Berilgan M0(x0,y0,z0) nuqtadan o‘tuvchi tekisliklar dastasining tenglamasi qayerda to’g’ri ifodalangan?
A) A( x+x0)+B( y+y0)+C( z+z0)=0 . B) A xx0+B yy0+C zz0=0.
C) x - x0
A
B
C
= 0 . D)
x + x0
A
B
C
= 0 .
*E) A( x–x0)+B( y–y0)+C( z–z0)=0 .
Fazoning M(1,2,-3) nuqtasidan o‘tuvchi tekisliklar dastasi tenglamasini ko‘rsating.
A) A x+2B y-3C z=0. *B) A( x-1)+B( y-2)+C( z+3)=0. C) A( x+1)+B( y+2)+C( z−3)=0. D) Ax+2 By-3 Cz+D=0.
E) A( x-1)+B( y-2)+C( z-3)+D=0.
M1(3,2,-1), M2(0,3,1) va M3(4,5,0) nuqtalardan o‘tuvchi tekislik tenglamasini ko‘rsating.
A) 2x-y+z−3=0.
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B)
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x-2y+z+2=0.
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*C) x-y+2z+1=0.
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D) x-y+z=0.
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E)
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x-2y+2z+3=0.
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M1(3,2,-1), M2(0,3,1) va M3(4,5,0) nuqtalardan o‘tuvchi tekislikda yotuvchi M0(x,−4, 7) nuqtaning abssissasini toping.
A) x0=−3 . B) x0=5 . C) x0=−1,5 . *D) x0=9 . E) x0=0 .
Berilgan M0(x0,y0,z0) nuqtadan o‘tuvchi va n=(A,B,C) vektorga perpendikulyar tekislik tenglamasini ko‘rsating.
A) A(x+x0)+B(y+y0)+C(z+z0)=0 . B) Axx0+Byy0+Czz0=0.
C) x - x0
A
B
C
= 0 . D)
x + x0
A
B
C
= 0 .
*E) A( x–x0)+B( y–y0)+C( z–z0)=0 .
M(1,2,3) nuqtadan o‘tuvchi va n=(3,2,1) normal vektorga ega tekislik tenglamasini yozing.
*A) 3 x+2 y+ z-10=0 . B) x+2 y+3 z-14=0 . C) 2 x+3 y+ z-11=0 .
D) x+3 y+2 z-13=0 . E) 3 x+ y+2 z-11=0.
Berilgan M0(3,−4,0) nuqtadan o‘tuvchi va n=(1,2,−3) normal vektorga ega bo’lgan tekislikda yotuvchi N(−3,8, z) nuqtaning aplikatasini toping.
A) z=0 . B) z=5 . C) z= −1 . *D) z=6 . E) z= −3,5 .
M0(x0,y0,z0) nuqtadan Ax+By+Cz+D=0 tekislikkacha bo‘lgan masofani topish formulasini ko’rsating.
A) d =
Ax0
+ D . B) d = .
E) d =
Ax0 + By0 + Cz0 + D .
A2 + B2 + C 2
9. 3x+4y-2 toping.
z+14=0 tekislikdan koordinata boshigacha bo‘lgan masofani
A) 14. B) 7. *C) 2. D) 1. E) 2 6 .
Ushbu 4x+3y-5z-8=0 va 4x+3y-5z-12=0 parallel tekisliklar orasidagi masofani toping.
A) 4 . B) 20 . C)
. *D) 2 2
5
. E) 2 2 .
x+y-z-1=0 va 2x-2y-2z+1=0 tekisliklar orasidagi burchak kosinusini toping.
A) 0. B) 1. C)
3 / 4 . *D) 1/3. E) 3/4.
x+y-18=0 va y+z-72=0 tenglamalar bilan berilgan tekisliklar orasidagi burchak topilsin.
A) 300 . B)
arccos
3 . C) 450 . D)
4
arccos
3 . *E) 600 .
5
Normal vektorlari n1=(-1, -1,0) va n2=(0, -1, -1) bo‘lgan tekisliklar orasidagi burchak topilsin.
A) 450 . B) 300 . C) arccos 2
3
. *D) 600 . E) 900.
A1x+B1y+C1z+D1=0 va A2x+B2y+C2z+D2=0 tekisliklarning parallellik sharti qayerda to‘gr’ri ko‘rsatilgan ?
