( )P
ﺍﻟﻤﺎﺩﺓ :ﺍﻟﺭﻴﺎﻀﻴﺎﺕ
ﺘﻤﺎﺭﻴﻥ ﻭ ﺃﻨﺸﻁﺔ ﺩﺭﺱ :
ﺍﻟﻤﺴﺘﻭﻯ :ﺍﻟﺜﺎﻨﻴﺔ ﻉ.ﺕ.ﻉ.ﺡ.ﺃ
ﺍﻷﺴﺘﺎﺫ :ﻋﻠﻲ ﺍﻟﺸﺭﻴﻑ
ﺍﻟﺠﺩﺍﺀ ﺍﻟﺴﻠﻤﻲ ﻓﻲ ﺍﻟﻔﻀﺎﺀ ﻭ ﺘﻁﺒﻴﻘﺎﺘﻪ
ﺜﺎ.ﺍﻟﻤﺨﺘﺎﺭ ﺍﻟﺴﻭﺴﻲ.ﻨﻴﺎﺒﺔ ﺍﻟﺨﻤﻴﺴﺎﺕ
أ ﻧﺸﻄﺔ اﻟﺘﺬآﻴﺮ
ﻧﺸﺎط ﺗﺬآﻴﺮي رﻗﻢ : 1
اﻟﻤﺴﺘﻮى ) (Pﻣﻨﺴﻮب ﻟﻤﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ . O; i; jﻧﻌﺘﺒﺮ
اﻟﻨﻘﻂ . C (5;−1) , B(− 1;−1) , A(1;3) :
)
(
(1أ -أﺣﺴﺐ اﻟﺠﺪاء اﻟﺴﻠﻤﻲ . CA.CB :
ب -اﺣﺴﺐ اﻟﻤﺴﺎﻓﺘﻴﻦ CBو . CA
ج -أﺣﺴﺐ . det CA; CB
)
(
)
(
(2ﻧﻌﺘﺒﺮ αﻗﻴﺎس اﻟﺰاوﻳﺔ , CA; CBأﺣﺴﺐ ) cos(α
و ) sin (αﺛﻢ ﺁﺳﺘﻨﺘﺞ . α
(3ﺑﻴﻦ أن x + y − 4 = 0 :ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ل ) . ( AC
(4ﺣﺪد ﻣﻌﺎدﻟﺔ ) (Dواﺳﻂ اﻟﻘﻄﻌﺔ ] . [AB
(5أ -أﺣﺴﺐ ﻣﺴﺎﻓﺔ اﻟﻨﻘﻄﺔ Bﻋﻦ اﻟﻤﺴﺘﻘﻴﻢ ) . ( AC
ب -ﺑﻴﻦ أن ﻣﺴﺎﺣﺔ اﻟﻤﺜﻠﺚ ABCهﻲ . 12
_____________________________________
ﻧﺸﺎط ﺗﺬآﻴﺮي رﻗﻢ : 2
ﻟﺘﻜﻦ) ζداﺋﺮة ﻣﻌﺎدﻟﺘﻬﺎ .x +y -2x-2y-3=0 :
‐ (1ﺗﺤﻘﻖ أن A(2, 3) :ﺗﻨﺘﻤﻲ اﻟﻰ) (ζﺛﻢ أوﺟﺪ ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس
) (ζل ) (ζﻓﻲ . A
-(2أدرس ﺗﻘﺎﻃﻊ اﻟﺪاﺋﺮة و اﻟﻤﺴﺘﻘﻴﻢ ) (Dدو اﻟﻤﻌﺎدﻟﺔ :
2x+y-1=0
_______________________________________
ﻧﺸﺎط ﺗﺬآﻴﺮي رﻗﻢ :3
2
2
( 1ﻓﻲ اﻟﻤﺴﺘﻮى ) , ( ABCأﺣﺴﺐ ﺑﺪﻻﻟﺔ aاﻟﺠﺪاءات اﻟﺴﻠﻤﻴﺔ
اﻟﺘﺎﻟﻴﺔ . AJ.CK , AC.KJ , AB. AK , AB. AJ :
( 2ﺑﺂﻋﺘﺒﺎر ﻣﺴﺘﻮى ﻣﻨﺎﺳﺐ ﻓﻲ آﻞ ﺣﺎﻟﺔ ﻣﻦ اﻟﺤﺎﻻت اﻟﺘﺎﻟﻴﺔ ,
أﺣﺴﺐ ﺑﺪﻻﻟﺔ aاﻟﺠﺪاءات اﻟﺴﻠﻤﻴﺔ اﻟﺘﺎﻟﻴﺔ :
. AB. AG , AC. AG , AB. AF
( 3أ – ﺑﻴﻦ أ ن اﻟﻤﺘﺠﻬﺎت BH :و BFو CGﻣﺴﺘﻮاﺋﻴﺔ .
