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Animating planes passing through two given points in a cube



The Next CEO of Stack OverflowHow to color several transparent planes in a cube?Line through two points with offset in TikZCurve through a sequence of points with Metapost and TikZPerpendicular line to two-points-line given the lengthline passing through a point and extendingTikZ: Drawing an ellipse through two pointsCube with louvre planes in LaTeXDraw a lightning bolt between two given points in tikz?Ellipse through defined pointsDraw two opposing planes setting










2















Is there a smarter way to animate a collection of possible planes passing through 2 given points P on ABFE and Q on DCGH such that we can visually prove the point R is the common point of ABCD and PQ?



enter image description here



documentclass[pstricks,border=12pt,12pt]standalone
usepackagepst-eucl

begindocument
% first frame
beginpspicture[showgrid=false](-1,-3)(13,14)
pstGeonode[PosAngle=180,0,0,135,180](0,0)A(10,0)B(12,3)C(2,3)D(0,10)E
pstTranslation[PosAngle=0,0,135]AEB,C,D[F,G,H]
psline(E)(H)(G)(C)(B)(F)(E)(A)(B)
psline(F)(G)
psline[linestyle=dashed](H)(D)(C)
psline[linestyle=dashed](A)(D)
% Extra
pstGeonode[PosAngle=180,-90](1,9)P(7,-2)Q
pstTranslation[PosAngle=180,PointName=none,PointSymbol=none]EAP
pstTranslation[PointName=none,PointSymbol=none]BFQ
pstInterLL[PosAngle=-90]ABPP'X
pstInterLL[PosAngle=45]CDQQ'Y
pstTranslation[PosAngle=90]AEX,Y
pspolygon[linestyle=none,fillstyle=solid,fillcolor=yellow,opacity=0.25](X)(X')(Y')(Y)
pspolygon[linestyle=none,fillstyle=solid,fillcolor=cyan,opacity=0.25](A)(B)(C)(D)
psline[linecolor=red,linestyle=dashed](X)(Y)
psline[linecolor=red,linestyle=dashed](P)(Q)
pstInterLL[PosAngle=0]PQXYR
endpspicture

% second frame
beginpspicture[showgrid=false](-1,-3)(13,14)
pstGeonode[PosAngle=180,0,0,135,180](0,0)A(10,0)B(12,3)C(2,3)D(0,10)E
pstTranslation[PosAngle=0,0,135]AEB,C,D[F,G,H]
psline(E)(H)(G)(C)(B)(F)(E)(A)(B)
psline(F)(G)
psline[linestyle=dashed](H)(D)(C)
psline[linestyle=dashed](A)(D)
% Extra
pstGeonode[PosAngle=180,-90](1,9)P(7,-2)Q
pstTranslation[PosAngle=180,PointName=none,PointSymbol=none]EHP
pstTranslation[PointName=none,PointSymbol=none]CBQ
pstInterLL[PosAngle=-90]ABPQ'X
pstInterLL[PosAngle=45]CDQP'Y
pstInterLL[PosAngle=-90]EFPQ'X'
pstInterLL[PosAngle=45]HGQP'Y'
pspolygon[linestyle=none,fillstyle=solid,fillcolor=yellow,opacity=0.25](X)(X')(Y')(Y)
pspolygon[linestyle=none,fillstyle=solid,fillcolor=cyan,opacity=0.25](A)(B)(C)(D)
psline[linecolor=red,linestyle=dashed](X)(Y)
psline[linecolor=red,linestyle=dashed](P)(Q)
pstInterLL[PosAngle=0]PQXYR
endpspicture
enddocument









share|improve this question


























    2















    Is there a smarter way to animate a collection of possible planes passing through 2 given points P on ABFE and Q on DCGH such that we can visually prove the point R is the common point of ABCD and PQ?



