Oliy matematika kafedrasi
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fazoda tekislik tenglamalari (1)
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- “Fazoda tekislik tenglamalari” mavzusi bo‘yicha test topshriqlari
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II darajali testlar 9. Tekislikning kesmalar bo’yicha tenglamasini toping. A) 1
+ +
z b y a x
В ) 0 ) ( ) ( ) ( 0 0 0 = − + − + − z z C y y B x x A
D) 0 = + + +
Cz By Ax
E) 0 ) ( ) ( ) ( 0 0 0 = + + + + + z z C y y B x x A
10. Ikki tekislikning parallellik shartini toping. A)
2 1 2 1 2 1 C C B B A A = =
В ) 0 2 1 2 1 2 1 = + +
C B B A A
D) 0 2 1 2 1 2 1 = + + C C B B A A
E) 2 1 2 1
B A A =
11. Ikki tekislikning perpendikulyarlik shartini toping. A)
0 2 1 2 1 2 1 = + + C C B B A A
В ) 2 1 2 1 2 1
C B B A A = = D)
0 2 1 2 1 2 1 = + + C C B B A A
E) 0 2 1 2 1 2 1 = − −
C B B A A
12. ) 0 , 5 , 2 ( A va
) 12 , 1 , 5 ( B nuqtalar orasidagi masofani toping. A) 13
) 169
D) 13
E) 189
13.
) 4 , 7 , 3 ( A va
) 3 , 2 , 8 ( B nuqtalarni tutashtiruvchi AB kesmani 3 :
= λ
nisbatda bo‘luvchi ) , , (
y x C nuqtani toping. 20
A) ) 6 , 3 ; 5 ; 5 ( C
B) ) 6 , 3 ; 5 ; 25 ( C D)
) 6 , 3 ; 25 ; 5 ( C E)
) 18 ; 5 ; 5 ( C
14. ) 2 ; 3 ; 2 ( −
nuqtadan o‘tib, ) 3 , 4 , 5 (
vektorga perpendikulyar bo‘lgan tekislik tenglamasini toping. A) 0
3 4 5 = − + + z y x
B) 0 10 3 4 5 = − + + z y x
D) 0 12 3 4 5 = + + + z y x
E) 0 16 3 4 5 = − + + z y x
15. 0 6 3 2 = − + −
y x tekislikning kesmalar bo‘yicha tenglamasini toping. A) 1
2 3 = + − + z y x
B) 1 6 2 3 = + + − z y x
D) 1 6 2 3 = + + z y x
E)
0 6 2 3 = + − +
y x
III darajali testlar 16. OX o‘qiga parallel va ) 2
0 ; 4 ( −
, )
, 1 , 5 (
nuqtalardan o‘tuvchi tekislik tenglamasini toping. A) 0
9 = − − z y
B) 0 2 7 = − − z y
D) 0 2 = − −
y
E)
0 2 2 9 = − − z y
17. OZ o‘qdan va ) 4 , 3 , 2 ( Ì nuqtadan o‘tuvchi tekislik tenglamasini toping. A) 0
3 = − y x
B) 0 2 3 = +
x
D) 0 = − By Ax
E) 0 = + By Ax
18. YOZ koordinat tekisligiga parallel va ) 4 ; 5 ; 2 ( −
nuqtadan o‘tuvchi tekislik tenglamasini toping.
A)
0 5 2 = −
B)
0 4 = − x
D) 0 2 = + x
E) 0 2 = − x
19. 0 4 2 2 = + − −
y x va
0 8 2 2 = − − −
y x parallel tekisliklar orasidagi masofani toping. A) 4
В ) -4
D) 3 8 E) 3 12 20.
