎء ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ‪ M‬اﻟﺘﻲ ﺗﺤﻘﻖ ‪ MA ⋅ MB = 0‬هﻲ‬
‫ﻓﻠﻜﺔ اﻟﺘﻲ أﺣﺪ اﻗﻄﺎرهﺎ ]‪[A;B‬‬
‫ﺧﺎﺻﻴﺔ‬
‫اذا آﺎﻧﺖ )‪ A(xA;yA;zA‬و )‪ B(xB;yB;zB‬ﻧﻘﻄﺘﻴﻦ ﻣﺨﺘﻠﻔﺘﻴﻦ ﻓﺎن ﻣﻌﺎدﻟﺔ اﻟﻔﻠﻜﺔ اﻟﺘﻲ أﺣﺪ اﻗﻄﺎرهﺎ ]‪[A;B‬‬
‫‪(x-x A )(x-x B )+(y-y A )(y-y B )+(z-z A )(z-z B )=0‬‬
‫هﻲ‬
‫ﺗﻤﺮﻳﻦ‬
‫ﻓﻲ‬

‫اﻟﻔﻀﺎء اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ )‬

‫(‬

‫‪ ، O ; i ; j ; k‬ﻧﻌﺘﺒﺮ )‪ Ω(1;2;-1‬و )‪ A(2;1;2‬و )‪B(4;1;2‬‬

‫‪ -1‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ ‪ S‬اﻟﺘﻲ ﻣﺮآﺰهﺎ ‪ Ω‬و اﻟﻤﺎر ﻣﻦ ‪A‬‬
‫‪ -2‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ ´‪ S‬اﻟﺘﻲ ﻗﻄﺮهﺎ ]‪[A;B‬‬
‫‪ -3‬دراﺳﺔ اﻟﻤﻌﺎدﻟﺔ ‪(1): x²+y²+z²+ax+by-+cz+d=0‬‬
‫‪Moustaouli Mohamed‬‬

‫‪4‬‬

‫‪http://arabmaths.ift.fr‬‬

‫ﻟﺘﻜﻦ ‪ E‬ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ) ‪ M ( x; y; z‬اﻟﺘﻲ ﺗﺤﻘﻖ اﻟﻤﻌﺎدﻟﺔ )‪(1‬‬
‫‪2‬‬

‫‪c‬‬
‫‪a 2 + b 2 + c 2 − 4d‬‬
‫=‬
‫‪‬‬
‫‪2‬‬
‫‪4‬‬
‫‪ a b c‬‬
‫ﻟﺘﻜﻦ ‪Ω  − ; − ; − ‬‬
‫‪ 2 2 2‬‬

‫‪2‬‬

‫‪2‬‬

‫‪a ‬‬
‫‪b ‬‬
‫‪‬‬
‫‪M ∈E ⇔ x+  +y+  +z+‬‬
‫‪2 ‬‬
‫‪2 ‬‬
‫‪‬‬

‫*‪ -‬إذا آﺎن ‪ a 2 + b 2 + c 2 − 4d ≺ 0‬ﻓﺎن‪E =Ø‬‬
‫*‪ -‬اذا آﺎن ‪ a 2 + b 2 + c 2 − 4d = 0‬ﻓﺎن }‪E={Ω‬‬
‫*‪ -‬اذا آﺎن‬

‫‪0‬‬

‫‪ a 2 + b 2 + c 2 − 4d‬ﻓﺎن )‪ E=S(Ω;r‬ﺣﻴﺚ‬

‫‪a 2 + b 2 + c 2 − 4d‬‬
‫‪= r2‬‬
‫‪4‬‬

‫ﻣﺒﺮهﻨﺔ‬
‫‪ a‬و ‪ b‬و ‪ c‬و ‪ d‬أﻋﺪاد ﺣﻘﻴﻘﻴﺔ‬
‫ﺗﻜﻮن ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ )‪ M(x;y;z‬اﻟﺘﻲ ﺗﺤﻘﻖ اﻟﻤﻌﺎدﻟﺔ ‪ x 2 + y 2 + z 2 + ax + by + cz + d = 0‬ﻓﻠﻜﺔ‬

