On Symmetry Through-put linking 3-4-5-6-7 Dimension's ,
Octo-Symmetry for the { 4-Dimension Quasi-Octahedron } ,
and 2 Identical { 8-D Structures } oriented Symmetrically

by Frank C. Fung ( 1st published in July, 2017. )

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Section XXVII - Identifying { 4 Symmetric Lines in 3-D }

Summary for the Section :

• In this Section , we shall identify :
  • the { 4 Symmetric Lines } residing in a { 3-Dimension Vector Space } ,
    • intersecting at the angle of [ 70.53 degrees ] .

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Bringing in the Cube :

• We take note at this point that the Cube , residing in a { 3-Dimension Vector Space } ,
  • has { 8 corners / nodal-points } and { 4 Cross-Diagonal Lines } .

    Cross-Diagonal Line I.D.	* * *	Position Vector for 1st end-point of the Cross-Diagonal Line				* * *	Position Vector for 2nd end-point of the Cross-Diagonal Line				* * *
    	*	Vector I.D.	Co-ordinates			*	Vector I.D.	Co-ordinates			*
    *******************	*	***********	****	****	****	*	***********	****	****	****	*
    Cross-Diagonal Line 1	*	Vector U-1	+1	+1	+1	*	Vector U-2	-1	-1	-1	*
    Cross-Diagonal Line 2	*	Vector U-3	+1	+1	-1	*	Vector U-4	-1	-1	+1	*
    Cross-Diagonal Line 3	*	Vector U-5	+1	-1	+1	*	Vector U-6	-1	+1	-1	*
    Cross-Diagonal Line 4	*	Vector U-7	-1	+1	+1	*	Vector U-8	+1	-1	-1	*

  And it happens that :

  • the 4 Cross-Diagonal Lines do intersect one-another at an angle of [ 70.53 degrees ] ,
    • i.e. :
      • arccos [ 1 / 3 ] = [ 70.53 degrees ] .

• Since we did identify in the last Section :
  • a set of { 6-Symmetric Lines } residing in a { 4-Dimension Vector Space }
    • intersecting one-another at the angle of [ 70.53 degrees ] ;

  there is now a possibility that we would be able to identify therefrom :

  • one or more subsets of { 4 Lines residing in a 3-Dimension Sub-Space }
    • also intersecting one-another at an angle of [ 70.53 degrees ] .

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Bringing-in the { 6 Symmetric Lines in 4-D } :

• Let us now bring-in the { 6 Symmetric Lines } residing in a { 4-Dimension Space } ,
  • from the last Section .

  Cross-Diagonal Line I.D.	Class	* * *	Position Vector for 1st end-point of the Cross-Diagonal Line							* * *	Position Vector for 2nd end-point of the Cross-Diagonal Line
  		*	Vector I.D.	Co-ordinates						*	Vector I.D.	Co-ordinates
  **************	**************	*	***********	***	***	***	***	***	***	*	***********	***	***	***	***	***	***
  Line T-1	R-1 plus R-10	*	Vector T-1	+2	0	+2	-2	0	-2	*	Vector T-2	-2	0	-2	+2	0	+2
  Line T-2	R-1 minus R-10	*	Vector T-3	0	+2	0	0	-2	0	*	Vector T-4	0	-2	0	0	+2	0
  **************	**************	*	***********	***	***	***	***	***	***	*	***********	***	***	***	***	***	***
  Line Zero	Class 6-0 / 0-6	*	Vector Z-1	+1	+1	+1	+1	+1	+1	*	Vector Z-2	-1	-1	-1	-1	-1	-1
  Line Q-11	Class 4-2 / 2-4	*	Vector Q-21	+1	-1	+1	+1	-1	+1	*	Vector Q-22	-1	+1	-1	-1	+1	-1
  Line Q-3	Class 4-2 / 2-4	*	Vector Q-5	+1	+1	-1	-1	+1	+1	*	Vector Q-6	-1	-1	+1	+1	-1	-1
  Line Q-15	Class 4-2 / 2-4	*	Vector Q-29	-1	+1	+1	+1	+1	-1	*	Vector Q-30	+1	-1	-1	-1	-1	+1
  Line Q-12	Class 4-2 / 2-4	*	Vector Q-23	-1	+1	+1	-1	+1	+1	*	Vector Q-24	+1	-1	-1	+1	-1	-1
  Line Q-10	Class 4-2 / 2-4	*	Vector Q-19	+1	+1	-1	+1	+1	-1	*	Vector Q-20	-1	-1	+1	-1	-1	+1

• And for our analysis to follow ,
  • it would more convenient for us to keep on using [ integers ] for the 6 co-ordinates ,
    • in lieu of bringing-in [ fractions ] for the same .

