Ukulingana kwendiza: indlela yokubhala? Iindidi zokulingana kweendiza

Isithuba moya inokuchazwa ngeendlela ezahlukeneyo (ichaphaza enye kumda wesangqa, kumda wesangqa kunye iingongoma ezimbini, amanqaku amathathu, njalo njalo). Ke oku engqondweni, i equation moya kunokuba iintlobo ezahlukeneyo. Kwakhona phantsi kweemeko ezithile moya inokuba parallel, nkqo, ezidibanayo, njl Kule kwaye ndiya kuthetha ngayo kweli nqaku. Siza kufunda ukwenza inxaki jikelele moya, hayi nje kuphela.

Ifom eqhelekileyo lenxaki

Masithi R _isithuba-3, nto leyo iba ebuxande nokulungelelanisa inkqubo xyz. Sichaza α kumda wesangqa, nto leyo iya kukhululwa kwindawo yokuqala O. Ngokusebenzisa ekupheleni α wesangqa kukha P-moya apho nkqo kuyo.

Libhekisela P kwi engenasizathu kwindawo Q = (x, y, z). Embindini wesangqa ye kwindawo Q unobumba uphawu p. Ubude kumda wesangqa ilingana p α = IαI kunye Ʋ = (cosα, cosβ, cosγ).

Le yunithi kumda wesangqa, ethe ngqo kwi kwicala njengoko esiyua α. α, β kunye γ - zizinto engile enziwa phakathi kumda wesangqa kunye imiyalelo ezintle Ʋ isithuba amazembe x, y, z ngokulandelelanayo. I ezayo kwindawo kumda wesangqa QεP Ʋ oqhubekayo leyo ilingana p (p, Ʋ) = p (r≥0).

Le equation ngasentla lebyi xa p = 0. Kuphela n moya kulo mzekelo, bawele indawo O (α = 0), nto leyo imvelaphi, kunye neyunithi yomzobo Ʋ, kukhululwa kwindawo O iya kuba aa Incopho P, nangona kwawo, nto leyo ethetha ukuba Ʋ kumda wesangqa sizimisele phezulu ukuya umqondiso. equation Previous lo moya P wethu, ochazwa ifomu kumda wesangqa. Kodwa ngenxa zilungelelanise yalo:

P mkhulu okanye ilingana no 0. Simfumene equation moya ngohlobo oluqhelekileyo.

Le equation jikelele

Ukuba lenxaki kwi uququzelelo phinda ngo naliphi na inani akalingani zero, siye sifumane inxaki elilingana le nto ichaza kanye moya. Iya kuba le fomu ilandelayo:

Apha, A, B, C - linani ngaxeshanye ezahlukeneyo ukusuka kwiqanda. Le equation kuthiwa inxaki kwefom jikelele moya.

Ezi zibalo ze iinqwelomoya. iimeko ezizodwa

Inxaki sikwazi ilungiswe kunye neemeko ezongezelelweyo. Cinga ezinye zazo.

Ukucinga ukuba inani A yi 0. Oku kubonisa ukuba le nqwelo-moya eyayingqalene esqwini inkabi kwangaphambili. Kulo mzekelo, uhlobo lenxaki utshintsha: Wu + Cz + D = 0.

Ngokufanayo, uhlobo quadratic kwaye ziya kwahluka nale miqathango ilandelayo:

• Okokuqala, ukuba B = 0, iinguqu alenxaki Ax + Cz + D = 0, nto leyo ingabonisa ukuba okukhona ukuya esqwini Oy.
• Okwesibini, ukuba C = 0, inxaki uguqulwe Ax + Ngu + D = 0, oko kukuthi malunga eyayingqalene esqwini kwangaphambili Oz.
• Okwesithathu, ukuba D = 0, kuya kuvela inxaki njengoko Axe + Ngu + Cz = 0, nto leyo iza kuthetha ukuba lo moya yekhampasi O (imvelaphi).
• Okwesine, ukuba A = B = 0, iinguqu alenxaki Cz + D = 0, nto leyo eya kuba okukhona Oxy.
• Okwesihlanu, ukuba B = C = 0, inxaki uba Axe + D = 0, nto leyo ethetha ukuba lo moya onxusene Oyz.
• Sixthly, ukuba A = C = 0, inxaki kuthatha uhlobo Wu + D = 0, ngamanye amazwi, kuza ingxelo Oxz okukhona.