‫إذا وﻓﻘﻂ إذا آﺎن ‪a 2 + b 2 + c 2 − 4d ≥ 0‬‬
‫‪a 2 + b 2 + c 2 − 4d‬‬
‫‪ a b c‬‬
‫‪ Ω  − ; − ; − ‬ﻣﺮآﺰ هﺬﻩ اﻟﻔﻠﻜﺔ و ‪= r‬‬
‫‪2‬‬
‫‪ 2 2 2‬‬

‫ﺷﻌﺎﻋﻬﺎ‬

‫ﻣﻼﺣﻈﺔ ﻳﻤﻜﻦ اﻋﺘﺒﺎر }‪ E={Ω‬ﻓﻠﻜﺔ ﻣﺮآﺰهﺎ ‪ Ω‬و ﺷﻌﺎﻋﻬﺎ ﻣﻨﻌﺪم‬
‫ﺗﻤﺮﻳﻦ ﻧﻌﺘﺒﺮ ‪ E‬ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ )‪ M(x;y;z‬اﻟﺘﻲ ﺗﺤﻘﻖ اﻟﻤﻌﺎدﻟﺔ ‪x²+y²+z²+4x-2y -6z+5=0‬‬
‫ﺑﻴﻦ إن ‪ E‬ﻓﻠﻜﺔ ﻣﺤﺪدا ﻋﻨﺎﺻﺮهﺎ اﻟﻤﻤﻴﺰة‬
‫ﺣﻴﺚ )‪ A(2;0;-1‬و )‪B(-1;1;-1‬‬
‫ﺗﻤﺮﻳﻦ ﺣﺪد ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ‪ M‬اﻟﺘﻲ ﺗﺤﻘﻖ ‪2MA²+3MB²=16‬‬
‫‪ –4‬ﺗﻘﺎﻃﻊ ﻣﺴﺘﻮى و ﻓﻠﻜﺔ‬
‫أ‪ -‬دراﺳﺔ ﺗﻘﺎﻃﻊ ﻟﻠﻔﻠﻜﺔ )‪ S(Ω;r‬و اﻟﻤﺴﺘﻮى )‪(P‬‬
‫ﻧﻨﺴﺐ اﻟﻔﻀﺎء‪ E‬إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ )‪ ( A;u ;v ;w‬ﺣﻴﺚ ‪ A‬اﻟﻤﺴﻘﻂ اﻟﻌﻤﻮدي ﻟـ ‪ Ω‬ﻋﻠﻰ اﻟﻤﺴﺘﻮى )‪(P‬‬
‫و ) ‪ ( A;u ;v‬م‪.‬م‪.‬م ﻟـ )‪(P‬‬
‫‪ w‬ﻣﺘﺠﻬﺔ واﺣﺪﻳﺔ ﻣﻮﺟﻬﺔ ﻟﻠﻤﺴﺘﻘﻴﻢ )‪(AΩ‬‬
‫‪ Ω‬ﺗﻨﺘﻤﻲ إﻟﻰ ﻣﺤﻮر اﻻﻧﺎﺳﻴﺐ وﻣﻨﻪ ﻳﻮﺟﺪ ‪ c‬ﺣﻴﺚ )‪Ω(0;0;c‬‬
‫ﻣﻌﺎدﻟﺔ اﻟﻤﺴﺘﻮى )‪ (P‬ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻌﻠﻢ )‪ ( A;u ;v ;w‬هﻲ ‪ z=0‬و ﻣﻌﺎدﻟﺔ اﻟﻔﻠﻜﺔ‪ S‬هﻲ ‪x²+y²+(z-c)²=r²‬‬
‫|‪d(Ω;(P))=|c‬‬
‫‪ x²+y²+(z-c)²=r ²‬و ‪z=0‬‬
‫ﻟﺪﻳﻨﺎ ‪⇔ M(x;y;z) ∈ (P)∩S‬‬
‫‪ x²+y²=r²-c²‬و ‪z=0‬‬
‫⇔‬
‫وﻣﻨﻪ ﺗﻘﺎﻃﻊ ‪ S‬و )‪ (P‬ﻣﺮﺗﺒﻂ ﺑﺤﻞ اﻟﻤﻌﺎدﻟﺔ ‪ x²+y²=r²-c²‬ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻌﻠﻢ ) ‪( A;u;v‬‬
‫*اﻟﺤﺎﻟﺔ‪ 1‬إذا آﺎن ‪ d(Ω;(P))>r‬ﻓﺎن ‪(P)∩S = Ø‬‬
‫*اﻟﺤﺎﻟﺔ‪ 2‬إذا آﺎن ‪ d(Ω;(P))= r‬ﻓﺎن } ‪(P)∩S = {A‬‬

