AM@space plane beam (plane system) equation pencil of planes

plane beam equation

• Let the straight line LLGeneral equation for$L\; :$

  • $$\begin{gathered} {Pi}_1:A_{1}x+B_{1}y+C_{1}z+D_1=0 \\ {Pi}_2:A_{2}x+B_{2}y+C_{2}z+D_2=0 \end{gathered}$$

  • Among them, $A_1,B_1,C_1$,$A_2,B_2,C_2$out of proportion

    • 即$\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}=k$ does not hold, (that is, the two lines are not parallel) or
    • There is no k such that $A_1=to{A}_2;B_1=k{B}_2;C_1=k{C}_2$

• Adding the general expressions of the two plane equations, we get

  • $$A_{1}x+B_{1}y+C_{1}z+D_1+(A_{2}x+B_{2}y+C_{2}z+D_2)=0$$

• More generally, with ${Pi}_2$Parallel planes can be expressed as ${Pi}_2^p:l(A_{2}x+B_{2}y+C_{2}z)+D^{\prime }=0$

  • It can also be written ${Pi}_2^p:l(A_{2}x+B_{2}y+C_{2}z+l^{-1}{D}^{\prime })=0$

    • 即${Pi}_2^p:(A_{2}x+B_{2}y+C_{2}z+l^{-1}{D}^{\prime })=0$ ; while$D^{\prime }$ means any constant,$l^{-1}{D}^{\text{'}}$ also means any constant, you can use$D^{\prime }$ slope$l^{-1}{D}^{\prime }$ , so one can write

      • $${Pi}_2^p:(A_{2}x+B_{2}y+C_{2}z+D^{\prime })=0$$

  • Plane ${Pi}_2$A more general form of $l{P}_2:l(A_{2}x+B_{2}y+C_{2}z+D^{\prime })=0$ , this equation and${Pi}_2$The planes represented are exactly the same (coincident), which is different from the general parallel but not coincident (because $l{P}_2$Divide both sides by $\lambda$ , we get${Pi}_2$)

    • This is like $4x\_=3$ and$8x=6$ , the two equations have the same solution and can be directly converted to each other

    $$Pi:A_{1}x+B_{1}y+C_{1}z+D_1+l(A_{2}x+B_{2}y+C_{2}z+D_2)=0$$

  • where $\lambda$ is an arbitrary constant, and the equation is transformed equivalently:

    • $$Pi:(A_1+\lambda A_2)x+(B_1+\lambda B_2)y+(C_1+\lambda C_2)z+(D_1+\lambda D_2)=0$$

    • x , y , zx,y,z in this equation$x,y,The\; coefficients\; of\; z$ are not 0 at the same time, because the necessary and sufficient condition for being 0 at the same time is the existence of$\lambda$ such that

      • $$\begin{gathered} A_1+\lambda A_2=0,A_1=-\lambda A_2 \\ {B}_1+\lambda B_2=0,B_1=-\lambda B_2 \\ {C}_1+\lambda C_2=0,C_1=-\lambda C_2 \end{gathered}$$

      • And the previous assumption points out that ${Pi}_1,{Pi}_2$are not parallel, that is, there cannot exist such $k=-\lambda$ , makex$x,y,The\; coefficient\; of\; z$ is 0 at the same time

    • Therefore the equation $\Pi$ represents a subject affected byλ \lambdaThe plane controlled by$\lambda$$The\; \lambda$ parameter can be recorded as$\Pi \left(\lambda \right)$ , can be calledthe plane family

• Located on line LLPoints on$L$${Pi}_1,{Pi}_2$The equation of , so it also satisfies the equation $P\left(l\right)$

  • $$\begin{gathered} A_1{x}_0+B_1{y}_0+C_1{z}_0+D_1=0 \\ {A}_2{x}_0+B_2{y}_0+C_2{z}_0+D_2=0 \\ {A}_1{x}_0+B_1{y}_0+C_1{z}_0+D_1+l(A_2{x}_0+B_2{y}_0+C_2{z}_0+D_2)=0 \end{gathered}$$

  • So all points on the line lie on the plane $\Pi$ , the straight line$L$ lies on the plane$P$ _