) 6 , 5 , 3 ( A va
) 4 , 7 , 5 ( −
nuqtalar berilgan. A nuqtadan o‘tib, ÀÂ vektorga perpendikulyar tekislik tenglamasini toping. A) 0
6 = + − −
y x
B) 0 66 6 = + − −
y x
D) 0 78 6 = + − − z y x
E) 0 33 6 = + + +
y x “Fazoda tekislik tenglamalari” mavzusi bo‘yicha test topshriqlari 21
I darajali testlar 1. Fazodagi berilgan ) ,
( 1 1 1 z y x A va
) , , ( 2 2 2 z y x B nuqtalar orasidagi d masofa qanday formula yordamida topiladi? A)
2 1 2 2 1 2 2 1 2 ) ( ) ( ) ( z z y y x x d − + − + − =
B) 2 1 2 2 1 2 2 1 2 ) ( ) ( ) ( z z y y x x d − − − + − =
D) 2 1 2 2 1 2 2 1 2 ) ( ) ( ) ( z z y y x x d + + + + + =
E) 2 1 2 2 1 2 2 1 2 ) ( ) ( ) ( z z y y x x d − + − − − =
2. Fazodagi nuqtalar berilgan ) , , ( 1 1 1
y x A va
) , , ( 2 2 2 z y x B berilgan. AB kesmani
= λ nisbatda bo’luvchi ) , , (
y x C nuqtaning koordinatlari qanday formulalar yordamida topiladi? A)
λ λ λ λ λ λ + + = + + = + + = 1 , 1 , 1 2 1 2 1 2 1
z z y y y x x x
B) λ λ λ λ λ λ + − = + − = + − = 1 , 1 , 1 2 1 2 1 2 1
z z y y y x x x
D) λ λ λ λ λ λ − + = − + = − + = 1 , 1 , 1 2 1 2 1 2 1
z z y y y x x x
E) λ λ λ λ λ λ 2 1 2 1 2 1 , ,
z z y y y x x x + = + = + =
3. )
, ( 0 0 0 0 z y x M nuqtadan o’tib, → →
→ + + = k C j B i A N vektorga perpendikulyar tekislikning tenglamasini toping. A)
0 ) ( ) ( ) ( 0 0 0 = − + − + − z z C y y B x x A
В ) 0 ) ( ) ( ) ( 0 0 0 = + + + + +
z C y y B x x A
D) 0 ) ( ) ( ) ( 0 0 0 = − − − − − z z C y y B x x A
E) 1 = + +
z b y a x
4. Tekislikning umumiy tenglamasini toping A) 0 = + + + D Cz By Ax
В ) 0 ) ( ) ( ) ( 0 0 0 = − + − + −
z C y y B x x A
D) 1 = + + c z b y a x
E) 0 ) ( ) ( ) ( 0 0 0 = + + + + + z z C y y B x x A
22
5. Tekislikning 0 = + + + D Cz By Ax umumiy tenglamasida 0 =
bo’lsa, uning fazodagi holati qanday bo’ladi?
A)
0 =
bo’lsa, 0 = + +
By Ax bo’lib, tekislik koordinatlar boshidan o’tadi В )
= D bo’lsa, 0 =
+ Cz By Ax bo’lib, tekislik koordinatlar boshidan o’tmaydi D) tekislik OY o’qiga parallel bo’ladi E) tekislik OX o’qiga parallel bo’ladi
6. Tekislikning 0 = + + +
Cz By Ax umumiy tenglamasida 0 =
bo’lsa, uning fazodagi holati qanday bo’ladi? A) 0
C bo’lsa, 0 =
+ D By Ax bo’lib, tekislik OZ o’qiga parallel bo’ladi В )
= C bo’lsa, 0 =
+ D By Ax bo’lib, tekislik OÓ o’qiga parallel bo’ladi D) 0
C bo’lsa, 0 =
+ D By Ax bo’lib, tekislik OÕ o’qiga parallel bo’ladi E) 0
C bo’lsa, 0 =
+ D By Ax bo’lib, tekislik OZ o’qiga perpendikulyar bo’ladi
7. Tekislikning 0 = + + +
Cz By Ax umumiy tenglamasida 0 =
C Â bo’lsa, uning fazodagi holati qanday bo’ladi? A)
0 = = Ñ B , bo’lsa, 0 =
D Ax bo’lib, tekislik YOZ koordinat tekisligiga parallel bo’ladi В ) 0 = = Ñ B , bo’lsa, 0 =
D Ax bo’lib, tekislik YOÕ koordinat tekisligiga parallel bo’ladi D)
0 = = Ñ B , bo’lsa, 0 =
D Ax bo’lib, tekislik ÕOZ koordinat tekisligiga parallel bo’ladi E)
0 = = Ñ B , bo’lsa, 0 =
D Ax bo’lib, tekislik ÓOZ koordinat tekisligiga perpendikulyar bo’ladi
8. Tekislikning 0 = + + +
Cz By Ax umumiy tenglamasida 0 =
= D C В
bo’lsa, uning fazodagi holati qanday bo’ladi? A)
0 = = = D C B bo’lsa, 0 =
bo’lib, YOZ koordinat tekisligi bilan ustma- ust tushadi, ya’ni 0 = x , YOZ koordinat tekisligining tenglamasi bo’ladi 23
В ) 0 = = = D C B bo’lsa, 0 =
bo’lib, YOZ koordinat tekisligi bilan ustma- ust tushadi, ya’ni А
= , YOZ koordinat tekisligining tenglamasi bo’ladi D) 0
x bo’lib, ХOZ koordinat tekisligining tenglamasi bo’ladi E) 0
x bo’lib, Х
koordinat tekisligining tenglamasi bo’ladi Download 195.84 Kb. Do'stlaringiz bilan baham: 
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