ب – أ ﺣﺴﺐ ﺑﺪﻻﻟﺔ اﻟﺠﺪاء اﻟﺴﻠﻤﻲ , BH.BF
ﺛﻢ ﺁﺳﺘﻨﺘﺞ BH.CG
( 4ﻟﻴﻜﻦ θﻗﻴﺎﺳﺎ ﻟﻠﺰاوﻳﺔ اﻟﻬﻨﺪﺳﻴﺔ JIBأ ﺣﺴﺐ اﻟﺠﺪاء اﻟﺴﻠﻤﻲ
IB.IJﺛﻢ ﺁﺳﺘﻨﺘﺞ ) . cos(θ
________________________________________
☺ ﻧﺸﺎط ﺑﻨﺎﺋﻲ رﻗﻢ ) : 2اﻷﺳﺎس و اﻟﻤﻌﻠﻢ اﻟﻤﺘﻌﺎﻣﺪان اﻟﻤﻤﻨﻈﻤﺎن(
ABCDEFGHﻣﻜﻌﺐ ﺣﻴﺚ . AB = 2 :ﻟﺘﻜﻦ اﻟﻨﻘﻂ Iو J
و Kﻋﻠﻰ اﻟﺘﻮاﻟﻲ ﻣﻨﺘﻀﻔﺎت اﻟﻘﻄﻊ ] [ABو ] [ADو ] . [AE
ﻓﻲ اﻟﻔﻀﺎء ) (ζاﻟﻤﻨﺴﻮب إ ﻟﻰ ﻣﻌﻠﻢ .م.م ) (O, i, j , kاﻟﻨﻘﻂ :
)D(2,0,4) ; C (−1,1,1) ; B(2,0,3) ; A(0,0,1
(1أ – ﺑﻴﻦ أ ن اﻟﻨﻘﻂ Aو Bو Cﻏﻴﺮ ﻣﺴﺘﻘﻴﻤﻴﺔ .
ب -ﺑﻴﻦ أ ن x + y − z + 1 = 0 :هﻲ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ
ﻟﻠﻤﺴﺘﻮى ) . ( ABC
(2ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى ) (Qاﻟﻤﺎر ﻣﻦ Aو اﻟﻤﻮﺟﻪ
)v(2,0,3
ﺑﺎﻟﻤﺘﺠﻬﺘﻴﻦ u (3,−1,2) :
و
(3أ – أ آﺘﺐ ﺗﻤﺜﻴﻼ ﺑﺎراﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ ) ∆( اﻟﻤﺎر ﻣﻦ D
و اﻟﻤﻮﺟﻪ ﺑﺎﻟﻤﺘﺠﻬﺔ ). w( 4,−2,1
ب -ﺣﺪد إ ﺣﺪاﺛﻴﺎت اﻟﻨﻘﻄﺔ Eﺗﻘﺎﻃﻊ ) ∆( و اﻟﻤﺴﺘﻮى
) . ( ABC
ج -ﺑﻴﻦ أ ن (∆) ⊂ (Q) :
(4ﺁﺳﺘﻨﺘﺞ ﺗﻘﺎﻃﻊ اﻟﻤﺴﺘﻮﻳﻴﻦ ) ( ABCو ). (Q
أﻧﺸﻄﺔ ﺑﻨﺎﺋﻴﺔ
☺ ﻧﺸﺎط ﺑﻨﺎﺋﻲ رﻗﻢ ) : 1ﺗﻘﺪﻳﻢ اﻟﺠﺪاء اﻟﺴﻠﻤﻲ ﻓﻲ اﻟﻔﻀﺎء (
ABCDEFGHﻣﻜﻌﺐ ﻃﻮل ﺿﻠﻌﻪ . A
ﻟﺘﻜﻦ اﻟﻨﻘﻂ Iو Jو Kﻋﻠﻰ اﻟﺘﻮاﻟﻲ ﻣﻨﺘﻀﻔﺎت اﻟﻘﻄﻊ ] [AEو
] [ABو ] . [BC
)
(
( 1ﺑﻴﻦ أ ن AI ; AJ ; AK :أﺳﺎس ﻟﻠﻔﻀﺎء .
