A lissafi na jirgin sama: yadda za a yi? Nau'in jirgin lissafai

A jirgin sama sarari za a iya bayyana a hanyoyi daban-daban (daya dot da vector, da vector da maki biyu, da maki uku, da dai sauransu). Yana da da wannan a zuciyarsa, da jirgin saman lissafi iya samun daban-daban. Har ila yau, a karkashin wani yanayi jirgin sama zai iya zama a layi daya, perpendicular, intersecting, da dai sauransu A wannan kuma zai yi magana a wannan labarin. Za mu koyi yin general lissafi daga cikin jirgin saman da ba kawai.

A al'ada irin na lissafi

Zata R ne sarari _3, wanda yana da rectangular daidaita tsarin XYZ. Mun ayyana wani vector α, wanda za a saki daga cikin masomin O. Ta ƙarshen vector α zana jirgin P wanda yake perpendicular zuwa gare shi.

Nuna a fakaice P a wani sabani batu Q = (x, y, z). A radius vector na ma'ana Q alamar harafi p. A tsawon vector daidai α p = IαI da Ʋ = (cosα, cosβ, cosγ).

Wannan naúrar vector, wanda aka directed a cikin shugabanci kamar yadda vector α. α, β da γ - ne kusassari da cewa an kafa tsakanin vector da kyau kwatance Ʋ sarari gatura x, y, z bi da bi. A tsinkaya daga wani batu a vector QεP Ʋ ne akai wanda shi ne daidai p (p, Ʋ) = p (r≥0).

A bisa lissafi tana da ma'ana idan p = 0. The kawai n jirgin sama a cikin wannan hali, zai haye batu Yã (α = 0), wanda yake shi ne asali, kuma naúrar vector Ʋ, saki daga batu Yã zai zama perpendicular zuwa P, ko da yake ta shugabanci, wanda ke nufin cewa vector Ʋ m har zuwa alama. Previous lissafi ne mu jirgin P, bayyana, a vector form. Amma a cikin ra'ayi na ta tsarawa ne:

P ne fi ko daidai to 0. Mun sami jirgin lissafi a al'ada form.

A general lissafi

Idan lissafi a cikin tsarawa ninka ta kowane lambar da cewa ba su daidaita da sifili, mun samu lissafi daidai da wannan da cewa ma'anar da sosai jirgin sama. Yana zai yi da wadannan nau'i:

A nan, A, B, C - shi ne yawan lokaci guda daban-daban daga sifili. Wannan lissafi ne da ake kira lissafi na general nau'i daga cikin jirgin saman.

A lissafai na jirage. lokuta na musamman

A lissafi iya kullum a modified tare da ƙarin yanayi. Ka yi la'akari da wasu daga cikinsu.

Zaton cewa coefficient A ne 0. Wannan ya nuna cewa jirgin saman layi daya da qaddara axis sa. A wannan yanayin, da tsari na lissafi ya canza: Wu + Cz + D = 0.

Hakazalika, irin lissafi kuma zai bambanta da wadannan yanayi:

• Da fari dai, idan B = 0, da lissafi canje-canje zuwa Axe + Cz + D = 0, wanda zai nuna parallelism ga axis Oy.
• Abu na biyu, idan C = 0, da lissafi an canza kama zuwa Axe + By + D = 0, wato a ce game da layi daya da qaddara axis Oz.
• Na uku, idan D = 0, da lissafi zai bayyana a matsayin Axe + By + Cz = 0, wanda zai nufin cewa, jirgin intersects Yã (asalin).
• Huxu, idan A = B = 0, da lissafi canje-canje zuwa Cz + D = 0, wanda zai tabbatar da parallelism Oxy.
• Biyar, idan B = C = 0, da lissafi zama Axe + D = 0, wanda ke nufin cewa, jirgin ne a layi daya to Oyz.
• Sixthly, idan A = C = 0, da lissafi daukan nau'i Wu + D = 0, Ina nufin, zai bayar da rahoton zuwa parallelism Oxz.

Form na lissafi a segments

A cikin akwati inda lambobin A, B, C, D daban-daban daga sifili, nau'i na lissafi (0) na iya zama kamar haka:

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

cikinsa da wani = -D / A, b = -D / B, C = -D / C.

Mun sami a sakamakon lissafi daga cikin jirgin sama a guda. Ya kamata a lura cewa wannan jirgin sama zai ratsa cikin x-axis a ma'ana tare da tsarawa (a, 0,0), Oy - (0, b, 0), da kuma Oz - (0,0, s).

Ganin lissafi x / mai + y / b + z / c = 1, shi ne, ba wuya a iya ganin jeri jirgin dangi zuwa wani qaddara tsara tsarin.