‫*اﻟﺤﺎﻟﺔ ‪ 3‬إذا آﺎن ‪ d(Ω;(P))< r‬ﻓﺎن )‪ (P)∩S = (C‬ﺣﻴﺚ )‪ (C‬اﻟﺪاﺋﺮة اﻟﺘﻲ ﻣﺮآﺰهﺎ ‪ A‬وﺷﻌﺎﻋﻬﺎ ‪r 2 − c 2‬‬

‫‪Moustaouli Mohamed‬‬

‫‪5‬‬

‫‪http://arabmaths.ift.fr‬‬

‫ﻣﺒﺮهﻨﺔ‬
‫ﻟﻴﻜﻦ )‪ (P‬ﻣﺴﺘﻮى ﻓﻲ اﻟﻔﻀﺎء و ‪ S‬ﻓﻠﻜﺔ ﻣﺮآﺰهﺎ ‪ Ω‬و ﺷﻌﺎﻋﻬﺎ ‪r‬‬
‫ﻳﻜﻮن ﺗﻘﺎﻃﻊ )‪ (P‬و ‪: S‬‬
‫* داﺋﺮة ﻣﺮآﺰهﺎ ‪ A‬اﻟﻤﺴﻘﻂ اﻟﻌﻤﻮدي ﻟـ ‪ Ω‬ﻋﻠﻰ اﻟﻤﺴﺘﻮى )‪ (P‬و ﺷﻌﺎﻋﻬﺎ ) ) ‪r 2 − d 2 ( Ω; ( P‬‬

‫اذا آﺎن ‪d(Ω;(P))< r‬‬
‫* داﺋﺮة ﻧﻘﻄﺔ اذا آﺎن ‪d(Ω;(P))= r‬‬
‫* اﻟﻤﺠﻤﻮﻋﺔ اﻟﻔﺎرﻏﺔ اذا آﺎن ‪d(Ω;(P))>r‬‬
‫ب‪ -‬ﻣﺴﺘﻮى ﻣﻤﺎس ﻟﻔﻠﻜﺔ ﻓﻲ أﺣﺪ ﻧﻘﻄﻬﺎ‬
‫ﺗﻌﺮﻳﻒ‬
‫ﻟﺘﻜﻦ ‪ A‬ﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻠﻜﺔ )‪S(Ω;r‬‬
‫ﻧﻘﻮل إن اﻟﻤﺴﺘﻮى )‪ (P‬ﻣﻤﺎس ﻟﻠﻔﻠﻜﺔ ‪ S‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ ‪ A‬ادا آﺎن )‪ (P‬ﻋﻤﻮدي ﻋﻠﻰ )‪ (ΩA‬ﻓﻲ ‪A‬‬
‫ﺧﺎﺻﻴﺔ‬
‫ﻟﺘﻜﻦ ‪ A‬ﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻠﻜﺔ )‪S(Ω;r‬‬
‫) ‪∀M∈(P‬‬
‫‪ΩA ⋅ AM = 0‬‬
‫⇔‬
‫)‪ (P‬ﻣﻤﺎس ﻋﻠﻰ )‪ S(Ω;r‬ﻓﻲ ‪A‬‬
‫ﺗﻤﺮﻳﻦ‬