( 2أ – ﺗﺤﻘﻖ ﻣﻦ أ ن AI . AJ = 0 :و AI . AK = 0
و AJ . AK = 0و . AI = AJ = AK = 1
)
(
ب-هﻞ اﻟﻤﻌﻠﻢ A; AI ; AJ ; AEﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ﻟﻠﻔﻀﺎء ؟
ج – ﺣﺪد ﻣﻌﻠﻤﺎ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ﻟﻠﻔﻀﺎء أﺻﻠﻪ . D
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☺ ﻧﺸﺎط ﺑﻨﺎﺋﻲ رﻗﻢ ) :3اﻟﺼﻴﻐﺔ اﻟﺘﺤﻠﻴﻠﻴﺔ ﻟﻠﺠﺪاء اﻟﺴﻠﻤﻲ و ﻣﻨﻈﻢ
ﻣﺘﺠﻬﺔ و ﻣﺴﺎﻓﺔ ﻧﻘﻄﺘﻴﻦ (
ﻓﻲ اﻟﻔﻀﺎء ) (ζاﻟﻤﻨﺴﻮب إ ﻟﻰ ﻣﻌﻠﻢ .م.م ) (O, i, j , kﻧﻌﺘﺒﺮ
اﻟﻤﺘﺠﻬﺘﻴﻦ u = x.i + y. j + z.kو v = x′.i + y′. j + z ′.k
( 1ﺑﻴﻦ أ ن . u.v = x.x′ + y. y′ + z.z ′ :
( 2أ ﺣﺴﺐ ﺛﻢ ﺁﺳﺘﻨﺘﺞ أ ن . u = x 2 + y 2 + z 2 :
(3ﻧﻌﺘﺒﺮ اﻟﻨﻘﻄﺘﻴﻦ A( x A ; y A ; z A ) :و ) . B ( xB ; y B ; z B
أ – ﺣﺪد ﻣﺜﻠﻮث إﺣﺪاﺛﻴﺎت اﻟﻤﺘﺠﻬﺔ . AB
ب – ﺁﺳﺘﻨﺘﺞ اﻟﻤﺴﺎﻓﺔ . AB
ﺍﻟﻤﺎﺩﺓ :ﺍﻟﺭﻴﺎﻀﻴﺎﺕ
ﺘﻤﺎﺭﻴﻥ ﻭ ﺃﻨﺸﻁﺔ ﺩﺭﺱ :
ﺍﻟﻤﺴﺘﻭﻯ :ﺍﻟﺜﺎﻨﻴﺔ ﻉ.ﺕ.ﻉ.ﺡ.ﺃ
ﺍﻷﺴﺘﺎﺫ :ﻋﻠﻲ ﺍﻟﺸﺭﻴﻑ
ﺍﻟﺠﺩﺍﺀ ﺍﻟﺴﻠﻤﻲ ﻓﻲ ﺍﻟﻔﻀﺎﺀ ﻭ ﺘﻁﺒﻴﻘﺎﺘﻪ
ﺜﺎ.ﺍﻟﻤﺨﺘﺎﺭ ﺍﻟﺴﻭﺴﻲ.ﻨﻴﺎﺒﺔ ﺍﻟﺨﻤﻴﺴﺎﺕ
☺ ﻧﺸﺎط ﺑﻨﺎﺋﻲ رﻗﻢ ) :4اﻟﻤﺘﺠﻬﺔ اﻟﻤﻨﻈﻤﻴﺔ ﻋﻠﻰ ﻟﻤﺴﺘﻮى –
اﻟﻤﻌﺎدﻟﺔ اﻟﺪﻳﻜﺎرﺗﻴﺔ ﻟﻤﺴﺘﻮى ﻣﺤﺪد ﺑﻨﻘﻄﺔ و ﻣﺘﺠﻬﺔ ﻣﻨﻈﻤﻴﺔ (
☺ ﻧﺸﺎط ﺑﻨﺎﺋﻲ رﻗﻢ ) :7دراﺳﺔ ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ) M ( x; y; z
ﺑﺤﻴﺚ ( x 2 + y 2 + z 2 + ax + by + cz + d = 0
ﻧﻌﺘﺒﺮ اﻟﻨﻘﻂ A(2;1;0) :و ) B (−1;1;2و )C (1;0;1
