fazoda to‘g‘ri chiziq

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mavzu: fazoda to’g’ri chiziq reja: 1. fazoda to’g’ri chiziq 2. to’g’ri chiziq bilan tekislik 3. sfera fazoda to’g’ri chiziq tug’ri chiziqni ikki tekislikning kesishish chizig’i deb ta‘riflash mumkin; shuning uchun u ikkita birinchi darajali tenglamalar to’plami bilan ifodalanadi: 814. а ( + 1 ; - 5; + 3 ) nuqtadan o’tuvchi va koordinata o’qlari bilan mos ravishda 60, 45va 120li burchaklar tuzuvchi to’g’ri chiziqning tenglamasi tuzilsin. quyidagi to’g’ri chiziqlar bilan tuzilgan burchak topilsin: ва 816. uchlari а ( + 3; - 1; 0), - 7; 3), с ( - 2; + 1; - 1), d ( + 3; + 2; + 6) nuqtalarda yotgan tetraedrning qarama – qarshi qirralari orasidagi burchaklar topilsin. 818. quyidagi to’g’ri chiziqning yunaltiruvchi kosinuslari hisoblab topilsin: 819. ikki to’g’ri chiziq orasidagi burchak topilsin: ва 1. ( + 2; - 5; + 3) nuqtadan а) z o’qiga parallel to’g’ri chiziq o’tkazilsin; b) to’g’ri chiziqqa parallel to’g’ri chiziq …

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ncha ixcham bajariladi. bu uch tenglamani birgalikda yechish kerak. agar ( 14 ) dagi uchta nisbat o’rniga unga teng parametr ishlatilsa, yechish ancha ixcham bajariladi. bu holda х = m+ а, у = n+ b, z =+ с bo’ladi; koordinatalarning bu qiymatlarini tekislikning ( 15) tenglamasiga quyib ning qiymatini hosil qilamiz, so’ngra izlangan koordinatalarni topamiz. ( 14) to’g’ri chiziq bilan ( 15 ) tekislik orasidagi burchak ushbu formula bilan hisoblanadi: sin= ( 14 ) to’g’ri chiziq bilan ( 15 ) tekislikning parallellik sharti: аm +bn+cp=0. to’g’ri chiziq bilan tekislikning perpendikulyarlik sharti: ( 14 ) to’g’ri chiziqning ( 15 ) tekislikda yotish sharti quyidagi ikki tenglik bilan ifodalanadi: с) to’g’ri chiziqqa parallel to’g’ri chiziq o’tkazilsin. 828. to’g’ri chiziq bilan 3х + 5у – z -2 = 0 tekislikning kesishish nuqtasi topilsin. 829. kesishish nuqtasi topilsin: а) to’g’ri chiziq bilan 3х – 3у + 2z – 5 = 0 tekislikning; b) to’g’ri …

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rga parallel tekislikning tenglamasi tuzilsin. 848. p( + 7; + 9; + 7 ) nuqtadan to’g’ri chiziqqacha bo’lgan masofa topilsin. sfera yasovchi ( х, у, z ) nuqtaning sfera markazi deb atalgan o’zgarmas ( a, b, s) nuqtadan sfera radiusiga teng o’zgarmas r masofada yotishini analitik usulda ifodalab sferik sirtning tenglamasini hosil qilamiz: ( х – а)+ ( у – b)2 + ( z- с )2 =r2, yoki qavslarni ochib yozilsa: х2+ у2+z2-2ах – 2bу -2сz + а2 + b2 + с2 – r2=0 shaklni oladi. sferaning tenglamasida to’rtta erkin parametr bor; ular: markazning koordinatalari va radiusdir. bu tenglama ikkinchi darajali bo’lgani uchun sfera ikkinchi tartibli sirtdir. sferaning ( 2 ) tenglamasi shunday xususiyatga egaki, unda ikkinchi darajali koordinatalarning koeftsentlari bir – biriga teng, koordinatalarning kupaytmalardan iborat bo’lgan xadlar yuq..aksincha, agar ikkinchi darajali tenglama shu ikki shartni qanoatlantirsa u sferani ifodalaydi, (2) tenglamaning bir xil koordinatali hadlari gruppalarini tula kvadratlarga …

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islikning tenglamasi bunday ko’rinishni oladi: хх1 + уу 1 + zz 1= r 2 image5.wmf image6.wmf image7.wmf image8.wmf image9.wmf image10.wmf image11.wmf image12.wmf image13.wmf image14.wmf image15.wmf image16.wmf image17.wmf image18.wmf image19.wmf image20.wmf image21.wmf image22.wmf image23.wmf image24.wmf image25.wmf image26.wmf image27.wmf image28.wmf image29.wmf image30.wmf image31.wmf image32.wmf image33.wmf image34.wmf image35.wmf image36.wmf image37.wmf image38.wmf image39.wmf image40.wmf image41.wmf image42.wmf image43.wmf image44.wmf image45.wmf image46.wmf image47.wmf image48.wmf image49.wmf image50.wmf image51.wmf image52.wmf image53.wmf image54.wmf image1.wmf image55.wmf image56.wmf image57.wmf image58.wmf image59.wmf image60.wmf image61.wmf image2.wmf image3.wmf image4.wmf 2 5 6 2 3 1 - = + = - z у х . 6 1 9 3 2 + = - = z у х î í ì = + - = + + - z х z у х . 0 3 , 0 21 2 6 5 î í ì = - + = - z у х z у х 0 2 2 , 0 2 4 3 î í …

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z у х = - = + } , 0 = + + + d сz b у ах 1 2 3 5 4 z у х = + = - 2 1 3 4 4 - + = - = z у х 3 2 6 - = = z у х 2 4 4 3 5 1 - = - = + z у х 2 3 1 4 2 z у х = - = - 2 ( ) ( ) ( ) ( ) 2 1 1 1 r c z b у а х а х = - + - + - - 0 0 0

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mavzu: fazoda to’g’ri chiziq reja: 1. fazoda to’g’ri chiziq 2. to’g’ri chiziq bilan tekislik 3. sfera fazoda to’g’ri chiziq tug’ri chiziqni ikki tekislikning kesishish chizig’i deb ta‘riflash mumkin; shuning uchun u ikkita birinchi darajali tenglamalar to’plami bilan ifodalanadi: 814. а ( + 1 ; - 5; + 3 ) nuqtadan o’tuvchi va koordinata o’qlari bilan mos ravishda 60, 45va 120li burchaklar tuzuvchi to’g’ri chiziqning tenglamasi tuzilsin. quyidagi to’g’ri chiziqlar bilan tuzilgan burchak topilsin: ва 816. uchlari а ( + 3; - 1; 0), - 7; 3), с ( - 2; + 1; - 1), d ( + 3; + 2; + 6) nuqtalarda yotgan tetraedrning qarama – qarshi qirralari orasidagi burchaklar topilsin. 818. quyidagi to’g’ri chiziqning yunaltiruvchi kosinuslari hisoblab topilsin: …

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