A lura na al'ada vector

A al'ada vector n da jirgin saman P yana tsarawa da cewa su ne coefficients na janar lissafi daga cikin jirgin sama, Ina nufin n (A, B, C).

A don ƙayyade da lura na al'ada n, shi ne isa ya san janar lissafi ba jirgin sama.

Lokacin amfani da lissafi a segments, wanda yana cikin irin x / mai + y / b + z / c = 1, kamar lokacin amfani da janar lissafi za a iya rubuta tsarawa daga wani al'ada vector wani ba jirgin sama: (1 / wani + 1 / b + 1 / c).

Ya kamata a lura da cewa al'ada vector na taimaka wa warware matsaloli daban-daban. Mafi na kowa matsaloli suna kunsha a hujja perpendicular ko layi daya jirage, da aikin gano da kusassari tsakanin jirage ko kusassari tsakanin jirage da kuma madaidaiciya Lines.

Rubuta bisa ga jirgin lissafi da kuma lura na ma'ana al'ada vector

A nonzero vector n, perpendicular zuwa ba jirgin sama, da ake kira al'ada (al'ada) zuwa ga makoma jirgin sama.

Yi tsammani cewa, a cikin tsara sarari (a rectangular daidaita tsarin) Oxyz saita:

• Mₒ nufi da tsarawa (hₒ, uₒ, zₒ).
• sifili vector n = A * i + B * j + C * k.

Kana bukatar ka yi lissafi daga cikin jirgin saman da ya wuce ta Mₒ batu perpendicular ga al'ada n.

A sarari muka zabi wani sabani batu da kuma nuna a fakaice M (x, y, z). Bari radius vector na kowane batu M (x, y, z) zai zama r = x * i + y * j + z * k, da kuma radius vector na wani batu Mₒ (hₒ, uₒ, zₒ) - rₒ = hₒ * i + uₒ * j + zₒ * k. Batun M zai kasance a wani ba jirgin sama, idan vector MₒM zama perpendicular zuwa vector n. Mun rubuta da yanayin orthogonality amfani da scalar samfurin:

[MₒM, n] = 0.

Tun MₒM = r-rₒ, da vector lissafi daga cikin jirgin sama zai yi kama da wannan:

[R - rₒ, n] = 0.

Wannan lissafi kuma iya samun wani siffar. A saboda wannan dalili, da kaddarorin da scalar samfurin, da kuma tuba gefen hagu na lissafi. [R - rₒ, n] = [r, n] - [rₒ, n]. Idan [rₒ, n] denoted kamar s, mu samu wadannan lissafi: [r, n] - mai = 0 ko [r, n] = s, wanda ya nuna da haƙuri da tsinkaya a kan al'ada vector na radius-vectors na ba da maki cewa suna cikin jirgin sama.

Yanzu za ka iya samun tsara irin rikodi jirgin sama mu vector lissafi [r - rₒ, n] = 0. Tun r-rₒ = (x-hₒ) * i + (y-uₒ) * j + (z-zₒ) * k, da kuma n = A * i + B * j + C * k, muna da:

Sai dai itace cewa muna da lissafi an kafa jirgin sama wucewa ta cikin aya perpendicular ga al'ada n:

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

Rubuta bisa ga jirgin lissafi da kuma tsarawa daga maki biyu daga cikin vector jirgin collinear

Mun ayyana biyu sabani da maki M '(x', y ', z') da kuma M "(x", y ", z"), kazalika da vector (a ', wani ", a ‴).

Yanzu za mu iya rubuta lissafi qaddara jirgin sama wanda ya wuce ta data kasance maki M 'kuma M ", da kuma kowane batu tare da tsarawa M (x, y, z) a layi daya zuwa wani ba vector.

Kamar wancan M'M vectors x = {x ', y-y'. ZZ '} da M "M = {x" -x', y 'y'. Z "-z '} kasance coplanar da vector mai = (a ', wani ", a ‴), wanda ke nufin cewa (M'M M" M, wani) = 0.