  Therefore ,

  • we shall now re-size the { 6 Symmetric Lines in 4-D } ,
    • but without changing their 6 line-directions ;
      • via this table below .

  Cross-Diagonal Line I.D.	Class	* * *	Position Vector for 1st end-point of the Cross-Diagonal Line							* * *	Position Vector for 2nd end-point of the Cross-Diagonal Line
  		*	Vector I.D.	Co-ordinates						*	Vector I.D.	Co-ordinates
  **************	**************	*	***********	***	***	***	***	***	***	*	***********	***	***	***	***	***	***
  Line T-1	R-1 plus R-10	*	Vector T-1	+2	0	+2	-2	0	-2	*	Vector T2	-2	0	-2	+2	0	+2
  Line T-2	R-1 minus R-10	*	Vector T-3	0	+2	0	0	-2	0	*	Vector T-4	0	-2	0	0	+2	0
  **************	**************	*	***********	***	***	***	***	***	***	*	***********	***	***	***	***	***	***
  Line A	Class 6-0 / 0-6	*	Vector A-1	+3	+3	+3	+3	+3	+3	*	Vector A-2	-3	-3	-3	-3	-3	-3
  Line B	Class 4-2 / 2-4	*	Vector B-1	+3	-3	+3	+3	-3	+3	*	Vector B-2	-3	+3	-3	-3	+3	-3
  Line C	Class 4-2 / 2-4	*	Vector C-1	+3	+3	-3	-3	+3	+3	*	Vector C-2	-3	-3	+3	+3	-3	-3
  Line D	Class 4-2 / 2-4	*	Vector D-1	-3	+3	+3	+3	+3	-3	*	Vector D-2	+3	-3	-3	-3	-3	+3
  Line E	Class 4-2 / 2-4	*	Vector E-1	-3	+3	+3	-3	+3	+3	*	Vector E-2	+3	-3	-3	+3	-3	-3
  Line F	Class 4-2 / 2-4	*	Vector F-1	+3	+3	-3	+3	+3	-3	*	Vector F-2	-3	-3	+3	-3	-3	+3

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Outline of our Analystic Procedure :

• For { 4 Symmetric Lines in 3-D } intersecting one-another at an angle of [ 70.53 degrees ] ,
  • the 4 Lines would be in the formation of the { 4 Cross-Diagonal Lines of a 3-D Cube } .

  And we take note here that :

  • IF :
    • we take any one of the 4 lines as our { Line-Of-Concern } for the time-being ,

    THEN :

    • the projection of { the remaining 3 lines } in the [ 2-Dimension Plane ] orthogonal to the { Line-Of-Concern }
      • would intersect one-another at the [ 60 degress / 120 degrees ] combination .

  It is therefore our intention to select , in sequence , each of the { 6 Symmetric Lines in 4-D } as our { Line-Of-Concern } for the time-being ,

  • to look at the projections of the remaining 5 lines in the [ 3-Dimension Sub-Space ] orthogonal to the { Line-Of-Concern } ;

  and

  • if the projections of any 3 of the remaining 5 lines do lie in a [ 2-Dimension Plane ] and also :
    • intersect one-another at the [ 60 degrees / 120 degrees ] combination ,

    we would have identified a set of { 4 Symmetric Lines in 3-D } intersecting one-another at the angle of [ 70.53 degrees ] .

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Selecting { Line A } as the { Line-Of-Concern } :

• Let us now select { Line A } as our { Line-Of-Concern } for the time-being ,
  • using { Vector A-1 } as the primary vector for our analysis .

