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01
اﻟﻜﻔﺎءات اﻟﻤﺴﺘﮭﺪﻓﺔ
ﺗﻌﻠﯿﻢ ﻧﻘﻂ أﻋﻄﯿﺖ إﺣﺪاﺛﯿﺎﺗﮭﺎ.
ﺗﻌﯿﯿﻦ ﻣﻌﺎدﻟﺔ ﻟﻤﺴﺘﻮ ﻣﻮاز ﻷﺣﺪ ﻣﺴﺘﻮﯾﺎت
اﻹﺣﺪاﺛﯿﺎت.
ﺗﻌﯿﯿﻦ ﻣﻌﺎدﻻت ﻣﺴﺘﻘﯿﻢ ﻣﻌﺮف ﺑﻨﻘﻄﺔ و
ﺷﻌﺎع ﺗﻮﺟﯿﮫ ﻟﮫ.
إﺛﺒﺎت أن أﺷﻌﺔ ﻣﻌﻄﺎة ﺗﻨﺘﻤﻲ إﻟﻰ ﻧﻔﺲ
اﻟﻤﺴﺘﻮي.
اﺳﺘﻌﻤﺎل ﻣﺒﺮھﻨﺔ ﻓﯿﺜﺎﻏﻮرس ﻹﯾﺠﺎد
اﻟﻤﺴﺎﻓﺔ ﺑﯿﻦ ﻧﻘﻄﺘﯿﻦ.
اﺳﺘﻌﻤﺎل دﺳﺘﻮر اﻟﻤﺴﺎﻓﺔ ﺑﯿﻦ ﻧﻘﻄﺘﯿﻦ
ﻟﺘﻌﯿﯿﻦ ﻣﺠﻤﻮﻋﺔ
ﻧﻘﻂ ﺗﺤﻘﻖ ﺧﺎﺻﯿﺔ ﻣﺎ.
ﯾﺘﻤﺤﻮر ھﺬا اﻟﻔﺼﻞ ﻋﻠﻰ ﺛﻼث ﻣﺤﺎور
أﺳﺎﺳﯿﺔ ھﻲ:
1 * ﺗﻌﻠﯿﻢ اﻟﻨﻘﻂ ﻓﻲ اﻟﻔﻀﺎء ﻣﻦ ﺧﻼل
إدراج ﻣﻔﮭﻮم اﻟﻤﻌﻠﻢ.
2 * اﺳﺘﻌﻤﺎل اﻹﺣﺪاﺛﯿﺎت ﻟﺤﻞ ﻣﺴﺎﺋﻞ
ﻣﺮﺗﺒﻄﺔ ﺑﺎﻻﺳﺘﻘﺎﻣﯿﺔ، اﻟﺘﻮازي، اﻷﺷﻌﺔ
ﻣﻦ ﻧﻔﺲ اﻟﻤﺴﺘﻮي...
3* ﺗﻌﯿﯿﻦ اﻟﻤﻌﺎدﻟﺔ اﻟﺪﯾﻜﺎرﺗﯿﺔ ﻟﻜﻞ ﻣﻦ ﺳﻄﺢ
اﻟﻜﺮة، اﻟﻤﺨﺮوط اﻟﺪوراﻧﻲ، اﻻﺳﻄﻮاﻧﺔ
اﻟﺪوراﻧﯿﺔ، اﻟﻤﺴﺘﻮي اﻟﻤﻮازي ﻷﺣﺪ ﻣﺴﺘﻮﯾﺎت
اﻹﺣﺪاﺛﯿﺎت...
1
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اﻟﻨﺸﺎط 4 : اﻷﻧﺸﻄﺔ
اﻟﮭﺪف :دراﺳﺔ ﺗﻘﺎﻃﻊ ﻣﺨﺮوط دوراﻧﻲ ﻣﻊ ﻣﺴﺘﻮ.