( 1ﺑﻴﻦ أ ن اﻟﻨﻘﻂ Aو Bو Cﺗﺤﺪد ﻣﺴﺘﻮى ) . (P
اﻟﻔﻀﺎء ) (ζاﻟﻤﻨﺴﻮب إ ﻟﻰ ﻣﻌﻠﻢ .م.م ) . (O, i, j , k
ﻟﺘﻜﻦ aو bو cو dأﻋﺪاد ﺣﻘﻴﻘﻴﺔ ﺑﺤﻴﺚ )(a; b; c) ≠ (0;0;0
ﻧﻌﺘﺒﺮ اﻟﻤﺠﻤﻮﻋﺔ ) (Eﺑﺤﻴﺚ :
ﻓﻲ اﻟﻔﻀﺎء ) (ζاﻟﻤﻨﺴﻮب إ ﻟﻰ ﻣﻌﻠﻢ .م.م ) (O, i, j , k
( 2ﻧﻌﺘﺒﺮ اﻟﻤﺘﺠﻬﺔ ). n(2;1;3
أ – ﺑﻴﻦ أ ن أ ن n. AB = 0و . n. AC = 0
ب – ﺁ ﺳﺘﻨﺘﺞ أ ن ﺟﻤﻴﻊ اﻟﻤﺴﺘﻘﻴﻤﺎت اﻟﻤﻮﺟﻬﺔ ﺑﺎﻟﻤﺠﻬﺔ nﻋﻤﻮدﻳﺔ
ﻋﻠﻰ اﻟﻤﺴﺘﻮى ) . (P
(3أ -ﻟﺘﻜﻦ Mﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻀﺎء .
ﺑﻴﻦ أ ن . M ∈ (P ) ⇔ n. AM = 0 :
ب – ﺁﺳﺘﻨﺘﺞ أ ن 2 x + y + 3 z − 5 = 0 :هﻲ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ
ﻟﻠﻤﺴﺘﻮى ) . (P
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☺ ﻧﺸﺎط ﺑﻨﺎﺋﻲ رﻗﻢ ) :5ﻣﺴﺎﻓﺔ ﻧﻘﻄﺔ ﻋﻦ ﻣﺴﺘﻮى (
اﻟﻔﻀﺎء ) (ζاﻟﻤﻨﺴﻮب إ ﻟﻰ ﻣﻌﻠﻢ .م.م ) . (O, i, j , k
ﻟﻴﻜﻦ ) (Pﻣﺴﺘﻮى ﻣﻦ اﻟﻔﻀﺎء و Bﻧﻘﻄﺔ ﻣﻦ ) (Pو nﻣﺘﺠﻬﺔ
ﻣﻨﻈﻤﻴﺔ ﻋﻠﻰ اﻟﻤﺴﺘﻮى ) . (Pﻟﺘﻜﻦ Aﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻀﺎء و
Hﻣﺴﻘﻄﻬﺎ اﻟﻌﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻮى ) . (P
( 1أ .ﺑﻴﻦ أ ن AB.n = AH .n :ﻻﺣﻆ )(AB = AH + HB
ب – ﺁﺳﺘﻨﺘﺞ أ ن :
AB.n
= . AH
n
ج – أﺣﺴﺐ ﻣﺴﺎﻓﺔ اﻟﻨﻘﻄﺔ ) A(− 3;1;−4ﻋﻦ اﻟﻤﺴﺘﻮى ) (P
اﻟﻤﺎر ﻣﻦ اﻟﻨﻘﻄﺔ ) B(1;−1;2و ) n(2;1;−2ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ .
( 2ﻟﻴﻜﻦ ) (xB ; y B ; z Bﻣﺜﻠﻮث إﺣﺪاﺛﻴﺎت اﻟﻨﻘﻄﺔ, B
و ax + by + cz + d = 0ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى ) . (P
أ – ﺗﺤﻘﻖ ﻣﻦ أ ن d = −(axB + by B + cz B + d ) :
ب – ﺑﻴﻦ أن ﻣﺴﺎﻓﺔ اﻟﻨﻘﻄﺔ ) A(x A ; y A ; z Aﻋﻦ اﻟﻤﺴﺘﻮى ) (P
axB + by B + cz B + d
= AH
هﻲ :
a2 + b2 + c2
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☺ ﻧﺸﺎط ﺑﻨﺎﺋﻲ رﻗﻢ ) :6ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ (
اﻟﻔﻀﺎء ) (ζاﻟﻤﻨﺴﻮب إ ﻟﻰ ﻣﻌﻠﻢ .م.م ) . (O, i, j , k
ﻧﻌﺘﺒﺮ اﻟﻔﻠﻜﺔ ) (Sاﻟﺘﻲ ﻣﺮآﺰهﺎ ) Ω(−1;1;2و ﺷﻌﺎﻋﻬﺎ . 3
( 1ﻣﻦ ﺑﻴﻦ اﻟﻨﻘﻂ اﻟﺘﺎﻟﻴﺔ ﺣﺪد اﻟﻨﻘﻂ اﻟﺘﻲ ﺗﻨﺘﻤﻲ إﻟﻰ اﻟﻔﻠﻜﺔ ) : (S
). D(1;3;3) , C (−1;0;4) , B(1;1;1) , A(2;1;2
(2ﻟﺘﻜﻦ ) M ( x; y; zﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻀﺎء .
أ – أﺣﺴﺐ اﻟﻤﺴﺎﻓﺔ ΩMﺑﺪﻻﻟﺔ xو yو. z
ب – ﺑﻴﻦ أ ن Mﺗﻨﺘﻤﻲ إﻟﻰ اﻟﻔﻠﻜﺔ ) (Sإذا وﻓﻘﻂ إذاآﺎن :
x2 + y 2 + z 2 + 2x − 2 y − 4z − 3 = 0
}(E) = {M ( x; y; z) / x2 + y 2 + z 2 + ax + by + cz + d = 0
( 1أ – ﺑﻴﻦ أ ن اﻟﻤﻌﺎدﻟﺔ ) (Eﺗﻜﺎﻗﺊ :
2
2
2
2
2
2
⎛ a ⎞ ⎛ b ⎞ ⎛ c ⎞ a + b + c − 4d
=0
⎜x + ⎟ +⎜ y + ⎟ +⎜z + ⎟ −
⎠2⎠ ⎝ 2
4
⎝ ⎠⎝ 2
a 2 + b 2 + c 2 − 4d
= kو ﺑﺂﻋﺘﺒﺎر اﻟﻨﻘﻄﺔ
ب – ﺑﻮﺿﻊ :
4
a b c
) . Ω(− ;− ;−ﺁﺳﺘﻨﺘﺞ أن ﻟﻜﻞ ﻧﻘﻄﺔ ) M ( x; y; zﻣﻦ
2 2 2
2
اﻟﻔﻀﺎء ﻟﺪﻳﻨﺎ . M ( x; y; z ) ∈ ( E ) ⇔ ΩM = k :
ج -ﺣﺪد ﺗﺒﻌﺎ ﻟﻘﻴﻢ إﺷﺎرة kﻃﺒﻴﻌﺔ اﻟﻤﺠﻤﻮﻋﺔ ) . (E
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ﺗﻤﺎرﻳﻦ ﺗﻄﺒﻴﻘﻴﺔ
اﻟﻔﻀﺎء ) (ζاﻟﻤﻨﺴﻮب إ ﻟﻰ ﻣﻌﻠﻢ .م.م ) . (O, i, j , k
ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ : 1
(Iأ ﺣﺴﺐ اﻟﺠﺪاء اﻟﺴﻠﻤﻲ u.vﻓﻲ اﻟﺤﺎﻻت اﻟﺘﺎﻟﻴﺔ :
u (−1;2;−3) ( 1و ). v(−2;2;2
(2
)u (5;1;−5
و
). v(1;−3;7
u ( 3; 2 ;1) ( 3و ). v(− 3; 2 ;1
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :2
ﺣﺪد ﻗﻴﻤﺔ xاﻟﺘﻲ ﻣﻦ أ ﺟﻠﻬﺎ اﻟﻤﺘﺠﻬﺘﻴﻦ uو vﻣﺘﻌﺎﻣﺪﺗﻴﻦ .