Ifomu lenxaki ngezigaba

Kwimeko apho amanani A, B, C, ezahlukeneyo D ukusuka kwiqanda, indlela equation (0) ukuze kwenziwe ngolu hlobo lulandelayo:

x / a + y / b + z / c = 1,

apho a = -D / A, b = -D / B, c = -D / C.

Sifumana ngenxa equation lo moya ziingceba. Kufuneka kuqatshelwe ukuba lo moya uya phambana x-axis kwi indawo nge zilungelelanise (a, 0,0), Oy - (0, b, 0), kunye Oz - (0,0, s).

Ngenxa equation x / a + y / b + z / c = 1, akukho nzima ukuba nomfanekiso isizalwane nokubekwa moya lwenkqubo kwangaphambili nokulawula.

Uququzelelo ye kumda wesangqa eqhelekileyo

Kumda wesangqa n yesiqhelo ukuya P moya uye zilungelelanise ukuba ezi okuza lenxaki jikelele we-moya, ngamanye n (A, B, C).

Ukuze kumiselwe uququzelelo ye n oqhelekileyo, oko kwanele ukwazi inxaki jikelele anikwe moya.

Xa usebenzisa quadratic ngezigaba, eba fomu x / a + y / b + z / c = 1, njengoko xa usebenzisa i equation jikelele ukuba zilungelelanise nawuphi kumda wesangqa eqhelekileyo ebhaliweyo-moya banikwa: (1 / a + 1 / b + 1 / c).

Kufuneka kuqatshelwe ukuba kumda wesangqa eqhelekileyo ukunceda ukusombulula iingxaki ezahlukahlukeneyo. Iingxaki kakhulu zixhaphakileyo owakhiwa moya ubungqina nkqo okanye parallel, umsebenzi ukufumana engile phakathi iinqwelomoya okanye engile phakathi iinqwelomoya kunye imigca ethe ngqo.

Chwetheza ngokuvumelana lenxaki moya kunye zilungelelanise lwendawo kumda wesangqa eqhelekileyo

A esiyua n nonzero, aa-moya elinikiweyo, ebizwa eqhelekileyo (eqhelekileyo) ukuya moya kwangaphambili.

Masithi kwi nokulungelelanisa isithuba (a uxande nokulungelelanisa inkqubo) Oxyz ukubeka:

• Mₒ indawo nge zilungelelanise (hₒ, uₒ, zₒ);
• zero vector n = A * i + B * j + C * k.

Kufuneka wenze equation lo moya ugqithayo kuwo Mₒ indawo nkqo ukuya n yesiqhelo.

Kwisithuba esikhetha nayiphi na indawo engenasizathu kwaye ingabonisa M (x, y, z). Makhe embindini wesangqa phuzu M nganye (x, y, z) iya kuba r = x * i + y * j + * k z, kwaye embindini wesangqa of a Mₒ indawo (hₒ, uₒ, zₒ) - rₒ = hₒ * i + uₒ * j + zₒ * k. Ingongoma M iya kuba moya elinikiweyo, ukuba MₒM kumda wesangqa kuba nkqo eya kumda wesangqa n. Sibhala imeko orthogonality usebenzisa imveliso scalar:

[MₒM, n] = 0.

Ekubeni MₒM = r-rₒ, inxaki kumda wesangqa ye-moya iza kukhangeleka ngolu hlobo:

[R - rₒ, n] = 0.

Le equation nako enye imilo. Ukulungiselela le njongo, iimpawu imveliso scalar, zaguqulwa icala lasekhohlo lenxaki. [R - rₒ, n] = [r, n] - [rₒ, n]. Ukuba [rₒ, n] zibonakaliswe njenge s, siya ukufumana inxaki ilandelayo: [r, n] - a = 0 okanye [r, n] = s, nto leyo ichaza ukuba rhoqo le noqikelelo kwizixa kumda wesangqa eqhelekileyo kwirediyasi-njengezazisi amanqaku anikwe ezila moya.