А1 А2
= В1 В2
= D1
D2
. B)
А1 = C1 А2 C2
= D1 D2
. *C)
А1 = В1 А2 В2
= C1 .
C2
В1 В2
= C1
C2
= D1
D2
. E)
A1 = В1 A2 В2
= C1
C2
= D1 .
D2
kx–2y+5z+10=0 va 6x– (1+k)y+10z–2=0 tekisliklar k parametrning qanday qiymatida parallel bo‘ladi ?
*A) k=3. B) k= – 4. C) k=2. D) k= –5. E) k=±1.
Umumiy tenglamalari A1x+B1y+C1z+D1=0 va A2x+B2y+C2z+D2=0 bilan berilgan tekisliklarning perpendikulyarlik shartini ko‘rsating.
A) A1A2+ B1B2+ C1C2+ D1D2=0. B) A1A2+ B1B2+ D1D2=0.
*C) A1A2+ B1B2+ C1C2=0. D) B1B2+ C1C2+ D1D2=0.
kx–2y−5z+10=0 va 6x–(1+k)y+10z–2=0 tekisliklar k parametrning qanday qiymatida parallel bo‘ladi ?
A) k=3. B) k= – 4. *C) k=6. D) k= –5. E) k=±1.
x-y-1=0, y+z=0 va x-z-1=0 tekisliklarning kesishish nuqtasi koordinatalarining yig’indisini toping.
A) 0 . *B) 1 . C) −1 . D) 4 . E) −4 .
2x-3y+4z-12=0 tenglama bilan berilgan tekislikning koordinata o‘qlari bilan kesishgan nuqtalarining koordinatalari topilsin.
A) ( -1,0,0), (0, -5,0), (0,0,4). B) (−6,0,0), (0, 3,0), (0,0,4).
C) (5,0,0), (0, -4,0), (0,0,−3). *D) (6,0,0), (0, -4,0), (0,0,3).
E) (1,0,0), (0, 3,0), (0,0,5).
M(2, -1,1) nuqtadan o‘tib, 3x+2y-z+4=0 tekislikka parallel bo‘lgan tekislikning koordinata o‘qlari bilan kesishish nuqtalarining koordinatalarini toping.
A) ( -1,0,0), (0, -5,0), (0,0,4). B) (−6,0,0), (0, 3,0), (0,0,4).
C) (5,0,0), (0, -4,0), (0,0,−3). D) (6,0,0), (0, -4,0), (0,0,3).
*E) (1,0,0), (0, 3/2,0), (0,0,−3).
bo‘lgan P tekislik va koordinata tekisliklari bilan chegaralangan piramida hajmini toping.
A) 90 B) 60. C) 45. D) 30. *E) 15.
Kesmalardagi tenglamasi
x + y +
a 4
z = 1
- 9
bo‘lgan P tekislik va koordinata tekisliklari bilan chegaralangan piramida hajmi 72 kub birlikka teng . Noma’lum a parametr qiymatini toping.
A) 2. B) −2. C) ±2. *D) ±12. E) 6.
M(3, -2,1) nuqtadan o‘tib, a1=(0,1,−2) va a2=(5,0,2) vektorlarga parallel bo’lgan tekislik tenglamasini toping.
A) 3x+2y-z−4=0.
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B) 3x−5y+2z−21=0.
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*C) 2x−10y-5z−21=0.
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D) 2x+5y-z+5=0.
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E) 5x−12y-3z−36=0.
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ADABIYOTLAR.
SOATOV YO.U. «Oliy matеmatika», I jild, Toshkеnt, O¢qituvchi, 1992 y.
PISKUNOV N.S. «Diffеrеntsial va intеgral hisob», 1-tom, Toshkеnt, O¢qituvchi, 1972 y.
MADRAXIMOV X.S., G’ANIЕV A.G., MO’MINOV N.S. «Analitik gеomеtriya va chiziqli algеbra», Toshkеnt, O¢qituvchi, 1988 y.
SARIMSOQOV T.А. «Haqiqiy o¢zgaruvchining funktsiyalari nazariyasi» Toshkеnt, O¢qituvchi, 1968 y.
T. YOQUBOV «Matеmatik logika elеmеntlari», Toshkеnt, O¢qituvchi, 1983y.
RAJABOV F., NURMЕTOV A. «Analitik gеomеtriya va chiziqli algеbra», Toshkеnt, O¢qituvchi, 1990 y.
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