) u ( x;1;−5و ). v( x;−3; x
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :3
ﻧﻌﺘﺒﺮ اﻟﻤﺘﺠﻬﺎت ) u (1;−2;5و ) v(0;1;2و ). w(4;−3;−2
( 1أ ﺣﺴﺐ u.vو . v.w
( 2أ ﺣﺴﺐ uو . v
( 3ﺑﻴﻦ أ ن :اﻟﻤﺘﺠﻬﺘﻴﻦ uو . v
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :4
( 1ﻧﻌﺘﺒﺮ اﻟﻨﻘﻂ اﻟﺘﺎﻟﻴﺔ ) A(3;3;−5و ) B(4;−1;3و )C (5;2;2
أ ﺣﺴﺐ AB. ACو ABو . AC
ﺍﻟﻤﺎﺩﺓ :ﺍﻟﺭﻴﺎﻀﻴﺎﺕ
ﺘﻤﺎﺭﻴﻥ ﻭ ﺃﻨﺸﻁﺔ ﺩﺭﺱ :
ﺍﻟﻤﺴﺘﻭﻯ :ﺍﻟﺜﺎﻨﻴﺔ ﻉ.ﺕ.ﻉ.ﺡ.ﺃ
ﺍﻷﺴﺘﺎﺫ :ﻋﻠﻲ ﺍﻟﺸﺭﻴﻑ
ﺍﻟﺠﺩﺍﺀ ﺍﻟﺴﻠﻤﻲ ﻓﻲ ﺍﻟﻔﻀﺎﺀ ﻭ ﺘﻁﺒﻴﻘﺎﺘﻪ
ﺜﺎ.ﺍﻟﻤﺨﺘﺎﺭ ﺍﻟﺴﻭﺴﻲ.ﻨﻴﺎﺒﺔ ﺍﻟﺨﻤﻴﺴﺎﺕ
ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :5
( 1ﻧﻌﺘﺒﺮ اﻟﻨﻘﻂ اﻟﺘﺎﻟﻴﺔ B (− 1;−2;−1) , A(5;−5;3) :
) . D(2;8;2 ) , C (8;5;6
أ – ﺑﻴﻦ أ ن . AC = BD :
ب – ﺑﻴﻦ أن اﻟﻤﺘﺠﻬﺘﻴﻦ ABو ACﻣﺘﻌﺎﻣﺪﺗﻴﻦ .
ج – ﺣﺪد ﻃﺒﻴﻌﺔ اﻟﺮﺑﺎﻋﻲ . ABCD
( 2ﻧﻌﺘﺒﺮ اﻟﻨﻘﻂ اﻟﺘﺎﻟﻴﺔ C (1;−2;2 ) , B(2;0;−2 ) , A(1;1;1) :
ﺑﻴﻦ أ ن اﻟﻤﺜﻠﺚ ABCﻗﺎﺋﻢ اﻟﺰاوﻳﺔ .