‫ﻓﻲ ﻓﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ )‬

‫(‬

‫‪ ، O ; i ; j ; k‬ﻧﻌﺘﺒﺮ‪ S1‬اﻟﻔﻠﻜﺔ اﻟﺘﻲ ﻣﻌﺎدﻟﺘﻬﺎ‬

‫‪ x²+y²+z²-4x+2y-2z-3=0‬و ‪ S2‬اﻟﻔﻠﻜﺔ اﻟﺘﻲ ﻣﺮآﺰهﺎ ‪ Ω2‬و ﺷﻌﺎﻋﻬﺎ ‪ , 2‬و )‪ (P‬اﻟﻤﺴﺘﻮى اﻟﺬي‬
‫و )´‪ (P‬اﻟﻤﺴﺘﻮى اﻟﺬي ﻣﻌﺎدﻟﺘﻪ ‪. 2x-y-2z-1=0‬‬
‫ﻣﻌﺎدﻟﺘﻪ ‪x-2y+z+1=0‬‬
‫‪ -1‬ﺗﺄآﺪ أن )‪ (P‬و ‪ S1‬ﻳﺘﻘﺎﻃﻌﺎن وﻓﻖ داﺋﺮة ﻣﺤﺪدا ﻋﻨﺎﺻﺮهﺎ اﻟﻤﻤﻴﺰة‪.‬‬
‫‪ -2‬أدرس ﺗﻘﺎﻃﻊ )´‪ (P‬و ‪. S2‬‬
‫‪ -3‬ﺣﺪد ﻣﻌﺎدﻟﺔ اﻟﻤﺴﺘﻮى اﻟﻤﻤﺎس ﻟﻠﻔﻠﻜﺔ ‪ S1‬ﻋﻨﺪ اﻟﻨﻘﻄﺔ )‪A(1;1; 3‬‬
‫إﺟﺎﺑﺔ‬
‫اذن )‪ S1 = S (Ω1;3‬ﺣﻴﺚ )‪Ω1(2;-1;1‬‬
‫‪S1 : (x-2)²+(y+1)²+(z-1)²=9‬‬
‫‪-1‬‬
‫‪2+ 2+1+1‬‬
‫=) )‪d(Ω1;(P‬‬
‫‪= 6 ≺3‬‬
‫‪1+ 4+1‬‬
‫)‪ (P‬و ‪ S1‬ﻳﺘﻘﺎﻃﻌﺎن وﻓﻖ داﺋﺮة ﻣﺮآﺰهﺎ ‪ B‬ﻣﺴﻘﻂ اﻟﻌﻤﻮدي ﻟـ ‪ Ω1‬ﻋﻠﻰ )‪ (P‬و ﺷﻌﺎﻋﻬﺎ ‪9−6 = 3‬‬
‫‪ B‬هﻮ ﺗﻘﺎﻃﻊ اﻟﻤﺴﺘﻮى )‪ (P‬و اﻟﻤﺴﺘﻘﻴﻢ )‪ (D‬اﻟﻤﺎر ﻣﻦ ‪ Ω1‬و اﻟﻌﻤﻮدي ﻋﻠﻰ )‪(P‬‬
‫ﻟﺪﻳﻨﺎ )‪ n(1;−2;1‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻰ )‪ (P‬و ﻣﻨﻪ ﻣﻮﺟﻬﺔ ﻟـ )‪ (D‬و ﺑﺎﻟﺘﺎﻟﻲ اﻟﺘﻤﺜﻴﻞ اﻟﺒﺎراﻣﺘﺮي ﻟـ )‪ (D‬هﻮ‬

‫∈‪t‬‬

‫‪ x = 2 +t‬‬
‫‪‬‬
‫‪ y = −1 − 2t‬‬
‫‪ z = 1+ t‬‬
‫‪‬‬

‫‪x − 2 y + z + 1 = 0‬‬
‫‪t = −1‬‬
‫‪‬‬
‫‪ x =1‬‬
‫‪x = 2+t‬‬
‫‪‬‬
‫‪‬‬
‫‪ B ∈ ( P ) ∩ ( D ) ⇔ ‬اذن ﺗﻘﺎﻃﻊ )‪ (P‬و ‪ S1‬هﻮ اﻟﺪاﺋﺮة ‪ C B; 3‬ﺣﻴﺚ )‪B(1;1;0‬‬
‫‪⇔‬‬
‫‪ y = −1 − 2t‬‬
‫‪ y =1‬‬
‫‪‬‬
‫‪ z = 0‬‬
‫‪z = 1+ t‬‬
‫‪ -2‬ﻟﺪﻳﻨﺎ ‪ d(Ω2;(P´))=2‬وﻣﻨﻪ ﺗﻘﺎﻃﻊ ‪ S2‬و )´‪ (P‬هﻮ اﻟﻨﻘﻄﺔ ‪ C‬ﺑﺎﺗﺒﺎع ﻧﻔﺲ اﻟﺨﻄﻮات اﻟﺴﺎﺑﻘﺔ ﻧﺤﺪد اﻟﻨﻘﻄﺔ ‪C‬‬
‫‪ -3‬ﻟﺪﻳﻨﺎ ‪ A ∈ S1‬ﻟﻴﻜﻦ )´´‪ (P‬ﻣﻤﺎس ﻟـ ‪ S1‬ﻋﻨﺪ ‪A‬‬