  • For different $\lambda$ , plane equation$\Pi \left(\lambda \right)$ means through$The\; different\; planes\; of\; L$ (analogous to the pole of the flag is the straight line$L$ , the flag sail represents the (half) plane, with the pole$L$ is the axis, rotate up)

  • through straight line $The\; plane\; of\; L$ (except${Pi}_2$other than ), you can use the equation $P\left(l\right)$ display

    • ${Pi}_2$cannot be $\Pi \left(\lambda \right)$ means that no matterλ \lambdaWhatever value$\lambda$${Pi}_2$Π $\Pi$ two equations are equal (or say, make${Pi}_2=\Pi \left(\lambda \right)$ λ$\lambda$ does not exist)

more general form

• $$P(m,l):m(A_1{x}_0+B_1{y}_0+C_1{z}_0+D_1)+l(A_2{x}_0+B_2{y}_0+C_2{z}_0+D_2)=0$$

• The plane bundle represented by this equation can represent $L\text{'}$ s plane equation

  • 当$m\ne 0$ ,$P(m,\lambda )$ divide both sides of the equation by$\mu$ ,interfaceΠ ( λ ) \Pi(\lambda)The form of$\Pi$$($$\lambda$$)$
  • 当$m=0$ ,$P(m,\lambda )$比$\Pi \left(\lambda \right)$ can express more than one surface, which is${Pi}_2$(take $m=0,l=1$)

• The equation Π ( μ , λ ) \Pi(\mu,\lambda) can be understood with the help of the normal vector of the plane$P(m,\lambda )$ construction

  • $$(\mu A_1+\lambda A_2)x+(\mu B_1+\lambda B_2)y+(\mu C{\_}_1+\lambda C_2)z+(\mu D_1+\lambda D_2)=0$$

  • planeΠ $P(m,\lambda )$ squares$(\mu A_1+\lambda A_2,\mu B_1+\lambda B_2,\mu C{\_}_1+\lambda C_2)$

• For a complete and systematic proof, refer to the relevant literature listed in the subsection at the end of the article

example

• Find the straight line $L_0$:
  $$\begin{gathered} x+y-z-1=0 \\ x-y+z+1=0 \end{gathered}$$
  in plane${Pi}_0:x+y+z=$Equation LLof the projected straight line on$0$$L$

• untie

  • Just ask for $L$得$\Pi$ gets the vertical plane equation${Pi}_1$, and then combine ${Pi}_0,{Pi}_1$Get the general formula equation (group) of the straight line

  • From the general equations of a straight line, the straight line system (bundle) of the straight line can be constructed.

  • Let the straight line $L_0$Targeted formula $(x+y-z-1)+\lambda (x-y+z+1)=0$ , ie

    • $${Pi}_1:(1+l)x+(1-l)y+(-1+\lambda )z+(-1+l)=0$$

    • From the plane vertical relationship: $(1+l)\cdot 1+(1-l)\cdot 1+(-1+l)\cdot 1=0$ , the solution is$l=-1$

    • Thus ${Pi}_1:2y-2z\_-2=0$ , ie$y-z-1=0$

  • Thus the straight line $L$

    • $$\begin{gathered} y-z-1=0 \\ x+y+z=0 \end{gathered}$$

refs

• Pencil (geometry) - Wikipedia
• Pencil (Pencil_of_planes) - Wikipedia
• Study on the Theorem Proving of Plance Pencil Equation | SpringerLink
  • The sci-hub mirror can download this booklet, and a chapter in it introduces the relevant proof
  • library.lol/main/51AE519BC38EABE162E4A439D030E872 this link may be time-sensitive
  • [pdf version](https://cloudflare-ipfs.com/ipfs/bafykbzaceb7ew25jjxjngeloinknz2ne6myd4ppy7fredhlddclr7wkkpo6ik?filename=(Communications in Computer and Information Science 243) Hai-E Zhang%2C Cheng Wang%2C Wen-Feng Huo%2C Guo-Ying Pang (auth.)%2C Chunfeng Liu%2C Jincai Chang%2C Aimin Yang (eds.) - Information Computing and Application.pdf)