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :6
أﻋﻂ ﻣﺘﺠﻬﺔ ﻣﻨﻈﻤﻴﺔ ﻋﻠﻰ اﻟﻤﺴﺘﻮى ) (Pﻓﻲ آﻞ ﺣﺎﻟﺔ :
(P ) : x − y + 2 = 0 ( 2 , (P ) : 2 x + y − z + 7 = 0 ( 1
(P ) : z = 2
(P ) : x = 0 ( 3
(4,
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :7
ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى ) (Pاﻟﻤﺎر ﻣﻦ اﻟﻨﻘﻄﺔ Aو اﻟﻤﺘﺠﻬﺔ
nﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ :
(1
)A(1;−2;3
;
) n(1;−3;−2
)A(2;−1;1
;
) n(1;0;2
(2
(3
) n(0;0;1) ; A(− 1;2;4
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :8
ﺣﺪد ﻓﻲ آﻞ ﺣﺎﻟﺔ ﺗﻤﺜﻴﻞ ﺑﺎراﻣﺘﺮي ﻟﻤﺴﺘﻘﻴﻢ اﻟﻤﺎر ﻣﻦ اﻟﻨﻘﻄﺔ Aو
اﻟﻌﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻮى ) . (P
)(P ) : 2 x − 3 y + z + 1 = 0 ; A(− 1;1;2
(1
(P ) : x + y − 1 = 0
) ; A(0;0;2
(2
(P ) : y − 4 = 0
) ; A(4;−3;2
(3
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :9
( 1ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى ) (Pﻓﻲ آﻞ ﺣﺎﻟﺔ :
أ (P ) -ﻣﺎر ﻣﻦ اﻟﻨﻘﻄﺔ ) A(4;−3;2و ﻋﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻘﻴﻢ ذي
⎧x − y + 2z − 3 = 0
اﻟﻤﻌﺎدﻟﺘﻴﻦ :
⎨
⎩2 x − y − 3 z = 0
ب ( ) (Pﻣﺎر ﻣﻦ اﻟﻨﻘﻄﺔ ) A(4;1;0و ﻋﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻘﻴﻢ
⎧ x = 2 − 3t
⎪
) (Dاﻟﻤﻌﺮف ﻳﺎﻟﺘﻤﺜﻴﻞ اﻟﺒﺎراﻣﺘﺮي ⎨ y = −3 + t (t ∈ IR ) :
⎪ z = 4 − 2t
⎩
ج ( ) (Pﻣﺎر ﻣﻦ اﻟﻨﻘﻄﺘﻴﻦ ) A(0;−2;0و ) B (2;2;2و ﻋﻤﻮدي
ﻋﻠﻰ اﻟﻤﺴﺘﻮى . (Q ): x − 2 y + 3 z − 7 = 0 :
( 2ﺣﺪد ﻣﺜﻠﻮث إﺣﺪاﺛﻴﺎت اﻟﻨﻘﻄﺔ Hاﻟﻤﺴﻘﻂ اﻟﻌﻤﻮدي ﻟﻠﻨﻘﻄﺔ
) A(0;1;2ﻋﻠﻰ اﻟﻤﺴﺘﻮى . (P ): 2 x + y − z + 2 = 0
( 3ﻧﻌﺘﺒﺮ اﻟﻨﻘﻂ ) A(1;0;−1و ) B (2;2;3و ) C (3;1;−2
أ – ﺗﺤﻘﻖ ﻣﻦ أ ن اﻟﻨﻘﻂ Aو Bو Cﻏﻴﺮ ﻣﺴﺘﻘﻴﻤﻴﺔ .
