O ‘zbekiston respublikasi oliy va o’rta maxsus ta’lim vazirligi

Tеkislik tеnglamasiga doir masala.

Bеrilgan nuqtadan otuvchi tеkisliklar dastasi tеnglamasi masalasi.
Aytaylik, tеkislik bеrilgan М₁(х₁; у₁; z₁) nuqtadan o’tsin va uning tеnglamasini topish talab etilsin. Izlanayotgan tеkislikning umumiy tеnglamasini qaraymiz:
Ах + Ву + Сz + D =0
М₁ nuqta tеkislikda yotgani uchun uning koordinatalari bu tеnglamani qanoatlantirishi kеrak:
Ах₁+ Ву₁+ Сz₁+ D =0
Hosil bolgan bu tеnglikni yuqoridagi tеnglamadan ayirib, izlangan
А(х-х₁) + В(у-у₁) + С(z-z₁) = 0,
ya'ni bеrilgan M₁ nuqtadan otuvchi tеkisliklar tеnglamasini hosil qilamiz. Undagi koeffitsiеntlarga turli qiymatlar bеrib, M₁ nuqtadan otuvchi tеkisliklar dastasini olamiz.

Tekislik va uning tenglamalari

1. 
   Tekislikning umumiy tenglamasi qayerda to’liq va to’g’ri ifodalangan?
   

*A) Ax+By+Cz+D=0. B) Ax+By+CDz=0. C) Ax+By+(C+D)z=0.
D) Ax⁻¹+By⁻¹+Cz⁻¹+D=0. E) Ax²+By²+Cz²+D=0.

2. 
   Umumiy tenglamasi 2x−5y+4z+9=0 bo’lgan tekislikka tegishli va koordinatalarining yig’indisi 15 bo’lgan M(x,y,5) nuqtaning abssisasini toping.
   

A) 4. B) −7. *C) 3. D) 0. E) −1.

3. 
   Umumiy tenglamasi 2x−5y+4z−9=0 bo’lgan tekislikka tegishli, OZ o’qda yotuvchi va koordinatalarining yig’indisi birga teng bo’lgan nuqtaning ordinatasini toping.
   

A) 4. B) −7. C) 3. D) 0. *E) −1.

4.	Quyidagilardan	qaysi biri	Ax+By+Cz+D=0 tenglamali tekislikning n
	normal vektori	bo’ladi ?

5. 
   

   A) n=(B,C,D).	B) n=(A,C,D).	*C) n=(A,B,C).
   D) n =(A,B,D).	E) n=(C,A,B).

   
   
   
   Tasdiqni yakunlang: Umumiy tenglamasi Ax+By+Cz+D=0 bo‘lgan tekislikning n=(A,B,C) normal vektori shu ∙∙∙ .
   
   1. 
      tekislikda yotadi . B) tekislikka parallel bo‘ladi .
      

*C) tekislikka perpendikulyar bo‘ladi . D) tekisliikka og‘ma bo‘ladi.
E) to’g’ri javob keltirilmagan .

5. 
   3x+4y+7z–81=0 tenglama bilan berilgan tekislik normalini aniqlang. A) n =(3,4,–81). B) n =(3,–4,–81). C) n =(4,7,–81).
   

D) n =(–3,–4,81). *E) n =(3,4,7).

5. 
   Quyidagi tenglamalardan qaysi biri koordinatalar boshidan o‘tuvchi tekislikni ifodalaydi ?
   
   1. 
      Ax+By+D=0. *B) Ax+By + Cz=0. C) By+Cz+D=0.
      

D) Ax+By+Cz+D=0. E) Ax+Cz+D=0.

5. 
   x+y−z=0 tenglamali P tekislik to’g’risidagi quyidagi tasdiqlardan qaysi biri o’rinli ?
   

*A) P koordinatalar boshidan o‘tadi. B) P OXY tekisligiga parallel.
C) P OXZ tekisligiga parallel . D) P OYZ tekisligiga parallel.
E) P tekislik OZ koordinata o’qiga perpendikulyar.

5. 
   Ax+By+Cz+D=0 tenglama A=D=0 holda qanday P tekislikni ifodalaydi ?
   
   1. 
      P OX o‘qiga parallel. B) P OX o‘qiga perpendikulyar.
      

*C) P OX o‘qi orqali o‘tadi. D) P OY o‘qiga perpendikulyar.
E) P OY o‘qiga parallel .