    enter image description here



    documentclass[pstricks,border=12pt,12pt]standalone
    usepackagepst-eucl

    begindocument
    % first frame
    beginpspicture[showgrid=false](-1,-3)(13,14)
    pstGeonode[PosAngle=180,0,0,135,180](0,0)A(10,0)B(12,3)C(2,3)D(0,10)E
    pstTranslation[PosAngle=0,0,135]AEB,C,D[F,G,H]
    psline(E)(H)(G)(C)(B)(F)(E)(A)(B)
    psline(F)(G)
    psline[linestyle=dashed](H)(D)(C)
    psline[linestyle=dashed](A)(D)
    % Extra
    pstGeonode[PosAngle=180,-90](1,9)P(7,-2)Q
    pstTranslation[PosAngle=180,PointName=none,PointSymbol=none]EAP
    pstTranslation[PointName=none,PointSymbol=none]BFQ
    pstInterLL[PosAngle=-90]ABPP'X
    pstInterLL[PosAngle=45]CDQQ'Y
    pstTranslation[PosAngle=90]AEX,Y
    pspolygon[linestyle=none,fillstyle=solid,fillcolor=yellow,opacity=0.25](X)(X')(Y')(Y)
    pspolygon[linestyle=none,fillstyle=solid,fillcolor=cyan,opacity=0.25](A)(B)(C)(D)
    psline[linecolor=red,linestyle=dashed](X)(Y)
    psline[linecolor=red,linestyle=dashed](P)(Q)
    pstInterLL[PosAngle=0]PQXYR
    endpspicture

    % second frame
    beginpspicture[showgrid=false](-1,-3)(13,14)
    pstGeonode[PosAngle=180,0,0,135,180](0,0)A(10,0)B(12,3)C(2,3)D(0,10)E
    pstTranslation[PosAngle=0,0,135]AEB,C,D[F,G,H]
    psline(E)(H)(G)(C)(B)(F)(E)(A)(B)
    psline(F)(G)
    psline[linestyle=dashed](H)(D)(C)
    psline[linestyle=dashed](A)(D)
    % Extra
    pstGeonode[PosAngle=180,-90](1,9)P(7,-2)Q
    pstTranslation[PosAngle=180,PointName=none,PointSymbol=none]EHP
    pstTranslation[PointName=none,PointSymbol=none]CBQ
    pstInterLL[PosAngle=-90]ABPQ'X
    pstInterLL[PosAngle=45]CDQP'Y
    pstInterLL[PosAngle=-90]EFPQ'X'
    pstInterLL[PosAngle=45]HGQP'Y'
    pspolygon[linestyle=none,fillstyle=solid,fillcolor=yellow,opacity=0.25](X)(X')(Y')(Y)
    pspolygon[linestyle=none,fillstyle=solid,fillcolor=cyan,opacity=0.25](A)(B)(C)(D)
    psline[linecolor=red,linestyle=dashed](X)(Y)
    psline[linecolor=red,linestyle=dashed](P)(Q)
    pstInterLL[PosAngle=0]PQXYR
    endpspicture
    enddocument









    share|improve this question
























      2












      2








      2








      Is there a smarter way to animate a collection of possible planes passing through 2 given points P on ABFE and Q on DCGH such that we can visually prove the point R is the common point of ABCD and PQ?



      enter image description here



      documentclass[pstricks,border=12pt,12pt]standalone
      usepackagepst-eucl

      begindocument
      % first frame
      beginpspicture[showgrid=false](-1,-3)(13,14)
      pstGeonode[PosAngle=180,0,0,135,180](0,0)A(10,0)B(12,3)C(2,3)D(0,10)E
      pstTranslation[PosAngle=0,0,135]AEB,C,D[F,G,H]
      psline(E)(H)(G)(C)(B)(F)(E)(A)(B)
      psline(F)(G)
      psline[linestyle=dashed](H)(D)(C)
      psline[linestyle=dashed](A)(D)
      % Extra
      pstGeonode[PosAngle=180,-90](1,9)P(7,-2)Q
      pstTranslation[PosAngle=180,PointName=none,PointSymbol=none]EAP
      pstTranslation[PointName=none,PointSymbol=none]BFQ
      pstInterLL[PosAngle=-90]ABPP'X
      pstInterLL[PosAngle=45]CDQQ'Y
      pstTranslation[PosAngle=90]AEX,Y
      pspolygon[linestyle=none,fillstyle=solid,fillcolor=yellow,opacity=0.25](X)(X')(Y')(Y)
      pspolygon[linestyle=none,fillstyle=solid,fillcolor=cyan,opacity=0.25](A)(B)(C)(D)
      psline[linecolor=red,linestyle=dashed](X)(Y)
      psline[linecolor=red,linestyle=dashed](P)(Q)
      pstInterLL[PosAngle=0]PQXYR
      endpspicture