Saboda haka mu Jadawalin wani jirgin saman a sararin samaniya za su yi kama da wannan:

Irin jirgin sama lissafi, tsallaka maki uku

Bari mu ce muna da maki uku: (x ', y', z '), (x', y ', z'), (x ‴ Have ‴, z ‴), wanda ba su kasance a cikin wannan layi. Wajibi ne a rubuta lissafi daga cikin jirgin sama wucewa ta cikin maki uku kayyade. lissafi ka'idar bayar da hujjar cewa, irin wannan jirgin sama ba wanzu, yana da guda ɗaya kawai kawai. Tun da wannan jirgin sama intersects batu (x ', y', z '), ta lissafi form zai zama:

A nan, A, B, C ne daban-daban daga sifili a lokaci guda. Har ila yau, ba jirgin sama intersects biyu mafi maki (x ", y", z ") da kuma (x ‴, y ‴, z ‴). A wannan dangane kamata a da za'ayi irin wannan yanayi:

Yanzu za mu iya ƙirƙirar uniform tsarin na lissafai (mikakke) tare da unknowns u, v, w:

A cikin yanayin x, y ko z tsaye sabani nufi da kosad da lissafi (1). Ganin lissafi (1) kuma a tsarin na lissafai (2) da kuma (3) da tsarin na lissafai nuna a cikin adadi sama, da vector kosad N (A, B, C) wanda yake nontrivial. Shi ne saboda determinant na tsarin ne sifili.

Lissafi (1) cewa mun samu, wannan ne lissafi daga cikin jirgin saman. 3 aya ta gaske ke, kuma yana da sauki a duba. Don yin wannan, mun fadada determinant da abubuwa a cikin ta farko jere. Na data kasance Properties determinant haka cewa mu jirgin lokaci guda intersects da uku asali qaddara batu (x ', y', z '), (x ", y", z "), (x ‴, y ‴, z ‴). Saboda haka muka yanke shawarar kallafã a gaban mu.

Dihedral kwana tsakanin jirage

Dihedral kwana ne a sarari na lissafi siffar kafa ta biyu rabin-jirage cewa fitowa daga wani madaidaiciya line. A wasu kalmomin, wani ɓangare na sarari da aka iyakance zuwa rabin-jirage.

Yi tsammani muna da biyu jirgin sama tare da wadannan lissafai:

Mun san cewa vector N = (A, B, C) da kuma N¹ = (A¹, H¹, S¹) bisa ga qaddara jirage ne perpendicular. A wannan batun, da kwana φ tsakanin vectors N kuma N¹ daidai kwana (dihedral), wanda aka located tsakanin wadannan jirage. A scalar samfurin da aka ba by:

NN¹ = | N || N¹ | cos φ,

daidai saboda

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

Shi ne isa zuwa ga la'akari da cewa 0≤φ≤π.

A gaskiya biyu jirage cewa rarraba, nau'i biyu kwana (dihedral): φ ₁ da kuma φ _2. Su ware Naira Miliyan Xari ne daidai π (φ ₁ + φ ₂ = π). Amma ga su cosines, su cikakkar dabi'u ne daidai, amma su ne daban-daban da ãyõyinMu, wato, cos φ ₁ = -cos φ _2. Idan a cikin lissafi (0) da aka maye gurbinsu da A, B da C da -A, -B da -C bi da bi, da lissafi, za mu samu, zai ƙayyade guda jirgin sama, kadai kwana φ a lissafi cos φ = NN ¹ / | N || N ¹ | Za a maye gurbinsu da π-φ.

A lissafi na perpendicular jirgin sama

Kira perpendicular jirgin sama, tsakanin wanda da kwana ne 90 digiri. Amfani da kayan gabatar a sama, za mu iya samun Jadawalin wani jirgin saman perpendicular zuwa wancan. Yi tsammani muna da biyu jirage: Axe + By + Cz + D = 0, kuma + A¹h V¹u S¹z + + D = 0. Za mu iya cewa su ne orthogonal idan cos = 0. Wannan yana nufin cewa NN¹ = AA¹ + VV¹ SS¹ + = 0.

A Jadawalin a layi daya jirgin sama

Yana kira zuwa biyu a layi daya jirage wanda dauke da wani maki a na kowa.

A yanayin na layi daya jirage (su lissafai ne guda kamar yadda a baya sakin layi) ne cewa vectors N kuma N¹, wanda suke perpendicular zuwa gare su, collinear. Wannan yana nufin cewa wadannan yanayi ne ya sadu proportionality:

A / A¹ = B / C = H¹ / S¹.

Idan gwargwado sharuddan aka fadada - A / A¹ = B / C = H¹ / S¹ = DD¹,

wannan ya nuna cewa data jirgin sama na wannan. Wannan yana nufin cewa lissafi Axe + By + Cz + D = 0 kuma + A¹h V¹u S¹z + + D¹ = 0 bayyana daya jirgin sama.

A nesa daga batu to jirgin sama

Zata da muke da wani jirgin saman P, wanda aka bai da (0). Wajibi ne a nemo nesa daga batu tare da tsarawa (hₒ, uₒ, zₒ) = Qₒ. , Kana bukatar ka kawo da lissafi a cikin jirgin sama II al'ada bayyanar yi shi:

(Ρ, v) = p (r≥0).