  We then bring in the 5 vectors intersecting { Vector A-1 } at the angle of [ 70.53 degrees ] ,

  • i.e. : { Vector B-1 } , { Vector C-1 } , { Vector D-1 } , { Vector E-1 } , and { Vector F-1 } , respectively ;
    • as shown in the table below .

• Since all 5 Vectors above are equal-in-size and intersect { Vector A-1 } at the angle of [ 70.53 degrees ] ,
  • their projections in the { Vector A-1 line-direction } are all the same , namely :
    • { Vector AQ } being [ 1 / 3 ] x { Vector A-1 } ;

      noting here :

      • cosine of [ 70.53 degrees ] = [ 1 / 3 ] .

  Thus , the projections of the 5 Vectors above in the [ 3-Dimension Sub-Space ] orthogonal ot { Vector A-1 } are given by :

  • { Projection of [ Vector B-1 ] in 3-D Sub-Space } = { Vector B-1 } minus { Vector AQ } ;
  • { Projection of [ Vector C-1 ] in 3-D Sub-Space } = { Vector C-1 } minus { Vector AQ } ;
  • { Projection of [ Vector D-1 ] in 3-D Sub-Space } = { Vector D-1 } minus { Vector AQ } ;
  • { Projection of [ Vector E-1 ] in 3-D Sub-Space } = { Vector E-1 } minus { Vector AQ } ;
  • { Projection of [ Vector F-1 ] in 3-D Sub-Space } = { Vector F-1 } minus { Vector AQ } ;

    as shown in the table below .

  Line I.D.	* * *	Vector for the Line							* * *	Comments
  	*	Vector I.D.	Co-ordinates						*
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line A	*	Vector A-1	+3	+3	+3	+3	+3	+3	*	Line-Of-Concern
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line B	*	Vector B-1	+3	-3	+3	+3	-3	+3	*	intersects { Vector A-1 } at [ 70.53 degrees ]
  Line C	*	Vector C-1	+3	+3	-3	-3	+3	+3	*	intersects { Vector A-1 } at [ 70.53 degrees ]
  Line D	*	Vector D-1	-3	+3	+3	+3	+3	-3	*	intersects { Vector A-1 } at [ 70.53 degrees ]
  Line E	*	Vector E-1	-3	+3	+3	-3	+3	+3	*	intersects { Vector A-1 } at [ 70.53 degrees ]
  Line F	*	Vector F-1	+3	+3	-3	+3	+3	-3	*	intersects { Vector A-1 } at [ 70.53 degrees ]
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  1/3 x Line-Of-Concern	*	Vector AQ	+1	+1	+1	+1	+1	+1	*	Projection in { Vector A-1 line-direction }
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Projection of Line B	*	Vector AB	+2	-4	+2	+2	-4	+2	*	{ Vector B-1 } minus { Vector AQ }
  Projection of Line C	*	Vector AC	+2	+2	-4	-4	+2	+2	*	{ Vector C-1 } minus { Vector AQ }
  Projection of Line D	*	Vector AD	-4	+2	+2	+2	+2	-4	*	{ Vector D-1 } minus { Vector AQ }
  Projection of Line E	*	Vector AE	-4	+2	+2	-4	+2	+2	*	{ Vector E-1 } minus { Vector AQ }
  Projection of Line F	*	Vector AF	+2	+2	-4	+2	+2	-4	*	{ Vector F-1 } minus { Vector AQ }

• We now makes these observations on the above table :
  • { Vector AC } and { Vector AD } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector AE } and { Vector AF } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector AB } intersects { Vector AC } / { Vector AD } / { Vector AE } / { Vector AF } at the angle of [ 120 degrees ] ;

  • { Vector AB } , { Vector AC } and { Vector AD } add up to [ zero ] ;

  • { Vector AB } , { Vector AE } and { Vector AF } add up to [ zero ] .

  As such :

  • { Vector AB } , { Vector AC } and { Vector AD } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

  • { Vector AB } , { Vector AE } and { Vector AF } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

• The 2 sets of { 4 Symmetric Lines in a 3-Dimension Vector Space } identified here are then :
  • { Line A / Line B / Line C / Line D } , and

  • { Line A / Line B / Line E / Line F } .