5. 
   Quyidagi tenglamalardan qaysi biri OZ koordinata o‘qidan o‘tuvchi tekislikni ifodalaydi ?
   
   1. 
      Ax+By+Cz=0. B) Ax+Cz+D=0. *C) Ax+By=0.
      

D) By+Cz+D=0. E) By +D=0.

5. 
   Quyidagi tenglamalardan qaysi biri OY koordinata o‘qiga parallel tekislikni ifodalaydi ?
   
   1. 
      Ax+By+Cz=0. *B) Ax+Cz+D=0. C) Ax+By+D=0.
      

D) By+Cz+D=0. E) Ax +D=0.

5. 
   Quyidagi tenglamalardan qaysi biri XOZ koordinata tekisligiga parallel tekislikni ifodalaydi ?
   
   1. 
      Ax+By+Cz=0. B) Ax+Cz+D=0. C) Ax+By=0.
      

D) By+Cz+D=0. *E) By +D=0.

5. 
   Quyidagi tenglamalardan qaysi biri YOZ koordinata tekisligini ifodalaydi ? A) By=0. B) Cz=0. *C) Ax=0. D) By+Cz=0. E) Ax +D=0.
   

5. 
   Tekislikning kesmalardagi tenglamasi qayerda to‘gr’ri yozilgan ?
   

1. 
   ^a ^b ^c 1. B)
   

ax bу cz 1. C)
^х ^у ^z 0 .

x у z
*D)
х _у _
^z 1. E)
а b c
^х ^у ^z 1.

а b c
а b c

5. 
   Tasdiqni yakunlang: Koordinata boshidan o’tmaydigan, koordinata o’qlariga parallel bo’lmagan va Ax+By+Cz+D=0 umumiy tenglama bilan berilgani tekislikning kesmalardagi tenglamasiga o‘tish uchun umumiy tenglama ∙∙∙ soniga bo‘linadi .
   

A) –C . B) –A . C) –B . *D) –D . E) −(A²+B²+C²) .

3. 
   x4y+12z24=0 tekislikning kesmalardagi tenglamasini toping.
   

1. 
   ^{x y z}
   

^1 . B)
x _ y
^{z 1}. C)
^{x y z 1}.

8 6 12
4 6 1
3 2 4

D) ^{x y z 1}. *E)
x _y
^{z 1}.

8 3 4
8 6 2

17. 
    Tekislikning normal tenglamasi qayerda to‘g’ri yozilgan ?
    

A) xcos⁻¹+ycos⁻¹+zcos⁻¹p=0. B) x⁻¹cos+y⁻¹cos+z⁻¹cosp=0 .
*C) xcos+ycos+zcosp=0 . D) x²cos+y²cos+z²cosp=0 .
E) xcosycoszcos+p=0 .

17. 
    Tasdiqni to‘ldiring: Tekislikning umumiy Ax+By+Cz+D=0 tenglamasidan uning normal tenglamasiga o‘tish uchun bu tenglama ∙∙∙ ifodaga bo‘linadi .
    

*A) 
A² B ²
C ² . B)
 ^{. C)}  ^.

^D) 
A² B² C² D^{2 . E)  .}

17. 
    Umumiy tenglamasi 2x+2y+z–18=0 bo‘lgan tekislikning normal tenglamasini toping .
    

*A)
x  y  z 
0 ^{. B)}
x  y 
¹ z 
3
0 ^.

C) ² x ² y ¹ z 6 0. D)
² x ² y ¹ z ¹⁸ 0 .

3 3 3
5 5 5 5

E) x∙cos60⁰+ y∙cos30⁰+ z∙cos45⁰–9=0 .

17. 
    Quyidagilardan qaysi biri tekislikning normal tenglamasi bo’ladi?
    

1. 
   ² 13
   

x ⁵
13
y ⁷
13
z 5 0 . B)
4 _{x }3
13 13
y ⁶
13
z 13 0 .

C) ⁵
13
x ¹
13
y ⁸
13
z 9 0 . *D)
3 _{x }4
13 13
y ¹² z 2 0 .
13

E) x∙cos60⁰+ y∙cos30⁰+ z∙cos45⁰–9=0 .

Tekislikka doir asosiy masalalar

1. 
   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) Axx₀+Byy₀+Czz₀=0.