      % second frame
      beginpspicture[showgrid=false](-1,-3)(13,14)
      pstGeonode[PosAngle=180,0,0,135,180](0,0)A(10,0)B(12,3)C(2,3)D(0,10)E
      pstTranslation[PosAngle=0,0,135]AEB,C,D[F,G,H]
      psline(E)(H)(G)(C)(B)(F)(E)(A)(B)
      psline(F)(G)
      psline[linestyle=dashed](H)(D)(C)
      psline[linestyle=dashed](A)(D)
      % Extra
      pstGeonode[PosAngle=180,-90](1,9)P(7,-2)Q
      pstTranslation[PosAngle=180,PointName=none,PointSymbol=none]EHP
      pstTranslation[PointName=none,PointSymbol=none]CBQ
      pstInterLL[PosAngle=-90]ABPQ'X
      pstInterLL[PosAngle=45]CDQP'Y
      pstInterLL[PosAngle=-90]EFPQ'X'
      pstInterLL[PosAngle=45]HGQP'Y'
      pspolygon[linestyle=none,fillstyle=solid,fillcolor=yellow,opacity=0.25](X)(X')(Y')(Y)
      pspolygon[linestyle=none,fillstyle=solid,fillcolor=cyan,opacity=0.25](A)(B)(C)(D)
      psline[linecolor=red,linestyle=dashed](X)(Y)
      psline[linecolor=red,linestyle=dashed](P)(Q)
      pstInterLL[PosAngle=0]PQXYR
      endpspicture
      enddocument









      share|improve this question














      Is there a smarter way to animate a collection of possible planes passing through 2 given points P on ABFE and Q on DCGH such that we can visually prove the point R is the common point of ABCD and PQ?



      enter image description here



      documentclass[pstricks,border=12pt,12pt]standalone
      usepackagepst-eucl

      begindocument
      % first frame
      beginpspicture[showgrid=false](-1,-3)(13,14)
      pstGeonode[PosAngle=180,0,0,135,180](0,0)A(10,0)B(12,3)C(2,3)D(0,10)E
      pstTranslation[PosAngle=0,0,135]AEB,C,D[F,G,H]
      psline(E)(H)(G)(C)(B)(F)(E)(A)(B)
      psline(F)(G)
      psline[linestyle=dashed](H)(D)(C)
      psline[linestyle=dashed](A)(D)
      % Extra
      pstGeonode[PosAngle=180,-90](1,9)P(7,-2)Q
      pstTranslation[PosAngle=180,PointName=none,PointSymbol=none]EAP
      pstTranslation[PointName=none,PointSymbol=none]BFQ
      pstInterLL[PosAngle=-90]ABPP'X
      pstInterLL[PosAngle=45]CDQQ'Y
      pstTranslation[PosAngle=90]AEX,Y
      pspolygon[linestyle=none,fillstyle=solid,fillcolor=yellow,opacity=0.25](X)(X')(Y')(Y)
      pspolygon[linestyle=none,fillstyle=solid,fillcolor=cyan,opacity=0.25](A)(B)(C)(D)
      psline[linecolor=red,linestyle=dashed](X)(Y)
      psline[linecolor=red,linestyle=dashed](P)(Q)
      pstInterLL[PosAngle=0]PQXYR
      endpspicture

      % second frame
      beginpspicture[showgrid=false](-1,-3)(13,14)
      pstGeonode[PosAngle=180,0,0,135,180](0,0)A(10,0)B(12,3)C(2,3)D(0,10)E
      pstTranslation[PosAngle=0,0,135]AEB,C,D[F,G,H]
      psline(E)(H)(G)(C)(B)(F)(E)(A)(B)
      psline(F)(G)
      psline[linestyle=dashed](H)(D)(C)
      psline[linestyle=dashed](A)(D)
      % Extra
      pstGeonode[PosAngle=180,-90](1,9)P(7,-2)Q
      pstTranslation[PosAngle=180,PointName=none,PointSymbol=none]EHP
      pstTranslation[PointName=none,PointSymbol=none]CBQ
      pstInterLL[PosAngle=-90]ABPQ'X
      pstInterLL[PosAngle=45]CDQP'Y
      pstInterLL[PosAngle=-90]EFPQ'X'
      pstInterLL[PosAngle=45]HGQP'Y'
      pspolygon[linestyle=none,fillstyle=solid,fillcolor=yellow,opacity=0.25](X)(X')(Y')(Y)
      pspolygon[linestyle=none,fillstyle=solid,fillcolor=cyan,opacity=0.25](A)(B)(C)(D)
      psline[linecolor=red,linestyle=dashed](X)(Y)
      psline[linecolor=red,linestyle=dashed](P)(Q)
      pstInterLL[PosAngle=0]PQXYR
      endpspicture
      enddocument