Ngoku ungakwazi ukufumana Uhlobo ukurekhodwa moya yomzobo equation yethu [r - rₒ, n] = 0. Ekubeni r-rₒ = (x-hₒ) * i + (y-uₒ) * j + (z-zₒ) * k, kwaye n = A * i + B * j + C * k, kufuneka:

Kubonakala ukuba sinazo equation yenziwe moya edlula kwinqanaba nkqo ukuya n yesiqhelo:

A * (x hₒ) + B * (y uₒ) S * (z-zₒ) = 0.

Chwetheza ngokuvumelana lenxaki moya kunye zilungelelanise zamanqaku ezimbini collinear esiyua moya

Kanjani amanqaku ezimbini engenasizathu M '(x', y ', z') kunye M "(x", y ", z"), kwakunye kumda wesangqa (a ', i ", a ‴).

Ngoku ke bhala equation kwangaphambili moya bekuza kwindawo M okhoyo 'kwaye M ", yaye ingongoma nganye kunye uququzelelo M (x, y, z) enxuseneyo eya kumda wesangqa elinikiweyo.

Ngoko zithwala M'M x = {x ', y-y'; my '} kunye M "M = {x" -x', y 'y'; z "-z '} kufuneka coplanar kunye kumda wesangqa a = (a ', i ", indawo ‴), nto leyo ethetha ukuba (M'M M" M, a) = 0.

Ngoko equation wethu-moya kwisithuba iza kukhangeleka ngolu hlobo:

Uhlobo moya lalenxaki, ewela iingongoma ezintathu

Makhe sithi amanqaku amathathu: (x ', y', z '), (x', y ', z'), (x ‴ Awunayo ‴, z ‴), nto leyo ke andingowawo kumgca efanayo. Kuyimfuneko ukuba ubhale equation lo moya edlula iingongoma ezintathu ezichaziweyo. theory geometry uthi ikhona olu hlobo moya, nto nje omnye kuphela. Ekubeni lo moya inqumla ngongoma (x ', y', z '), uhlobo yayo equation iya kuba:

Apha, A, B, C ezahlukileyo zero ngexesha elifanayo. Kwakhona moya banikwa yekhampasi iingongoma ezimbini (x ", y", z ") kunye (x ‴, y ‴, z ‴). Kule uxhulumaniso kufuneka kwenziwe olu hlobo iimeko:

Ngoku sikwazi ukudala inkqubo efanayo kwizibalo (linear) kunye angaziwa u, v, w:

Kwimeko yethu x, y okanye z limi ingongoma engenasizathu oyifaneleyo equation (1). Xa ucinga equation (1) kunye nenkqubo kwizibalo (2) kunye (3) inkqubo kwizibalo kubonisiwe kulo mzobo ngasentla, kumda wesangqa Wakuhluthisa N (A, B, C) leyo nontrivial. Kungenxa yokuba onefuthe lwale nkqubo zero.

Equation (1) ukuba siyifumene, oku inxaki lo moya. 3 point yena ngenene ingena, kwaye kulula ukuba ukuhlola. Ukwenza oku, sandise onefuthe ngayo izinto kumqolo wokuqala. Of kwiipropati ezikhoyo onefuthe hlobo ukuba moya yethu ngaxeshanye inqumla ingongoma ezintathu ekuqaleni kwangaphambili (x ', y', z '), (x ", y", z "), (x ‴, y ‴, z ‴). Ngoko sagqiba ekubeni umsebenzi phambi kwethu.

angle Dihedral phakathi iinqwelomoya

angle Dihedral i imilo lwendawo zejiyometri abunjwe ngesiqingatha-moya ezimbini ezivela umgca othe ngqo. Ngamanye amazwi, inxalenye isithuba elinganiselwe kwisiqingatha-moya.