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Selecting { Line B } as the { Line-Of-Concern } :

• Let us now select { Line B } as our { Line-Of-Concern } for the time-being ,
  • using { Vector B-1 } as the primary vector for our analysis .

  We then bring in the 5 vectors intersecting { Vector B-1 } at the angle of [ 70.53 degrees ] ,

  • i.e. : { Vector A-1 } , { Vector C-2 } , { Vector D-2 } , { Vector E-2 } , and { Vector F-2 } , respectively ;
    • as shown in the table below .

• We then follow the same analystic procedure as in { Line A } above :
  • to arrive at the projections in the { 3-D sub-space } ,
    • as shown below .

  Line I.D.	* * *	Vector for the Line							* * *	Comments
  	*	Vector I.D.	Co-ordinates						*
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line B	*	Vector B-1	+3	-3	+3	+3	-3	+3	*	Line-Of-Concern
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line A	*	Vector A-1	+3	+3	+3	+3	+3	+3	*	intersects { Vector B-1 } at [ 70.53 degrees ]
  Line C	*	Vector C-2	-3	-3	+3	+3	-3	-3	*	intersects { Vector B-1 } at [ 70.53 degrees ]
  Line D	*	Vector D-2	+3	-3	-3	-3	-3	+3	*	intersects { Vector B-1 } at [ 70.53 degrees ]
  Line E	*	Vector E-2	+3	-3	-3	+3	-3	-3	*	intersects { Vector B-1 } at [ 70.53 degrees ]
  Line F	*	Vector F-2	-3	-3	+3	-3	-3	+3	*	intersects { Vector B-1 } at [ 70.53 degrees ]
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  1/3 x Line-Of-Concern	*	Vector BQ	+1	-1	+1	+1	-1	+1	*	projection in { Vector B-1 line-direction }
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Projection of Line A	*	Vector BA	+2	+4	+2	+2	+4	+2	*	{ Vector A-1 } minus { Vector BQ }
  Projection of Line C	*	Vector BC	-4	-2	+2	+2	-2	-4	*	{ Vector C-2 } minus { Vector BQ }
  Projection of Line D	*	Vector BD	+2	-2	-4	-4	-2	+2	*	{ Vector D-2 } minus { Vector BQ }
  Projection of Line E	*	Vector BE	+2	-2	-4	+2	-2	-4	*	{ Vector E-2 } minus { Vector BQ }
  Projection of Line F	*	Vector BF	-4	-2	+2	-4	-2	+2	*	{ Vector F-2 } minus { Vector BQ }

• We now makes these observations on the above table :
  • { Vector BC } and { Vector BD } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector BE } and { Vector BF } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector BA } intersects { Vector BC } / { Vector BD } / { Vector BE } / { Vector BF } at the angle of [ 120 degrees ] ;

  • { Vector BA } , { Vector BC } and { Vector BD } add up to [ zero ] ;

  • { Vector BA } , { Vector BE } and { Vector BF } add up to [ zero ] .

  As such :

  • { Vector BA } , { Vector BC } and { Vector BD } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

  • { Vector BA } , { Vector BE } and { Vector BF } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

• The 2 sets of { 4 Symmetric Lines in a 3-Dimension Vector Space } identified here are then :
  • { Line A / Line B / Line C / Line D } , and

  • { Line A / Line B / Line E / Line F } .

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Selecting { Line C } as the { Line-Of-Concern } :

• Let us now select { Line C } as our { Line-Of-Concern } for the time-being ,
  • using { Vector C-1 } as the primary vector for our analysis .

  We then bring in the 5 vectors intersecting { Vector C-1 } at the angle of [ 70.53 degrees ] ,

  • i.e. : { Vector A-1 } , { Vector B-2 } , { Vector D-2 } , { Vector E-1 } , and { Vector F-1 } , respectively ;
    • as shown in the table below .

• We then follow the same analystic procedure as in { Line A } above :
  • to arrive at the projections in the { 3-D sub-space } ,
    • as shown below .

