Ad argvmentvm transi
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Rectae et plana in spatio

Svmmarivm

Haec lectio rectam per pvnctvm et vectorem directionis datam in spativm transfert, et planvm describit forma vectoriali et parametrica, per pvnctvm et dvos vectores directionis, et forma Cartesiana, per aeqvationem ax+by+cz=dax + by + cz = d cvivs coefficientes vectorem normalem efficivnt. Prodvctvm vectoriale vtramqve formam conivngit, et demonstratvr omne planvm aeqvationem Cartesianam habere, vnicam praeter factorem. Positiones mvtvae planorvm et rectarvm in systemata linearia vertvntvr et theoremate Rovché-Frobenii in classes digervntvr; prodvctvm mixtvm decernit vtrvm dvae rectae se secent an non coplanares sint. Lectio fasce planorvm conclvditvr et problematis de itineribvs aeronavivm, de inclinatione soli et de tetraedris.

Proposita discendi

  1. Rectam spatii forma vectoriali, parametrica et symmetrica scribere, etiam cvm aliqva componens vectoris directionis nvlla est, et demonstrare per dvo pvncta distincta vnam solam rectam transire.
  2. Planvm forma vectoriali, parametrica et Cartesiana scribere, ab vna forma ad aliam prodvcto vectoriali transire et planvm per tria pvncta non collinearia constrvere.
  3. Demonstrare omne planvm aeqvationem Cartesianam habere, vnicam praeter factorem non nvllvm, et omnem aeqvationem linearem in xx, yy et zz cvm aliqvo coefficiente non nvllo planvm describere.
  4. Positionem mvtvam dvorvm vel trivm planorvm, rectae et plani, dvarvm rectarvm per rangos systematis linearis et per prodvctvm mixtvm decernere, etiam cvm data a parametro pendent.
  5. Fasce planorvm constrvere planvm qvod rectam continet et aliam condicionem implet, et rectis planisqve res e navigatione aeria, ex aedificatione et ex designatione exprimere.

Recta in spatio

In lectione cvrsvs Algebra et Geometria I de vectoribvs et aeqvationibvs parametricis rectae, recta plani, qvae vt delineatio aeqvationis primi gradvs nota erat inde a lectione de aeqvatione rectae, descripta est vt P0+td⃗P_0 + t \vec{d}. In spatio declivitas non est, et aeqvatio primi gradvs in xx, yy et zz describit, vt videbitvr, planvm. Qvod si ita est, ordinem invertere convenit: descriptio vectorialis definitio erit, et ex ea ceterae formae dedvcentvr.

Vectores spatii eorvmqve prodvctvm scalare tractata svnt in lectione de vectoribvs in spatio et prodvcto scalari; prodvctvm vectoriale et prodvctvm mixtvm, in lectione de prodvcto vectoriali et prodvcto mixto. Vt in plano, Q=P+v⃗Q = P + \vec{v} significat PQ⃗=v⃗\vec{PQ} = \vec{v}, et dvo vectores non nvlli paralleli svnt cvm alter mvltiplex alterivs est. In figvris axis yy dextrorsvm vergit, axis zz svrsvm et axis xx ad spectatorem.

Sint AA pvnctvm et v⃗=(v1,v2,v3)\vec{v} = \left( v_1, v_2, v_3 \right) vector non nvllvs. Recta per AA cvm vectore directionis v⃗\vec{v} est copia pvnctorvm PP pro qvibvs nvmervs realis tt exstat ita vt P=A+tv⃗P = A + t \vec{v}; nvmervs tt parametrvm pvncti PP appellatvr.
P=A+tv⃗,t∈RP = A + t \vec{v}, \quad t \in \mathbb{R}

Parametrvm cvivsqve pvncti vnicvm est: si A+tv⃗=A+sv⃗A + t \vec{v} = A + s \vec{v}, tvm (t−s)v⃗=0⃗\left( t - s \right) \vec{v} = \vec{0}, et qvoniam v⃗\vec{v} nvllvs non est, t=st = s. Pro A=(x0,y0,z0)A = (x_0, y_0, z_0) et P=(x,y,z)P = (x, y, z), aeqvatio vectorialis, per componentes scripta, aeqvationes parametricas rectae praebet:

{x=x0+tv1y=y0+tv2z=z0+tv3\begin{cases} x = x_0 + t v_1 \\ y = y_0 + t v_2 \\ z = z_0 + t v_3 \end{cases}

Si parametrvm tempvs est, hae aeqvationes motvm aeqvabilem velocitatis v⃗\vec{v} describvnt, vt in plano. Eadem recta mvlta pvncta initialia mvltosqve vectores directionis admittit; theorema qvod infra scribitvr ostendit dvo eivs pvncta eam determinare.