Masithi esinayo moya ezimbini kunye kwizibalo zilandelayo:

Siyazi ukuba kumda wesangqa N = (A, B, C) kunye N¹ = (A¹, H¹, S¹) ngokutsho moya kwangaphambili ukuba nkqo. Kule nkalo, i-engile φ phakathi bezifo N kunye N¹ angle alinganayo (dihedral), nto leyo ifumaneka phakathi kwezi iinqwelomoya. Le mveliso scalar siyinikwe:

NN¹ = | N || N¹ | cos φ,

kucacile kuba

cosφ = NN¹ / | N || N¹ | = (AA¹ + VV¹ SS¹ +) / ((√ (A² + s² + V²)) * (√ (A¹) ² + (H¹) + ² (S¹) ²)).

Kwanele ukuba abheke 0≤φ≤π.

Eneneni ezimbini moya phambana, uhlobo engile ezimbini (dihedral): φ ₁ kunye φ _2. sum yabo ilingana π (φ ₁ + φ ₂ = π). Ke cosines zawo, imilinganiselo yawo ingundoqo bayalingana, kodwa ke imiqondiso ezahlukeneyo, oko kukuthi, cos φ ₁ = -cos φ _2. Ukuba equation (0) kufakwa ngu A, B no-C -Ngaba, -B kunye -C ngokulandelelana, inxaki, sinokuyifumana, siya kumisela moya mnye,-engile kuphela φ kwi equation cos φ = NN ¹ / | N || N ¹ | Oku kuya kufakwa π-φ.

Inxaki lo moya nkqo

Ebizwa nkqo moya, phakathi apho angle degrees 90. Ukusebenzisa umbandela uchazwe apha ngasentla, sinokufumana inxaki-moya nkqo ukuya kwelinye. Masithi kufuneka moya ezimbini: Ax + Ngu + Cz + D = 0, kwaye + A¹h V¹u S¹z + + D = 0. Sinokuthi ukuba orthogonal ukuba cos = 0. Oku kuthetha ukuba NN¹ = AA¹ + VV¹ SS¹ + = 0.

Inxaki-moya enxuseneyo

It wabhekisela moya ezimbini ezinxuseneyo eziqulathe akukho manqaku ngazo.

Imeko -moya okunxuseneyo (quadratic zabo ziyafana kumhlathi ongaphambili) kukuba zithwala N kunye N¹, nto leyo nkqo kubo, collinear. Oku kuthetha ukuba le miqathango ilandelayo iyafezekiswa ukulingana:

A / A¹ = B / C = H¹ / S¹.

Ukuba imiqathango umlinganiselo-qhinga - A / A¹ = B / C = H¹ / S¹ = DD¹,

oku kubonisa ukuba lo moya idatha efanayo. Oku kuthetha ukuba equation Ax + Ngu + Cz + D = 0 no + A¹h V¹u S¹z + + D¹ = 0 ukuchaza moya mnye.

Umgama ukusuka akhombe moya

Masithi sibe P moya, nto leyo siyinikwe (0). Kuyimfuneko ukufumana umgama ukusuka indawo nge zilungelelanise (hₒ, uₒ, zₒ) = Qₒ. , Kufuneka ukuba inxaki kwi-moya II ukubonakala eqhelekileyo ukwenza oko;

(Ρ, v) = p (r≥0).

Kulo mzekelo, ρ (x, y, z) na kumda wesangqa owela ngaphakathi kwama-point Q wethu, ibekwe kwi n p - n ke ubude nkqo, nto leyo wakhululwa ukusuka kwindawo kwiqanda, v - lo kumda wesangqa iyunithi, nto leyo zicwangciswe icala.

Umahluko ρ-ρº radius kumda wesangqa ye Q indawo = (x, y, z), ukubayinxalenye n kunye embindini wesangqa yencopho elinikiweyo Q ₀ = (hₒ, uₒ, zₒ) i kumda wesangqa enjalo, ixabiso elililo ezayo okuthe v lilingana d umgama, apho kuyimfuneko ukufumana ukusuka Q = ₀ (hₒ, uₒ, zₒ) ukuya P:

D = | (ρ-ρ ^0, v) |, kodwa

(Ρ-ρ ^0, v) = (ρ, v ) - (ρ ^0, v) = p (ρ ^0, v).

Ngoko ke kuvela,

d = | (ρ ^0, v) p |.