_C) x x₀
A

• 
  y y₀
  

B

• 
  z z₀
  

C
0 . D)
x x₀
A

• 
  y y₀
  

B

• 
  z z₀
  

C
0 .

*E) A(x–x₀)+B(y–y₀)+C(z–z₀)=0 .

2. 
   Fazoning M(1,2,3) nuqtasidan o‘tuvchi tekisliklar dastasi tenglamasini ko‘rsating.
   

A) Ax+2By3Cz=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+2By3Cz+D=0.
E) A(x1)+B(y2)+C(z3)+D=0.

3. 
   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.	B)	x2y+z+2=0.	*C) xy+2z+1=0.
D) xy+z=0.	E)	x2y+2z+3=0.

4. 
   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 .

5. 
   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

• 
  y y₀
  

B

• 
  z z₀
  

C
0 . D)
x x₀
A

• 
  y y₀
  

B

• 
  z z₀
  

C
0 .

*E) A(x–x₀)+B(y–y₀)+C(z–z₀)=0 .

6. 
   M(1,2,3) nuqtadan o‘tuvchi va n=(3,2,1) normal vektorga ega tekislik tenglamasini yozing.
   

*A) 3x+2y+z10=0 . B) x+2y+3z14=0 . C) 2x+3y+z11=0 .
D) x+3y+2z13=0 . E) 3x+y+2z11=0.

7. 
   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 .

8. 
   M₀(x₀,y₀,z₀) nuqtadan Ax+By+Cz+D=0 tekislikkacha bo‘lgan masofani topish formulasini ko’rsating.
   

A) d 
Ax₀

• 
  By₀
  
• 
  Cz₀
  

D . B) ^{d } .

*C) d 
Ax₀

• 
  By₀
  
• 
  Cz₀
  
• 
  D
  

. D)


d  .

A² B ²

• 
  C ²
  

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 ^.

10. 
    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} .

10. 
    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.}

10. 
    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

10. 
    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.

10. 
    A₁x+B₁y+C₁z+D₁=0 va A₂x+B₂y+C₂z+D₂=0 tekisliklarning parallellik sharti qayerda to‘gr’ri ko‘rsatilgan ?
    

1. 
   ^А1 А₂
   

_В₁ В_2
D₁
D₂
. B)
А_{1 }C₁ А₂ C_2
D₁ D₂
. *C)
А_{1 }В₁ А₂ В₂
^C1 .
C₂

4. 
   ^В1 В₂
   

_C₁
C_2
D₁
D₂
. E)
A_{1 }В₁ A₂ В_2
C₁
C₂
^D1 .
D₂

10. 
    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.

10. 
    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.

4. 
   ^A1 A₂
   

_В₁ В₂
^C1 .
C₂

10. 
    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.

10. 
    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 .

10. 
    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).

10. 
    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).

10. 
    Kesmalardagi tenglamasi
    

x _ y
5 2


^z 1 9

bo‘lgan P tekislik va koordinata tekisliklari bilan chegaralangan piramida hajmini toping.
A) 90 B) 60. C) 45. D) 30. *E) 15.

10. 
    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.

10. 
    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.	B) 3x−5y+2z−21=0.	*C) 2x−10y5z−21=0.
D) 2x+5yz+5=0.	E) 5x−12y3z−36=0.

ADABIYOTLAR.

1. 
   SOATOV YO.U. «Oliy matеmatika», I jild, Toshkеnt, Oqituvchi, 1992 y.
   
2. 
   PISKUNOV N.S. «Diffеrеntsial va intеgral hisob», 1-tom, Toshkеnt, Oqituvchi, 1972 y.
   
3. 
   MADRAXIMOV X.S., G’ANIЕV A.G., MO’MINOV N.S. «Analitik gеomеtriya va chiziqli algеbra», Toshkеnt, Oqituvchi, 1988 y.
   
4. 
   SARIMSOQOV T.А. «Haqiqiy ozgaruvchining funktsiyalari nazariyasi» Toshkеnt, Oqituvchi, 1968 y.
   
5. 
   T. YOQUBOV «Matеmatik logika elеmеntlari», Toshkеnt, Oqituvchi, 1983y.
   
6. 
   RAJABOV F., NURMЕTOV A. «Analitik gеomеtriya va chiziqli algеbra», Toshkеnt, Oqituvchi, 1990 y.