ب -ﺑﻴﻦ أ ن اﻟﻤﺘﺠﻬﺔ ) n(2;−3;1ﻣﻨﻈﻤﻴﺔ ﻋﻠﻰ اﻟﻤﺴﺘﻮى ) ( ABC
ﺛﻢ ﺁﺳﺘﻨﺘﺞ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى ) . ( ABC
ج – ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻤﺴﺘﻮى ) (Pاﻟﻤﺎر ﻣﻦ اﻟﻨﻘﻄﺔ
) D(− 2;2;−1و اﻟﻤﻮازي ﻟﻠﻤﺴﺘﻮى ) . ( ABC
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :10
( 1أﺣﺴﺐ ﻣﺴﺎﻓﺔ اﻟﻨﻘﻄﺔ ) A(0;−2;−1ﻋﻦ اﻟﻤﺴﺘﻮى
(P ): x + y − z + 1 = 0
( 2ﻧﻌﺘﺒﺮ اﻟﻨﻘﻄﺘﻴﻦ ) A(1;−2;1و ) B(1;2;3و اﻟﻤﺘﺠﻬﺔ
) n(− 1;1;2أ ﺣﺴﺐ ﻣﺴﺎﻓﺔ اﻟﻨﻘﻄﺔ Aﻋﻦ اﻟﻤﺴﺘﻮى ) (Pاﻟﻤﺎر ﻣﻦ
اﻟﻨﻘﻄﺔ Bو nﻣﺘﺠﻬﺔ ﻣﻨﻈﻤﻴﺔ ﻋﻠﻴﻪ .
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :11
( 1أآﺘﺐ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ ) (Sاﻟﺘﻲ ﻣﺮآﺰهﺎ ) Ω(1;−2;1و
ﺷﻌﺎﻋﻬﺎ . R = 6
( 2أآﺘﺐ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ ) (Sاﻟﺘﻲ ﻣﺮآﺰهﺎ )Ω(− 1;4;5
و ﺗﻤﺮ ﻣﻦ اﻟﻨﻘﻄﺔ ) . A(3;5;2
( 3أآﺘﺐ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ ) (Sاﻟﺘﻲ أ ﺣﺪ أﻗﻄﺎرهﺎ ] [AB
ﺑﺤﻴﺚ ) A(− 1;0;3و ). B(1;2;5
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :12
ﺣﺪد ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ ) M ( x; y; zاﻟﺘﻲ ﺗﺤﻘﻖ اﻟﻤﻌﺎدﻟﺔ ,إذا آﺎﻧﺖ
ﻓﻠﻜﺔ ﺣﺪد ﺷﻌﺎﻋﻬﺎ و ﻣﺮآﺰهﺎ .
2
2
2
(1
x + y + z − 2x − 4 y + 4 = 0
11
x2 + y 2 + z 2 − x + 2 y − 2z + = 0 ( 2
4
2
2
2
(3
x + y + z + 4x + 2z + 5 = 0
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ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ رﻗﻢ :13
( 1ﺑﻴﻦ أن اﻟﻤﺴﺘﻮى ) (Pﻣﻤﺎس ﻟﻠﻔﻠﻜﺔ ) S (Ω; Rﺑﺤﻴﺚ :
(P ) : x + 2 y + 2 z − 1 = 0و ) Ω(2;1;0و R = 1
( 2ﺗﺤﻘﻖ ﻣﻦ أ ن اﻟﻨﻘﻄﺔ ) A(− 3;1;0ﺗﻨﺘﻤﻲ إﻟﻰ اﻟﻔﻠﻜﺔ
(S ): x 2 + y 2 + z 2 + 6 x − y + 9 = 0ﺛﻢ ﺣﺪد ﻣﻌﺎدﻟﺔ اﻟﻤﺴﺘﻮى
اﻟﻤﻤﺎس ﻟﻠﻔﻠﻜﺔ ) (Sﻓﻲ اﻟﻨﻘﻄﺔ . A
( 3أدرس اﻟﻮﺿﻊ اﻟﻨﺴﺒﻲ ﻟﻠﻔﻠﻜﺔ ) (Sاﻟﺘﻲ ﻣﻌﺎدﻟﺘﻬﺎ اﻟﺪﻳﻜﺎرﺗﻴﺔ هﻲ
(S ): x 2 + y 2 + z 2 − 4 x − 2 y + 4 = 0ﻣﻊ آﻞ ﻣﻦ اﻟﻤﺴﺘﻮﻳﺎت
اﻟﺘﺎﻟﻴﺔ :
(P1 ) : x + y + z − 4 = 0
أ-
(P2 ) : 2 x + + y + z + 1 = 0
ب-
(P3 ) : 2 x + y − 2 z − 8 = 0
ج-
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