TheoremaRecta per dvo pvncta

Per dvo pvncta distincta AA et BB vna sola recta transit, ea cvivs aeqvatio est P=A+tAB⃗P = A + t \vec{AB}. Itaqve aeqvationes P=A+tv⃗P = A + t \vec{v} et P=C+sw⃗P = C + s \vec{w} eandem rectam describvnt si et solvm si CC in prima iacet et w⃗\vec{w} parallelvs est vectori v⃗\vec{v}.

In figvra A=(2,0,1)A = (2, 0, 1) et B=(1,2,2)B = (1, 2, 2), ita vt AB⃗=(−1,2,1)\vec{AB} = \left( -1, 2, 1 \right), et altera recta per C=(0,4,3)C = (0, 4, 3) transit cvm vectore directionis v⃗=(1,−2,−1)\vec{v} = \left( 1, -2, -1 \right).

Demonstratio

  1. A+0⋅AB⃗=A,A+1⋅AB⃗=BA + \resaltar{0} \cdot \vec{AB} = A, \qquad A + \resaltar{1} \cdot \vec{AB} = B

    Exsistentia: vector AB⃗\vec{AB} nvllvs non est, qvia A≠BA \neq B, et recta per AA cvm hoc vectore directionis continet AA, parametro 00, et BB, parametro 11.

  2. A=C+t1v⃗,B=C+t2v⃗\dato{ab}{A = C + t_1 \vec{v}, \qquad B = C + t_2 \vec{v}}

    Vnicitas: sit rr recta qvaelibet, P=C+tv⃗P = C + t \vec{v}, qvae AA et BB contineat, et sint t1t_1 et t2t_2 eorvm parametra. In figvra t1=2t_1 = 2 et t2=1t_2 = 1.

  3. AB⃗=t2v⃗−t1v⃗=(t2−t1)v⃗\vec{AB} = t_2 \vec{v} - t_1 \vec{v} = \dato{d}{\resaltar{\left( t_2 - t_1 \right)} \vec{v}}

    Per relationem Chaslesianam, AB⃗=CB⃗−CA⃗\vec{AB} = \vec{CB} - \vec{CA}. Parametra distincta svnt, qvia A≠BA \neq B; ergo AB⃗\vec{AB} parallelvs est vectori v⃗\vec{v}. In figvra AB⃗=−v⃗\vec{AB} = -\vec{v}.

  4. C+tv⃗=A+(t−t1)v⃗=A+t−t1t2−t1AB⃗C + t \vec{v} = A + \left( t - t_1 \right) \vec{v} = A + \resaltar{\frac{t - t_1}{t_2 - t_1}} \vec{AB}

    Omne pvnctvm rectae rr in recta ABAB iacet: a pvncto AA loco pvncti CC incipimvs et pro v⃗\vec{v} eivs expressionem per AB⃗\vec{AB} svbstitvimvs. In figvra pvnctvm rectae rr parametro 12\frac{1}{2} est pvnctvm rectae ABAB parametro 32\frac{3}{2}.

  5. A+sAB⃗=A+s(t2−t1)v⃗=C+(t1+s(t2−t1))v⃗A + s \vec{AB} = A + s \left( t_2 - t_1 \right) \vec{v} = C + \resaltar{\left( t_1 + s \left( t_2 - t_1 \right) \right)} \vec{v}

    Vicissim pvnctvm parametro ss in recta ABAB est pvnctvm parametro t1+s(t2−t1)t_1 + s \left( t_2 - t_1 \right) in rr: dvae rectae eadem pvncta habent. Ad alteram affirmationem: si aeqvationes eandem rectam describvnt, CC et C+w⃗C + \vec{w} in ea iacent, et per id qvod tertio loco ostensvm est w⃗\vec{w} mvltiplex est vectoris v⃗\vec{v}; si CC in prima iacet et w⃗=kv⃗\vec{w} = k \vec{v}, prima continet CC et C+w⃗C + \vec{w}, et est recta per ea transiens.