اﻟﻨﺸﺎط 1 :
3 61 اﻟﮭﺪف : ﺗﻌﯿﯿﻦ إﺣﺪاﺛﯿﺎت ﻧﻘﻂ ﻓﻲ ﻣﻌﻠﻢ ﻟﻠﻔﻀﺎء
=x ﺗﺼﺤﯿﺢ: 2 z 2 − b
4 9 1( ﻟﺪﯾﻨﺎ ) 0 ,0 ,0 ( ، C ( 3, 0, 2 ) ، B ( 3, 0, 0 ) ، A
1( • ) ∑ ( داﺋﺮة ﻣﺮﻛﺰھﺎ اﻟﻨﻘﻄﺔ Cو ﻧﺼﻒ ﻗﻄﺮھﺎ . R
) 2 ,0 ,0 ( H ( 0, 4, 2 ) ، F ( 3, 4, 0 ) ، E ( 0, 4, 0 ) ، D
3
• ﺑﺘﻄﺒﯿﻖ ﻣﺒﺮھﻨﺔ ﻃﺎﻟﺲ ﻧﺠﺪ: R = cو ﻣﻨﮫ ﻣﻌﺎدﻟﺔ 2( ) 0 ,0 ,1( . K ( 0, 0,1) ، J ( 0,1, 0 ) ، Iاﻟﻨﻘﻄﺔ Aھﻲ
4
9 ﻣﺒﺪأ اﻟﻤﻌﻠﻢ.
) ∑ ( ھﻲ: 2 . x 2 + y 2 = c 3( ) 2 ,2 ,3 ( Lو ) 2 ,4 ,2 ( . M
61
• اﻟﻤﺴﺘﻮي ) ( Pﯾﻮﻟﺪ اﻟﻤﺨﺮوط اﻟﺪوراﻧﻲ ﻟﻤﺎ ﺗﺘﻐﯿﺮ cﻓﻲ
اﻟﻤﺠﺎل ]4 ,0 [ و ﻣﻨﮫ اﻟﻤﻌﺎدﻟﺔ.
اﻟﻨﺸﺎط 2 :
اﻟﮭﺪف : ﺗﻌﯿﯿﻦ ﻣﻌﺎدﻻت ﻣﺴﺘﻮﯾﺎت و ﻣﺴﺘﻘﯿﻤﺎت.
2( • ﻣﻌﺎدﻟﺔ اﻟﻤﺴﺘﻮي ) (Qھﻲ: . y = b 1( ) 0 ,0 ,0 ( ، C ( 0,1, 0 ) ، B (1, 0, 0 ) ، A
• ﺑﻜﺘﺎﺑﺔ ﺟﻤﻠﺔ اﻟﺘﻘﺎﻃﻊ ﻧﺘﺤﺼﻞ ﻋﻠﻰ اﻟﻤﻄﻠﻮب. )1,0 ,0 ( E ( 0,1,1) ، G (1, 0,1) ، F (1,1, 0 ) ، D
و )1,1,1( . H
2( اﻟﻤﺴﺘﻮي ) x ، z = 1 : (GDEو yﻛﯿﻔﯿﺎن.
اﻟﻤﺴﺘﻮي ) x ، z = 0 : ( ABCو yﻛﯿﻔﯿﺎن.
اﻟﻤﺴﺘﻮي ) x ، y = 1 : ( EHFو zﻛﯿﻔﯿﺎن.
اﻟﻤﺴﺘﻘﯿﻢ ) y = 0 : ( ABو 0 = x ، zﻛﯿﻔﻲ.
اﻟﻤﺴﺘﻘﯿﻢ ) x = 0 : ( ACو 0 = y ، zﻛﯿﻔﻲ.
اﻷﻋﻤﺎل اﻟﻤﻮﺟﮭﺔ
اﻟﮭﺪف ﻣﻦ اﻷﻋﻤﺎل اﻟﻤﻮﺟﮭﺔ اﻟﺨﺎﺻﺔ ﺑﮭﺬا اﻟﻔﺼﻞ ھﻮ ﺗﻌﯿﯿﻦ اﻟﻤﺴﺘﻘﯿﻢ ) y = 1 : ( HEو 1 = x ، zﻛﯿﻔﻲ.