‫)‬

‫(‬

‫‪M ( x ; y ; z ) ∈ ( P ") ⇔ AM ⋅ ΩA ⇔ ..........................‬‬

‫‪ -4‬ﺗﻘﺎﻃﻊ ﻣﺴﺘﻘﻴﻢ و ﻓﻠﻜﺔ‬
‫ﻧﻌﺘﺒﺮ ‪S : x²+y²+z²-2y +4z+4=0‬‬
‫ﻣﺜﺎل‬

‫‪Moustaouli Mohamed‬‬

‫‪6‬‬

‫‪http://arabmaths.ift.fr‬‬

‫∈‪t‬‬

‫∈ ‪t‬‬

‫‪ x = 1 + 2t‬‬
‫‪( D 1 ) :  y = 1 + t‬‬
‫‪ z = −3 + t‬‬
‫‪‬‬

‫∈‪t‬‬

‫‪ x = 3t‬‬
‫‪( D 2 ) :  y = 2‬‬
‫‪ z = −2 + t‬‬
‫‪‬‬

‫‪−1‬‬
‫‪‬‬
‫=‬
‫‪+ 2t‬‬
‫‪x‬‬
‫‪‬‬
‫‪2‬‬
‫‪‬‬
‫‪1‬‬
‫‪( D 3 ) :  y = + 3t‬‬
‫‪3‬‬
‫‪‬‬
‫‪ z = −2‬‬
‫‪‬‬
‫‪‬‬

‫ﺣﺪد ﺗﻘﺎﻃﻊ ‪ S‬ﻣﻊ آﻞ ﻣﻦ )‪ (D1‬و )‪ (D2‬و )‪(D3‬‬
‫ﺗﻤﺎرﻳﻦ‬
‫ﺗﻤﺮﻳﻦ‪1‬‬

‫ﻓﻲ ﻓﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) ‪(O ; i ; j ; k‬‬

‫ﻧﻌﺘﺒﺮ )‪ A(1;0;1‬و )‪ B(0;0;1‬و )‪ C(0;-1;1‬و اﻟﻤﺴﺘﻘﻴﻢ )‪ (D‬اﻟﻤﺎر ﻣﻦ ‪ C‬واﻟﻤﻮﺟﻪ ﺑـ )‪u(−1;2;1‬‬
‫‪ -1‬ﺑﻴﻦ أن ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ‪ M‬ﺣﻴﺚ ‪ MA=MB=MC‬ﻣﺴﺘﻘﻴﻢ وﺣﺪد ﺗﻤﺜﻴﻼ ﺑﺎرا ﻣﺘﺮﻳﺎ ﻟﻪ‬
‫‪ -2‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )‪ (P‬اﻟﻌﻤﻮدي ﻋﻠﻰ )‪ (D‬ﻓﻲ ‪C‬‬
‫‪ -3‬اﺳﺘﻨﺘﺞ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ ‪ S‬اﻟﻤﺎرة ﻣﻦ ‪A‬و ‪ B‬و اﻟﻤﻤﺎﺳﺔ ﻟـ )‪ (D‬ﻓﻲ ‪C‬‬