      tikz-pgf diagrams pstricks asymptote metapost






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 7 hours ago









      Artificial Hairless ArmpitArtificial Hairless Armpit

      5,03711142




      5,03711142




















          1 Answer
          1






          active

          oldest

          votes


















          1














          I may not do an animation and just use a plane with a nontrivial opacity such that one can see that the dashed red line hits the plane. My take is that you want to




          visually prove the point R is the common point of ABCD and PQ




          because this is what is written in the question. If that's not what the question is about, consider editing it to make it clearer. (Sorry, I have quit pstricks a while ago, but I am sure you will be able to redo this with pstricks.)



          documentclass[tikz,border=3.14mm]standalone 
          usepackagetikz-3dplot
          usetikzlibrarybackgrounds
          newcountercoord
          begindocument
          tdplotsetmaincoords70200
          begintikzpicture[tdplot_main_coords,scale=3,bullet/.style=fill,circle,inner sep=1pt]
          foreach Z in -1,1
          foreach Y in -1,1
          foreach X in -1,1
          stepcountercoord
          path (Y*X,-1*Y,Z) coordinate (Alphcoord) -- ++ (0.2*Y*X,0,0)
          node$Alphcoord$ ;
          fill[cyan,opacity=0.5] (A) -- (B) -- (C) -- (D) -- cycle;
          draw[dashed] (A) -- (D) -- (C) (D) -- (H);
          draw (E) -- (F) -- (G) -- (H) -- (E) -- (A) -- (B) -- (C) -- (G) (B) -- (F);
          node[bullet,label=left:$P$] (P) at (1,0,0.5);
          node[bullet,label=above right:$R$] (R) at (1/3,0,-1);
          draw[red,dashed,thick] (P) -- (R)
          node[pos=1.5,black,bullet,label=[black]below:$Q$] (Q);
          beginscope[on background layer]
          draw[red,dashed,thick] (R) -- (Q);
          endscope
          endtikzpicture
          enddocument


          enter image description here






          share|improve this answer

























          • Where is the collection of infinitely many planes passing through PQ?

            – Artificial Hairless Armpit
            6 hours ago











          • @ArtificialHairlessArmpit These planes are IMHO not necessary to "visually prove the point R is the common point of ABCD and PQ".

            – marmot
            6 hours ago












          • The planes are required for my question.

            – Artificial Hairless Armpit
            6 hours ago











          • @ArtificialHairlessArmpit But not for my answer. ;-) And sorry to say that, but your screen shot is confusing precisely because below R the line seems to run in front of the plane.

            – marmot
            6 hours ago












          Your Answer








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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          I may not do an animation and just use a plane with a nontrivial opacity such that one can see that the dashed red line hits the plane. My take is that you want to




          visually prove the point R is the common point of ABCD and PQ




          because this is what is written in the question. If that's not what the question is about, consider editing it to make it clearer. (Sorry, I have quit pstricks a while ago, but I am sure you will be able to redo this with pstricks.)



          documentclass[tikz,border=3.14mm]standalone 
          usepackagetikz-3dplot
          usetikzlibrarybackgrounds
          newcountercoord
          begindocument
          tdplotsetmaincoords70200
          begintikzpicture[tdplot_main_coords,scale=3,bullet/.style=fill,circle,inner sep=1pt]
          foreach Z in -1,1
          foreach Y in -1,1
          foreach X in -1,1
          stepcountercoord
          path (Y*X,-1*Y,Z) coordinate (Alphcoord) -- ++ (0.2*Y*X,0,0)
          node$Alphcoord$ ;
          fill[cyan,opacity=0.5] (A) -- (B) -- (C) -- (D) -- cycle;
          draw[dashed] (A) -- (D) -- (C) (D) -- (H);
          draw (E) -- (F) -- (G) -- (H) -- (E) -- (A) -- (B) -- (C) -- (G) (B) -- (F);
          node[bullet,label=left:$P$] (P) at (1,0,0.5);
          node[bullet,label=above right:$R$] (R) at (1/3,0,-1);
          draw[red,dashed,thick] (P) -- (R)
          node[pos=1.5,black,bullet,label=[black]below:$Q$] (Q);
          beginscope[on background layer]
          draw[red,dashed,thick] (R) -- (Q);
          endscope
          endtikzpicture
          enddocument


          enter image description here






          share|improve this answer

























          • Where is the collection of infinitely many planes passing through PQ?