اﻟﻤﻌﺎدﻻت اﻟﺪﯾﻜﺎرﺗﯿﺔ ﻟﺒﻌﺾ اﻟﻤﺠﻤﻮﻋﺎت اﻟﻤﻨﺼﻮص ﻋﻠﯿﮭﺎ 1
ﻓﻲ اﻟﺒﺮﻧﺎﻣﺞ و ﺑﺎﻟﺘﺎﻟﻲ ﻓﻜﻞ اﻟﻨﺘﺎﺋﺞ اﻷﺳﺎﺳﯿﺔ اﻟﺨﺎﺻﺔ ﺑﮭﺬه 3( إﺣﺪاﺛﯿﺎت ﻣﻨﺘﺼﻒ ] [ ABھﻲ . , 0, 0
اﻟﻤﻌﺎدﻻت ﻗﺪ أﻋﻄﯿﺖ و ﻻ ﻧﺮى أي داع ﻹﻋﺎدة ﻛﺘﺎﺑﺘﮭﺎ. 2
1
إﺣﺪاﺛﯿﺎت ﻣﻨﺘﺼﻒ ] [CEھﻲ . 0,1,
2
ﺗــﻤـــﺎرﯾــــﻦ اﻟﻨﺸﺎط 3 :
اﻟﮭﺪف : ﺗﻌﯿﯿﻦ اﻟﻤﺴﺎﻓﺔ ﺑﯿﻦ ﻧﻘﻄﺘﯿﻦ.
1 1( ﺧﻄﺄ. 2( ﺻﺤﯿﺢ . 3( ﺧﻄﺄ. 1( ) 0 ,0 ,0 ( ، C ( 3, 4, 0 ) ، B ( 3, 0, 0 ) ، A
) 0 ,4 ,0 ( G ( 3, 4, 2 ) ، F ( 3, 0, 2 ) ، E ( 0, 0, 2 ) ، D
2 1( ﺧﻄﺄ . 2( ﺧﻄﺄ. 3( ﺧﻄﺄ.
و ) 2 ,4 ,0 ( . H
3 1( ﺻﺤﯿﺢ. 2( ﺧﻄﺄ. 3( ﺧﻄﺄ. 2( ﺑﺘﻄﺒﯿﻖ ﻣﺒﺮھﻨﺔ ﻓﯿﺜﺎﻏﻮرث ﻓﻲ اﻟﻤﺜﻠﺚ ACGو ﻋﻠﻤﺎ أن
. CG = AEﯾﻜﻮن ﻟﺪﯾﻨﺎ: 2 . AG 2 = AC 2 + AE
4 1( ﺧﻄﺄ. 2( ﺻﺤﯿﺢ . 3( ﺧﻄﺄ.
ﺑﺘﻄﺒﯿﻖ ﻧﻔﺲ اﻟﻤﺒﺮھﻨﺔ ﻓﻲ اﻟﻤﺜﻠﺚ ABCو ﻋﻠﻤﺎ
1( ﺧﻄﺄ. 2( ﺻﺤﯿﺢ . 3( ﺧﻄﺄ أن BC = ADﯾﻜﻮن ﻟﺪﯾﻨﺎ: 2 . AC = AB 2 + ADﻣﻦ
2
5
اﻟﻌﻼﻗﺘﯿﻦ اﻟﺴﺎﺑﻘﺘﯿﻦ ﻧﺴﺘﻨﺘﺞ اﻟﻤﻄﻠﻮب.