‫ﺗﻤﺮﻳﻦ‪2‬‬

‫ﻓﻲ ﻓﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ )‬

‫(‬

‫‪ O ; i ; j ; k‬ﻧﻌﺘﺒﺮ )‪ A(0;3;-5‬و )‪ B(0;7;-3‬و )‪C(1;5;-3‬‬

‫‪ -1‬أﻋﻂ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )‪(ABC‬‬
‫‪ -2‬أﻋﻂ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )‪ (Q‬اﻟﻤﺎر ﻣﻦ ‪ A‬ﺣﻴﺚ )‪ u(−1;2;1‬ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ‬
‫‪ -3‬ﻟﻴﻜﻦ )‪ (P‬اﻟﻤﺴﺘﻮى اﻟﻤﺤﺪد ﺑﺎﻟﻤﻌﺎدﻟﺔ ‪x+y+z=0‬‬
‫أ‪ -‬ﺗﺄآﺪ أن )‪(P‬و )‪ (ABC‬ﻳﺘﻘﺎﻃﻌﺎن وﻓﻖ ﻣﺴﻨﻘﻴﻢ )‪(D‬‬
‫ب‪ -‬ﺣﺪد ﺗﻤﺜﻴﻼ ﺑﺎرا ﻣﺘﺮﻳﺎ ﻟـ )‪(D‬‬
‫‪2 + z 2 +10z +9 = 0‬‬
‫‪x‬‬
‫‪‬‬
‫‪ -4‬ﻧﻌﺘﺒﺮ ﻓﻲ اﻟﻔﻀﺎء اﻟﺪاﺋﺮة )‪ (C‬اﻟﺘﻲ اﻟﻤﺤﺪدة ﺑـ‬
‫‪‬‬
‫‪y =0‬‬
‫‪‬‬
‫أ‪ -‬ﺣﺪد ﻣﻌﺎدﻟﺔ ﻟﻠﻔﻜﺔ ‪ S‬اﻟﺘﻲ ﺗﺘﻀﻤﻦ اﻟﺪاﺋﺮة )‪ (C‬و ﻳﻨﺘﻤﻲ ﻣﺮآﺰهﺎ إﻟﻰ )‪(ABC‬‬
‫ب ﺣﺪد ﺗﻘﺎﻃﻊ ‪ S‬و )‪(AC‬‬
‫ﺗﻤﺮﻳﻦ‪3‬‬
‫ﻓﻲ ﻓﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ﻣﺒﺎﺷﺮ ﻧﻌﺘﺒﺮ )‪ A(1;-1;1‬و )‪ B(3;1;-1‬و )‪ (P‬اﻟﻤﺴﺘﻮى ذا‬
‫‪ x = 3t‬‬
‫‪‬‬
‫اﻟﻤﻌﺎدﻟﺔ ‪ (D) 2x-3y+2z=0‬اﻟﻤﺴﺘﻘﻴﻢ اﻟﻤﻤﺜﻞ ﺑﺎرا ﻣﺘﺮﻳﺎ ﺑـ‬
‫∈ ‪ x = −2 − 3t t‬‬
‫‪ z = 2 + 4t‬‬
‫‪‬‬
‫‪ -1‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )‪ (Q‬اﻟﻤﺎر ﻣﻦ ‪ A‬و ‪ B‬واﻟﻌﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻘﻴﻢ )‪(D‬‬
‫‪ -2‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )´‪ (Q‬اﻟﻤﺎر ﻣﻦ ‪ A‬و ‪ B‬واﻟﻌﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻮى )‪(P‬‬
‫‪ -3‬أﺣﺴﺐ ))‪ d(A;(P‬و ))‪d(A;(D‬‬
‫‪ -4‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )´´‪ (Q‬اﻟﻤﺎر ﻣﻦ ‪ B‬و اﻟﻤﻮازي ﻟﻠﻤﺴﺘﻮى )‪(P‬‬
‫ﺗﻤﺮﻳﻦ‪4‬‬
‫ﻓﻲ ﻓﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ﻧﻌﺘﺒﺮ اﻟﻤﺴﺘﻮى )‪ (P‬ذا اﻟﻤﻌﺎدﻟﺔ ‪3x+2y-z-5=0‬‬