A wannan yanayin, ρ (x, y, z) ne radius vector na mu auna Q, located a kan n p - n shi ne da tsawon da perpendicular, wadda aka saki daga sifili batu, v - sashi ne vector, wanda aka shirya a cikin shugabanci a.

Bambanci ρ-ρº radius vector na wani batu Q = (x, y, z), na zuwa n da radius vector na wani ba batu Q ₀ = (hₒ, uₒ, zₒ) shi ne irin wannan vector, da cikakkar darajar da tsinkaya daga wanda a v daidai da nesa d, wanda wajibi ne a sami daga Q = ₀ (hₒ, uₒ, zₒ) zuwa P:

D = | (ρ-ρ ^0, v) |, amma

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

Sabõda haka, shi dai itace,

d = | (ρ ^0, v) p |.

Yanzu a sarari yake cewa yin lissafi da nesa d daga ₀ zuwa Q jirgin P, shi wajibi ne a yi amfani da al'ada view jirgin lissafi, da miƙa zuwa hagu na p, kuma na karshe wuri na x, y, z canza (hₒ, uₒ, zₒ).

Saboda haka, za mu sami cikakkar darajar da sakamakon magana da ake bukata d.

Amfani da sigogi na harshe, muna samun bayyane:

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

Idan kayyade batu Q ₀ ne, a hayin daga cikin jirgin saman P a matsayin asalin, sa'an nan tsakanin vector ρ-ρ ⁰ kuma v ne da wani obtuse kwana, kamar haka:

d = - (ρ-ρ ^0, v) = (ρ ^0, v) -p> 0.

A cikin akwati a lokacin da ma'ana Q ₀ a tare da tare da asalin located a kan wannan gefe na U, da m kwana da aka halitta, cewa shi ne:

d = (ρ-ρ ^0, v) = p - (ρ ^0, v)> 0.

A sakamakon haka ne cewa a cikin tsohon harka (ρ ^0, v)> p, a karo na biyu (ρ ^0, v)

Kuma ta tangent jirgin lissafi

Game da jirgin sama zuwa surface a batu na tangency Mº - wani jirgin sama dauke da duk yiwu tangent zuwa kwana kõma ta hanyar wannan batu a farfajiya.

Tare da wannan surface nau'i na lissafi F (x, y, z) = 0 a cikin lissafi na tangent jirgin tangent batu Mº (hº, uº, zº) zai zama:

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

Idan surface an saita baro-baro z = f (x, y), sa'an nan da tangent jirgin sama da aka bayyana ta da lissafi:

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

The mahada biyu jirage

A uku-girma sarari ne mai tsara tsarin (rectangular) Oxyz, ba biyu jirage P 'da kuma P' cewa zoba, kuma kada ku daidaita. Tun da wani jirgin, wanda yake a cikin wani rectangular daidaita tsarin ayyana ta da janar lissafi, muna zaton cewa n 'da kuma n "an tsare ta da lissafai A'x + V'u S'z + + D' = 0 kuma A" + B x '+ y tare da "z + D" = 0. A wannan yanayin da muke da al'ada n '(A', B ', C') daga cikin jirgin saman P 'da kuma al'ada n "(A", B ", C") daga cikin jirgin saman P'. Kamar yadda mu jirgin sama ba a layi daya, kuma kada ku daidaita, to, wadannan vectors ba collinear. Amfani da harshen lissafi, muna da wannan yanayin za a iya rubuta kamar: n '≠ n "↔ (A', B ', C') ≠ (λ * Kuma", λ * A ", λ * C"), λεR. Bari mike layin da ya ta'allaka ne a mahada P 'da kuma P ", za a denoted da wasika a, cikin wannan harka mai = P' ∩ P".

da kuma - a layin kunshi wani jam'i na da maki (kowa) jirage P 'da kuma P ". Wannan yana nufin cewa da lura na wani batu na zuwa layin wani, dole ne a lokaci guda gamsar da lissafi A'x + V'u S'z + + D '= 0 kuma A "x + B' + C y" z + D "= 0. Wannan yana nufin cewa da lura na ma'ana zai zama wani musamman bayani na wadannan lissafai:

A sakamakon haka ne cewa bayani (sauran) na wannan tsarin na lissafai zai ƙayyade tsarawa daga kowane maki a kan layi wanda zai yi aiki a matsayin batu na mahada P 'da kuma P ", da kuma sanin wani layi a cikin wani daidaita tsarin Oxyz (rectangular) sarari.