            – Artificial Hairless Armpit
            6 hours ago











          • @ArtificialHairlessArmpit These planes are IMHO not necessary to "visually prove the point R is the common point of ABCD and PQ".

            – marmot
            6 hours ago












          • The planes are required for my question.

            – Artificial Hairless Armpit
            6 hours ago











          • @ArtificialHairlessArmpit But not for my answer. ;-) And sorry to say that, but your screen shot is confusing precisely because below R the line seems to run in front of the plane.

            – marmot
            6 hours ago
















          1














          I may not do an animation and just use a plane with a nontrivial opacity such that one can see that the dashed red line hits the plane. My take is that you want to




          visually prove the point R is the common point of ABCD and PQ




          because this is what is written in the question. If that's not what the question is about, consider editing it to make it clearer. (Sorry, I have quit pstricks a while ago, but I am sure you will be able to redo this with pstricks.)



          documentclass[tikz,border=3.14mm]standalone 
          usepackagetikz-3dplot
          usetikzlibrarybackgrounds
          newcountercoord
          begindocument
          tdplotsetmaincoords70200
          begintikzpicture[tdplot_main_coords,scale=3,bullet/.style=fill,circle,inner sep=1pt]
          foreach Z in -1,1
          foreach Y in -1,1
          foreach X in -1,1
          stepcountercoord
          path (Y*X,-1*Y,Z) coordinate (Alphcoord) -- ++ (0.2*Y*X,0,0)
          node$Alphcoord$ ;
          fill[cyan,opacity=0.5] (A) -- (B) -- (C) -- (D) -- cycle;
          draw[dashed] (A) -- (D) -- (C) (D) -- (H);
          draw (E) -- (F) -- (G) -- (H) -- (E) -- (A) -- (B) -- (C) -- (G) (B) -- (F);
          node[bullet,label=left:$P$] (P) at (1,0,0.5);
          node[bullet,label=above right:$R$] (R) at (1/3,0,-1);
          draw[red,dashed,thick] (P) -- (R)
          node[pos=1.5,black,bullet,label=[black]below:$Q$] (Q);
          beginscope[on background layer]
          draw[red,dashed,thick] (R) -- (Q);
          endscope
          endtikzpicture
          enddocument


          enter image description here






          share|improve this answer

























          • Where is the collection of infinitely many planes passing through PQ?

            – Artificial Hairless Armpit
            6 hours ago











          • @ArtificialHairlessArmpit These planes are IMHO not necessary to "visually prove the point R is the common point of ABCD and PQ".

            – marmot
            6 hours ago












          • The planes are required for my question.

            – Artificial Hairless Armpit
            6 hours ago











          • @ArtificialHairlessArmpit But not for my answer. ;-) And sorry to say that, but your screen shot is confusing precisely because below R the line seems to run in front of the plane.

            – marmot
            6 hours ago














          1












          1








          1







          I may not do an animation and just use a plane with a nontrivial opacity such that one can see that the dashed red line hits the plane. My take is that you want to




          visually prove the point R is the common point of ABCD and PQ




          because this is what is written in the question. If that's not what the question is about, consider editing it to make it clearer. (Sorry, I have quit pstricks a while ago, but I am sure you will be able to redo this with pstricks.)



          documentclass[tikz,border=3.14mm]standalone 
          usepackagetikz-3dplot
          usetikzlibrarybackgrounds
          newcountercoord
          begindocument
          tdplotsetmaincoords70200
          begintikzpicture[tdplot_main_coords,scale=3,bullet/.style=fill,circle,inner sep=1pt]
          foreach Z in -1,1
          foreach Y in -1,1
          foreach X in -1,1
          stepcountercoord
          path (Y*X,-1*Y,Z) coordinate (Alphcoord) -- ++ (0.2*Y*X,0,0)
          node$Alphcoord$ ;
          fill[cyan,opacity=0.5] (A) -- (B) -- (C) -- (D) -- cycle;
          draw[dashed] (A) -- (D) -- (C) (D) -- (H);
          draw (E) -- (F) -- (G) -- (H) -- (E) -- (A) -- (B) -- (C) -- (G) (B) -- (F);
          node[bullet,label=left:$P$] (P) at (1,0,0.5);
          node[bullet,label=above right:$R$] (R) at (1/3,0,-1);
          draw[red,dashed,thick] (P) -- (R)
          node[pos=1.5,black,bullet,label=[black]below:$Q$] (Q);
          beginscope[on background layer]
          draw[red,dashed,thick] (R) -- (Q);
          endscope
          endtikzpicture
          enddocument