‫‪Moustaouli Mohamed‬‬

‫‪7‬‬

‫‪http://arabmaths.ift.fr‬‬

‫‪x − 2 y + z − 3 = 0‬‬
‫و )‪ (D‬اﻟﻤﺴﺘﻘﻴﻢ اﻟﻤﻌﺮف ﺑـ‬
‫‪‬‬
‫‪ x− y−z+2=0‬‬
‫‪ -1‬ﺣﺪد ﺗﻤﺜﻴﻼ ﺑﺎرا ﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ )‪(D‬‬
‫‪ -2‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى )´‪ (P‬اﻟﺬي ﻳﺘﻀﻤﻦ )‪ (D‬و اﻟﻌﻤﻮدي ﻋﻠﻰ )‪.(P‬‬
‫ﺗﻤﺮﻳﻦ‪5‬‬
‫ﻓﻲ ﻓﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ﻧﻌﺘﺒﺮ اﻟﻤﺴﺘﻮى )‪ (P‬ذا اﻟﻤﻌﺎدﻟﺔ ‪x+y+z+1=0‬‬
‫و اﻟﻤﺴﺘﻮى )‪ (Q‬ذا اﻟﻤﻌﺎدﻟﺔ ‪2x-2y-5=0‬‬
‫و )‪ (S‬ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ )‪ M(x;y;z‬اﻟﺘﻲ ﺗﺤﻘﻖ‪x²+y²+z²-2x+4y+6z+11=0‬‬
‫‪ -1‬ﺑﻴﻦ أن )‪ (S‬ﻓﻠﻜﺔ ﻣﺤﺪدا ﻣﺮآﺰهﺎ و ﺷﻌﺎﻋﻬﺎ‬
‫‪ -2‬ﺗﺄآﺪ أن )‪ (P‬ﻣﻤﺎس ﻟﻠﻔﻠﻜﺔ و ﺣﺪد ﺗﻘﺎﻃﻌﻬﻤﺎ‬
‫‪ -3‬ﺣﺪد ﺗﻤﺜﻴﻼ ﺑﺎراﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ )‪ (D‬اﻟﻤﺎر ﻣﻦ )‪ A(0;1;2‬و اﻟﻌﻤﻮدي ﻋﻠﻰ )‪(P‬‬
‫‪ -4‬ﺗﺤﻘﻖ أن ) ‪ ( P ) ⊥ ( Q‬و أﻋﻂ ﺗﻤﺜﻴﻼ ﺑﺎراﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ )´‪(D‬ﺗﻘﺎﻃﻊ )‪ (P‬و)‪(Q‬‬
‫ﺗﻤﺮﻳﻦ‪6‬‬
‫ﻓﻲ ﻓﻀﺎء ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ﻧﻌﺘﺒﺮ اﻟﻨﻘﻄﺔ )‪A(-2;3;4‬‬
‫اﻟﻤﺴﺘﻮى )‪ (P‬ذا اﻟﻤﻌﺎدﻟﺔ ‪ (S) x+2y-2z+15=0‬ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ )‪ M(x;y;z‬اﻟﺘﻴﺘﺤﻘﻖ‬
‫‪2‬‬
‫‪2‬‬
‫‪x + y − 2x − 8 = 0‬‬
‫‪ x²+y²+z²-2x+6y+10z-26=0‬و )‪ (C‬اﻟﺪاﺋﺮة اﻟﺘﻲ ﻣﻌﺎدﻟﺘﻬﺎ‬
‫‪‬‬
‫‪z =0‬‬
‫‪-1‬‬
‫‪-2‬‬
‫‪-3‬‬
‫‪-4‬‬

‫ﺑﻴﻦ أن )‪ (S‬ﻓﻠﻜﺔ ﻣﺤﺪدا ﻋﻨﺎﺻﺮهﺎ اﻟﻤﻤﻴﺰة‬
‫ﺑﻴﻦ أن )‪ (P‬و )‪ (S‬ﻳﺘﻘﺎﻃﻌﺎن وﻓﻖ داﺋﺮة آﺒﺮى )'‪ (C‬و ﺣﺪدهﺎ‬
‫ﺣﺪد ﻣﻌﺎدﻟﺘﻲ اﻟﻤﺴﺘﻮﻳﻦ اﻟﻤﻤﺎﺳﻴﻦ ﻟﻠﻔﻠﻜﺔ )‪ (S‬و اﻟﻤﻮازﻳﻴﻦ ﻟـ )‪(P‬‬
‫أآﺘﺐ ﻣﻌﺎدﻟﺔ اﻟﻔﻠﻜﺔ )'‪ (S‬اﻟﻤﺎر ﻣﻦ ‪ A‬اﻟﻤﺘﻀﻤﻦ ﻟﻠﺪاﺋﺮة )‪(C‬‬

‫‪Moustaouli Mohamed‬‬

‫‪8‬‬

‫‪http://arabmaths.ift.fr‬‬




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