Ngoku ke kucacile ukuba ukubala umgama d ukusuka _kwi-0 ukuya Q moya P, kuyimfuneko ukusebenzisa kweyesiqhelo imboniselo moya lenxaki, nokutshintshela ekhohlo p, indawo yokugqibela ye x, y, z obambeleyo (hₒ, uₒ, zₒ).

Ngenxa yoko, sifumana ixabiso elililo ibinzana onesiphumo efunekayo d.

Usebenzisa imida ye ulwimi, sifumana ezicacileyo:

d = | Ahₒ Vuₒ + + Czₒ | / √ (A² + V² + s²).

Ukuba indawo Q elikhankanyiweyo ₀ yi kwelinye icala P moya njengoko imvelaphi, ngoko phakathi kumda wesangqa ρ-ρ ⁰ kwaye v i -engile obtuse, ukuthi:

d = - (ρ-ρ ^0, v) = (ρ ^0, v) -p> 0.

Kwimeko xa ingongoma Q ₀ ngokusebenzisana kunye imvelaphi ibekwe kwicala elifanayo U, iengile etsolo idaliwe, oko kukuthi:

d = (ρ-ρ ^0, v) = p - (ρ ^0, v)> 0.

Isiphumo kukuba kwimeko yangaphambili (ρ ^0, v)> p, elinesibini (ρ ^0, v)

Kwaye yayo tangent moya equation

Ngokuphathelele indiza komhlaba kwindawo tangency Mº --moya equlethe konke ebaxiweyo kunokwenzeka ukuba kwigophe azotywe ngayo ngelo xesha phezu komhlaba.

Nale fomu komhlaba lenxaki F (x, y, z) = 0 inxaki lo Mº tanjenti moya tanjenti ngongoma (hº, uº, zº) iya kuba:

F _x (hº, uº, zº) (hº x) + F _x (hº, uº, zº) (uº y) + F _x (hº, uº, zº) (z-zº) = 0.

Ukuba ngaphezulu ibekwe ngokucacileyo z = f (x, y), ngoko moya tanjenti sichazwa lenxaki:

z-zº = f (hº, uº) (hº x) + f (hº, uº) (y uº).

Kuhlangana moya ezimbini

In isithuba-ntathu i nokulungelelanisa inkqubo (buxande) Oxyz, anikwe moya ezimbini P 'kwaye P' ukuba ziyangenana kwaye musa idibana. Ukususela nayiphi moya, nto leyo engunxantathu nokulungelelanisa inkqubo echazwe quadratic ngokubanzi, sicingela ukuba n 'kwaye n "ichazwe ngu quadratic A'x + V'u S'z + + D' = 0 no A" + B x '+ y with "z + D" = 0. Kule meko siya kuba n oqhelekileyo '(A', B ', C') loMthetho P moya 'kwaye n eqhelekileyo "(A", B ", C") of the P moya'. Njengoko moya yethu nqo kwaye musa idibana, ngoko ezi zithwala nje collinear. Ukusebenzisa ulwimi lwemathematika, saba nale meko angabhalwa ngolu hlobo: n '≠ n "↔ (A', B ', C') ≠ (λ * Yaye", λ * In ", λ * C"), λεR. Ngamana umgca ngqo edulusele kwi P ekudibaneni 'kunye P ", ziya zibonakaliswe yi ezinonobumba ua, kule meko a = P' ∩ P".

kunye - umgca kuqulethe ezininzi iingongoma moya (common) P 'kunye P ". Oku kuthetha ukuba uququzelelo nawuphi na indawo yelo ukuya kumgca a, kufuneka ngaxeshanye wawuhluthisa equation A'x + V'u S'z + + D '= 0 no A "x + B' + C y" z + D "= 0. Oku kuthetha ukuba uququzelelo lwendawo uya kuba nesisombululo ethile kwizibalo zilandelayo:

Isiphumo kukuba isisombululo (ngokubanzi) yale nkqubo kwizibalo uya kumisela uququzelelo ngamnye amanqaku kumgca leyo aya kusebenza kwindawo yonqumlo P 'kunye P ", aqingqa umgca kwindawo nokulungelelanisa inkqubo Oxyz (eluxande) isithuba.