6 1( ﺧﻄﺄ . 2( ﺧﻄﺄ. 3( ﺻﺤﯿﺢ. 3( 92 = 2 AGو ﻣﻨﮫ 92 = AG
) ( xG − x A 4( + ... = 29 = AG
2
7 ﺧﻄﺄ
5( ﺑﺎﺳﺘﻌﻤﺎل ﻣﻦ ﺟﮭﺔ اﻟﻨﺘﯿﺠﺔ اﻟﺴﺎﺑﻘﺔ و ﺑﺎﺳﺘﻌﻤﺎل اﻟﻌﻼﻗﺔ
8 اﻟﺠﻮاب ﺟـ( MN = EGﻓﻲ اﻟﻤﺜﻠﺚ EHGﻣﻦ ﺟﮭﺔ ﺛﺎﻧﯿﺔ ﻧﺠﺪ:
1
2
اﻟﺠﻮاب ب( 9 2 / 5 = MN
2
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x = 2k + 2 (01 اﻟﺠﻮاب ﺟـ
3x − 2 y − 8 = 0
، y = 3k − 1
y − 3z + 2 = 0 z = −k + 1 (11 اﻟﺠﻮاب ب
( 2,1, 3) ﻧﻘﻄﺔ اﻟﺘﻘﺎﻃﻊ ھﻲ 78 (21 اﻟﺠﻮاب ب
79 (31 اﻟﺠﻮاب ب
x = 2 + k
و y = 1 − k ﯾﻜﻮن ﻟﺪﯾﻨﺎz = k ﺑﻮﺿﻊ ﻣﺜﻼ (41 اﻟﺠﻮاب ﺟـ
z = k
r r r r
uuu uu uuuu uuuu uuu uuu uuu
r r r
LA = JI = EH = DG ، IK = GL = CE 15
(1, −1,1) ﺑﺎﻟﺘﺎﻟﻲ ﻓﺎﻟﻨﻘﻄﺔ ھﻲ ) 0 ,1,2 ( و اﻟﺸﻌﺎع وھﻮ
r r
uuu uuur uuu uuur uuuu uuuu
r r
x 2 + y 2 + z 2 = 9 83 LB + BF = LF ، AB + AD = AC 19
x 2 + y 2 + z 2 = 9 84 r r
uuuu uuuu r
uuuu
AC + BD = 2AD 25
ﻧﺘﺤﺼﻞ ﻣﺜﻼ ﻋﻠﻰ اﻟﻤﻌﺎدﻟﺔx − y = 1
z = 2 uuur uuuur uuu r
و ﻣﻨﮫ ﻓﺎﻷﺷﻌﺔ ﻣﻦ ﻧﻔﺲEG = 3AD + 2DI 29
. x 2 − x − 2 = 0 : x ذات اﻟﻤﺠﮭﻮل
uuu uuu uuu اﻟﻤﺴﺘﻮي
r r r
86 ﻟﯿﺴﺖ ﻣﻦ ﻧﻔﺲ اﻟﻤﺴﺘﻮيSD وSB ، SA .03 اﻷﺷﻌﺔ
ﺗﻘﺎﻃﻊ ﺳﻄﺢ اﻟﻜﺮة ﻣﻊ اﻟﻤﺴﺘﻮي ھﻲ اﻟﺪاﺋﺮة اﻟﺘﻲ uuu uuur uuu
r r
ﻟﯿﺴﺖ ﻣﻦ ﻧﻔﺲ اﻟﻤﺴﺘﻮيSD وAB ، SA اﻷﺷﻌﺔ
ﻣﺮﻛﺰھﺎ ) 0 ,0 ,3 ( و ﻧﺼﻒ ﻗﻄﺮھﺎ4 و ھﻲ ﻣﻌﺮﻓﺔ r r
uuur uuu uuu r r
uuu uuu uuur
CD وSB ، SA و ﻣﻨﮫ ﻓﺎﻷﺷﻌﺔSA = SB − CD ﻟﺪﯾﻨﺎ
y 2 + z 2 = 16 .ﻣﻦ ﻧﻔﺲ اﻟﻤﺴﺘﻮي
:ﺑﺎﻟﺠﻤﻠﺔ
x =3
uuur r
2 uuur 2 uuuu
88 AE = − AB + AC 32
uur uuur 1 uuuu uuu
r r r