Ex qvo seqvitvr dvas rectas distinctas svmmvm vnvm pvnctvm commvne habere: si dvo commvnia haberent, vtraqve esset vnica recta per ea transiens. Tria pvncta collinearia svnt cvm aliqva recta ea continet. Si A≠BA \neq B, pvnctvm CC cvm eis collineare est si et solvm si C=A+tAB⃗C = A + t \vec{AB} pro aliqvo tt, id est si et solvm si AC⃗\vec{AC} mvltiplex est vectoris AB⃗\vec{AB}.

Forma symmetrica

Parametro explicato et resvltatis aeqvatis, aeqvationes sine parametro obtinentvr, cvm cavtione circa componentes nvllas vectoris directionis.

PropositioForma symmetrica rectae

Sit rr recta per A=(x0,y0,z0)A = (x_0, y_0, z_0) cvm vectore directionis v⃗=(v1,v2,v3)\vec{v} = \left( v_1, v_2, v_3 \right). Si tres componentes vectoris v⃗\vec{v} a nihilo differvnt, pvnctvm (x,y,z)(x, y, z) in rr iacet si et solvm si aeqvalitates infra scriptae valent. Si sola v2v_2 nvlla est, condicio est x−x0v1=z−z0v3\frac{x - x_0}{v_1} = \frac{z - z_0}{v_3} et y=y0y = y_0; si v2v_2 et v3v_3 nvllae svnt, condicio est y=y0y = y_0 et z=z0z = z_0. Casvs qvibvs aliae componentes evanescvnt obtinentvr mvnera coordinatarvm permvtando.

x−x0v1=y−y0v2=z−z0v3\frac{x - x_0}{v_1} = \frac{y - y_0}{v_2} = \frac{z - z_0}{v_3}

In figvra A=(1,2,−1)A = (1, 2, -1) et v⃗=(2,0,4)\vec{v} = \left( 2, 0, 4 \right), vector qvi a AA ad B=(3,2,3)B = (3, 2, 3) dvcit.

Demonstratio

  1. x−x0v1=t,y−y0v2=t,z−z0v3=t\frac{x - x_0}{v_1} = \resaltar{t}, \quad \frac{y - y_0}{v_2} = \resaltar{t}, \quad \frac{z - z_0}{v_3} = \resaltar{t}

    Si tres componentes a nihilo differvnt et P=A+tv⃗P = A + t \vec{v}, ex vnaqvaqve aeqvatione parametrica tt explicamvs: tres qvotientes valent tt.

  2. x=x0+tv1,y=y0+tv2,z=z0+tv3x = x_0 + \resaltar{t} v_1, \quad y = y_0 + \resaltar{t} v_2, \quad z = z_0 + \resaltar{t} v_3

    Vicissim, si tres qvotientes aeqvales svnt, valorem eorvm commvnem tt appellamvs; vnaqvaqve aeqvalitate per svvm denominatorem mvltiplicata, aeqvationes parametricae recvperantvr, et P=A+tv⃗P = A + t \vec{v} in rr iacet.

  3. x−12=z+14,y=2\frac{x - 1}{2} = \frac{z + 1}{4}, \quad \resaltar{y = 2}

    Si sola v2v_2 nvlla est, secvnda aeqvatio est y=y0y = y_0 pro omni tt, et seorsvm scribitvr, qvia dividere per 00 sensv caret. In figvra recta in plano y=2y = 2 iacet.

  4. y=2,z=−1\resaltar{y = 2}, \quad \resaltar{z = -1}

    Si v2v_2 et v3v_3 nvllae svnt, dvae vltimae aeqvationes yy et zz figvnt, et pro omni xx parametrvm t=x−x0v1t = \frac{x - x_0}{v_1} pvnctvm dat: recta est parallela axi xx per AA dvcta, qvae in figvra in recessv apparet.

  5. 4−12=5+14=32,2−12≠2+14\frac{4 - 1}{2} = \frac{5 + 1}{4} = \frac{3}{2}, \qquad \frac{2 - 1}{2} \neq \frac{2 + 1}{4}

    Confirmatio in figvra: C=(4,2,5)C = (4, 2, 5) in recta iacet, parametro 32\frac{3}{2}, et D=(2,2,2)D = (2, 2, 2) non iacet, qvamqvam y=2y = 2 implet.

Qvaeqve aeqvalitas formae symmetricae aeqvatio primi gradvs est: recta figvrae est copia solvtionvm aeqvationvm 2x−z=32x - z = 3 et y=2y = 2. Cvm visvm erit qvamqve talem aeqvationem planvm describere, hoc dicet rectam esse intersectionem dvorvm planorvm, eivs formam implicitam.