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  1. 1. www.Achamel.net : ' ' ' 1 2 6 3 2 3 3 3 3 2 t t t t t t ⎧ 1-‫إﺣﺪاﺛﻴﺎت‬ ‫ﻣﺜﻠﻮث‬ ‫ﺗﺤﺪﻳﺪ‬A − ‫اﻟﻨﻈﻤﺔ‬ ‫ﻧﺤﻞ‬ ‫أن‬ ‫ﻳﺠﺐ‬ + = − ⎪ − + = +⎨ ⎪ − = −⎩ : ' ' ' 2 3 7 3 5 2 0 t t t t t t ⎧ + ‫ﺗﻜﺎﻓﺊ‬ ‫اﻟﻨﻈﻤﺔ‬ ‫هﺬﻩ‬ = ⎪ − =⎨ ⎪ − =⎩ ' ' 2 3 7 2 0 t t t t + = ‫اﻟﻨﻈﻤﺔ‬ ‫ﺣﻞ‬‫هﻮ‬ ⎧⎪ ⎨( − =⎪⎩ )'21 ‫ج‬'21' 3 5t t ‫اﻟﺰو‬ ‫أن‬ ‫وﺑﻤﺎ‬( )‫ﻟ‬ ‫ﺣﻞ‬‫ﻠﻤﻌﺎدﻟﺔ‬− = ‫ﻓﺈن‬t = 2‫و‬t’= 1 ‫إﺣﺪاﺛﻴﺎت‬ ‫ﻣﺜﻠﻮث‬ ‫ﻓﺈن‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬A‫اﻟﻤﺴﺘﻘﻴﻤﻴﻦ‬ ‫ﺗﻘﺎﻃﻊ‬ ‫ﻧﻘﻄﺔ‬)D(‫و‬)D’(‫هﻮ‬)‫اﻟﻤﺜﻠﻮث‬ ‫هﺬا‬ ‫ﻋﻠﻰ‬ ‫ﺣﺼﻠﻨﺎ‬ ‫ﺑﺘﻌﻮﻳﺾ‬t‫ﺑﺎﻟﻘﻴﻤﺔ‬2‫ﻟﻠﻤﺴﺘﻘﻴﻢ‬ ‫اﻟﺒﺎراﻣﺘﺮي‬ ‫اﻟﺘﻤﺜﻴﻞ‬ ‫ﻓﻲ‬)D(‫ﺑﺘﻌﻮﻳﺾ‬ ‫أو‬t’‫ﺑﺎﻟﻘﻴﻤﺔ‬1‫اﻟﺘ‬ ‫ﻓﻲ‬‫ﻟﻠﻤﺴﺘﻘﻴﻢ‬ ‫اﻟﺒﺎراﻣﺘﺮي‬ ‫ﻤﺜﻴﻞ‬ )D’( ( )3,4,1 2–‫ﻣﻌﺎدﻟﺔ‬‫ﻟﻠﻤﺴﺘﻮى‬ ‫دﻳﻜﺎرﺗﻴﺔ‬)p( ‫اﻟﻤﺴﺘﻮى‬)p(‫ﺑﺎﻟﻨﻘﻄﺔ‬ ‫ﻣﺤﺪد‬‫وﺑﺎﻟﻤﺘﺠﻬﺘﻴﻦ‬u)‫ﻟﻠﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻮﺟﻬﺔ‬ ( )3,4,1A( )2,3, 1−)D((‫و‬( )3,1, 2− −' u ( ), ,M )‫ﻟﻠﻤﺴﺘﻘﻴﻢ‬ ‫ﻣﻮﺟﻬﺔ‬)D’((‫ﻟﺘﻜﻦ‬x y z ( ) ‫اﻟﻔﻀﺎء‬ ‫ﻣﻦ‬ ‫ﻧﻘﻄﺔ‬. ‫ﻟﺪﻳﻨﺎ‬:( )det , , ' 0M P AM u u∈ ⇔ = 3 2 1 0 − = − 2 3 1− 3 4 1 x y z − ⇔ − − 3 0 1 − =( ) 2 1 3 z+ − − 3 2 − − ( ) 2 4 . 1 y − − - 1 2− ( ) 3 3 . 1 x⇔ − − ( ) ( ) ( )5 3 7 4 11 1 0x y z⇔ − − + − + − = 5 7 11 24x y z⇔ − + + − = 0
  2. 2. www.Achamel.net 5 7 11 24 0x y z⇔ − − + = 5 7 11 24x y z− − + = ‫ﻟﻠﻤﺴﺘﻮى‬ ‫دﻳﻜﺎرﺗﻴﺔ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫إذن‬)p(‫ﺑﺎﻟﻔﻌﻞ‬ ‫هﻲ‬: 0 3–‫أ‬–‫ﻟﻠﻤﺴﺘﻘﻴﻢ‬ ‫ﺑﺎراﻣﺘﺮي‬ ‫ﺗﻤﺜﻴﻞ‬( )Δ ‫اﻟﻨﻈﻤﺔ‬: ⎪ ⎨ 6 14 0 2 4 0 x y z x y z 1 2 ⎧ + + − = − + − =⎪⎩ 2 8 18 0 2 4 x z x y z + − =⎧ ⎨ ‫ﺗﻜﺎﻓﺊ‬: 0− + − =⎩ 9 4 9 4 2 4 0 x z z y z = −⎧ ⎨ 2+1 ‫أي‬: − − + − =⎩ 9 4 5 2 ‫أي‬: x z= y −⎧ ⎨ z= −⎩ ‫ﻟﻠﻤﺴﺘﻘﻴﻢ‬ ‫ﺑﺎراﻣﺘﺮي‬ ‫ﺗﻤﺜﻴﻞ‬ ‫إذن‬‫هﻮ‬: ( )Δ ( ) 9 4 5 2 x t y t t z t = −⎧ ⎪ = − ∈⎨ ⎪ =⎩ ‫ب‬–‫إﺣﺪاﺛﻴﺎت‬ ‫ﻣﺜﻠﻮث‬B ( ):5 7 11 24 0P x y z− ‫ﻟﺪﻳﻨﺎ‬:− + = ( ) ( ) 9 4 : 5 2 x t y t t z t = −⎧ ⎪ Δ = − ∈⎨ ⎪ =⎩ ‫و‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫وﻟﺪﻳﻨﺎ‬5( ) ( :)9 4 7 5 2 11 24 0t t t− − − − + = 14 11 34 0t t t ‫ﺗﻜﺎﻓﺊ‬:20− + − + =17 34t ‫أي‬− = − ‫أي‬:t = 2 ( ( )Δ‫هﻮ‬: ‫إﺣﺪاﺛﻴﺎت‬ ‫ﻣﺜﻠﻮث‬ ‫إذن‬B‫ﺗﻘﺎﻃﻊ‬ ‫ﻧﻘﻄﺔ‬)P(‫و‬)' '112

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