uuur 1 uuuu r 3 uu 3r
r
3
FJ = − AB + AD ، IK = − AB + AD u = 5w − 3v (33 ب
4 4 uur uuu
3r r
uu r r
. ﻣﻦ ﻧﻔﺲ اﻟﻤﺴﺘﻮيFJ وIK ﯾﻨﺘﺞ أن . إذن اﻷﺷﻌﺔ ﻣﻦ ﻧﻔﺲ اﻟﻤﺴﺘﻮيw = u + v 36
89 r
uu r r
w = αu + β v : ﺑﺤﯿﺚβ وα 73 ھﻞ ﯾﻮﺟﺪ
r
uuu uuur uuur r
GA + GB + GC = 0
42
ﺛﻢ ﺑﺎﺳﺘﻌﻤﺎل ﻋﻼﻗﺔuuuur uuuur uuuuu r r
G ′A′ + G ′B ′ + G ′C ′ = 0
.ﺷﺎل ﻧﺘﻮﺻﻞ إﻟﻰ اﻟﻨﺘﯿﺠﺔ
r r
v = 2u 47
uuur r
uuuu
. . اﻟﻨﻘﻂ ﻓﻲ اﺳﺘﻘﺎﻣﯿﺔAB = 2AC 51
uuur uuur
. ( AB ) // (CD ) و ﻣﻨﮫk = 1 ﻣﻊAB = kCD 54
r
uuuu 3 uuur 2 uuur D ( 8, −4, 6 ) 63
. إذن اﻷﺷﻌﺔ ﻣﻦ ﻧﻔﺲAC = AE − AB 90
5 5 4 2 4
.اﻟﻤﺴﺘﻮي G , , 66
uu 1 uuur
r 3 3 3
r r
. و ﻣﻨﮫ اﻟﺸﻌﺎﻋﺎن ﻣﺘﻮازﯾﺎنIJ = EG 92 . و ﻣﻨﮫ اﻟﻨﻘﻂ ﻣﻦ ﻧﻔﺲ اﻟﻤﺴﺘﻮيu = 0 69
uu
r uuur 3 uuuu
r
( IJ ) // ( ABC ) و ﻣﻨﮫIJ = −AB + 2AC 93 73
3
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r
uuur 4 uuuu 1 uuur
49 AE = AC + CDو ﺑﺎﻟﺘﺎﻟﻲ ﻓﺎﻷﺷﻌﺔ ﻣﻦ ﻧﻔﺲ
3 3
اﻟﻤﺴﺘﻮي.
r r
uuur uuuu
59 u = 2AB − 3AC
r
69 uuuur uuuu
AM = BC
79 ﻻ ﺗﻮﺟﺪ ﻧﻘﻄﺔ Mﺗﺤﻘﻖ اﻟﺸﺮط ﻷن اﻟﺸﻌﺎع ﻣﺴﺘﻘﻞ
ﻋﻦ اﻟﻨﻘﻄﺔ . M
r r
uuu 1 uuuu
و ﻣﻨﮫ ) ( EF ) // ( BC 89 EF = BC
2
r r r
uuuu uuuu
(
ا( u = 2 ME + MF ) 99
ھﻲ ﻣﻨﺘﺼﻒ اﻟﻘﻄﻌﺔ ] [ EF ب( اﻟﻨﻘﻄﺔ I
اﻟﺘﻘﺎﻃﻊ ھﻲ اﻟﻨﻘﻄﺔ )3 ,1,2 ( 001
2 2 1 301
0 = x + z − y
2
أو ... 4
6 ≤ 0 ≤ y
401 9 = 2 y 2 + z
5 ≤ 2 ≤ x
ﻓﻲ ﺣﺎﻟﺔ ﺳﻄﺢ ﻏﯿﺮ ﻣﻨﺘﮫ ﺗﻜﺘﺐ اﻟﻤﻌﺎدﻟﺔ ﻋﻠﻰ
اﻟﺸﻜﻞ: 9 = 2 . y 2 + z
4
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