          enter image description here






          share|improve this answer















          I may not do an animation and just use a plane with a nontrivial opacity such that one can see that the dashed red line hits the plane. My take is that you want to




          visually prove the point R is the common point of ABCD and PQ




          because this is what is written in the question. If that's not what the question is about, consider editing it to make it clearer. (Sorry, I have quit pstricks a while ago, but I am sure you will be able to redo this with pstricks.)



          documentclass[tikz,border=3.14mm]standalone 
          usepackagetikz-3dplot
          usetikzlibrarybackgrounds
          newcountercoord
          begindocument
          tdplotsetmaincoords70200
          begintikzpicture[tdplot_main_coords,scale=3,bullet/.style=fill,circle,inner sep=1pt]
          foreach Z in -1,1
          foreach Y in -1,1
          foreach X in -1,1
          stepcountercoord
          path (Y*X,-1*Y,Z) coordinate (Alphcoord) -- ++ (0.2*Y*X,0,0)
          node$Alphcoord$ ;
          fill[cyan,opacity=0.5] (A) -- (B) -- (C) -- (D) -- cycle;
          draw[dashed] (A) -- (D) -- (C) (D) -- (H);
          draw (E) -- (F) -- (G) -- (H) -- (E) -- (A) -- (B) -- (C) -- (G) (B) -- (F);
          node[bullet,label=left:$P$] (P) at (1,0,0.5);
          node[bullet,label=above right:$R$] (R) at (1/3,0,-1);
          draw[red,dashed,thick] (P) -- (R)
          node[pos=1.5,black,bullet,label=[black]below:$Q$] (Q);
          beginscope[on background layer]
          draw[red,dashed,thick] (R) -- (Q);
          endscope
          endtikzpicture
          enddocument


          enter image description here







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 6 hours ago

























          answered 6 hours ago









          marmotmarmot

          113k5145274




          113k5145274












          • Where is the collection of infinitely many planes passing through PQ?

            – Artificial Hairless Armpit
            6 hours ago











          • @ArtificialHairlessArmpit These planes are IMHO not necessary to "visually prove the point R is the common point of ABCD and PQ".

            – marmot
            6 hours ago












          • The planes are required for my question.

            – Artificial Hairless Armpit
            6 hours ago











          • @ArtificialHairlessArmpit But not for my answer. ;-) And sorry to say that, but your screen shot is confusing precisely because below R the line seems to run in front of the plane.

            – marmot
            6 hours ago


















          • Where is the collection of infinitely many planes passing through PQ?

            – Artificial Hairless Armpit
            6 hours ago











          • @ArtificialHairlessArmpit These planes are IMHO not necessary to "visually prove the point R is the common point of ABCD and PQ".

            – marmot
            6 hours ago












          • The planes are required for my question.

            – Artificial Hairless Armpit
            6 hours ago











          • @ArtificialHairlessArmpit But not for my answer. ;-) And sorry to say that, but your screen shot is confusing precisely because below R the line seems to run in front of the plane.

            – marmot
            6 hours ago

















          Where is the collection of infinitely many planes passing through PQ?

          – Artificial Hairless Armpit
          6 hours ago





          Where is the collection of infinitely many planes passing through PQ?

          – Artificial Hairless Armpit
          6 hours ago













          @ArtificialHairlessArmpit These planes are IMHO not necessary to "visually prove the point R is the common point of ABCD and PQ".

          – marmot
          6 hours ago






          @ArtificialHairlessArmpit These planes are IMHO not necessary to "visually prove the point R is the common point of ABCD and PQ".

          – marmot
          6 hours ago














          The planes are required for my question.

          – Artificial Hairless Armpit
          6 hours ago





          The planes are required for my question.

          – Artificial Hairless Armpit
          6 hours ago













          @ArtificialHairlessArmpit But not for my answer. ;-) And sorry to say that, but your screen shot is confusing precisely because below R the line seems to run in front of the plane.

          – marmot
          6 hours ago






          @ArtificialHairlessArmpit But not for my answer. ;-) And sorry to say that, but your screen shot is confusing precisely because below R the line seems to run in front of the plane.

          – marmot
          6 hours ago


















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