  Line I.D.	* * *	Vector for the Line							* * *	Comments
  	*	Vector I.D.	Co-ordinates						*
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line C	*	Vector C-1	+3	+3	-3	-3	+3	+3	*	Line-Of-Concern
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line A	*	Vector A-1	+3	+3	+3	+3	+3	+3	*	intersects { Vector C-1 } at [ 70.53 degrees ]
  Line B	*	Vector B-2	-3	+3	-3	-3	+3	-3	*	intersects { Vector C-1 } at [ 70.53 degrees ]
  Line D	*	Vector D-2	+3	-3	-3	-3	-3	+3	*	intersects { Vector C-1 } at [ 70.53 degrees ]
  Line E	*	Vector E-1	-3	+3	+3	-3	+3	+3	*	intersects { Vector C-1 } at [ 70.53 degrees ]
  Line F	*	Vector F-1	+3	+3	-3	+3	+3	-3	*	intersects { Vector C-1 } at [ 70.53 degrees ]
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  1/3 x Line-Of-Concern	*	Vector CQ	+1	+1	-1	-1	+1	+1	*	projection in { Vector C-1 line-direction }
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Projection of Line A	*	Vector CA	+2	+2	+4	+4	+2	+2	*	{ Vector A-1 } minus { Vector CQ }
  Projection of Line B	*	Vector CB	-4	+2	-2	-2	+2	-4	*	{ Vector B-2 } minus { Vector CQ }
  Projection of Line D	*	Vector CD	+2	-4	-2	-2	-4	+2	*	{ Vector D-2 } minus { Vector CQ }
  Projection of Line E	*	Vector CE	-4	+2	+4	-2	+2	+2	*	{ Vector E-1 } minus { Vector CQ }
  Projection of Line F	*	Vector CF	+2	+2	-2	+4	+2	-4	*	{ Vector F-1 } minus { Vector CQ }

• We now makes these observations on the above table :
  • { Vector CA } and { Vector CB } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector CE } and { Vector CF } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector CD } intersects { Vector CA } / { Vector CB } / { Vector CE } / { Vector CF } at the angle of [ 120 degrees ] ;

  • { Vector CA } , { Vector CB } and { Vector CD } add up to [ zero ] ;

  • { Vector CD } , { Vector CE } and { Vector CF } add up to [ zero ] .

  As such :

  • { Vector CA } , { Vector CB } and { Vector CD } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

  • { Vector CD } , { Vector CE } and { Vector CF } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

• The 2 sets of { 4 Symmetric Lines in a 3-Dimension Vector Space } identified here are then :
  • { Line A / Line B / Line C / Line D } , and

  • { Line C / Line D / Line E / Line F } .

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Selecting { Line D } as the { Line-Of-Concern } :

• Let us now select { Line D } as our { Line-Of-Concern } for the time-being ,
  • using { Vector D-1 } as the primary vector for our analysis .

  We then bring in the 5 vectors intersecting { Vector D-1 } at the angle of [ 70.53 degrees ] ,

  • i.e. : { Vector A-1 } , { Vector B-2 } , { Vector C-2 } , { Vector E-1 } , and { Vector F-1 } , respectively ;
    • as shown in the table below .

• We then follow the same analystic procedure as in { Line A } above :
  • to arrive at the projections in the { 3-D sub-space } ,
    • as shown below .

  Line I.D.	* * *	Vector for the Line							* * *	Comments
  	*	Vector I.D.	Co-ordinates						*
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line D	*	Vector D-1	-3	+3	+3	+3	+3	-3	*	Line-Of-Concern
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line A	*	Vector A-1	+3	+3	+3	+3	+3	+3	*	intersects { Vector D-1 } at [ 70.53 degrees ]
  Line B	*	Vector B-2	-3	+3	-3	-3	+3	-3	*	intersects { Vector D-1 } at [ 70.53 degrees ]
  Line C	*	Vector C-2	-3	-3	+3	+3	-3	-3	*	intersects { Vector D-1 } at [ 70.53 degrees ]
  Line E	*	Vector E-1	-3	+3	+3	-3	+3	+3	*	intersects { Vector D-1 } at [ 70.53 degrees ]
  Line F	*	Vector F-1	+3	+3	-3	+3	+3	-3	*	intersects { Vector D-1 } at [ 70.53 degrees ]
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  1/3 x Line-Of-Concern	*	Vector DQ	-1	+1	+1	+1	+1	-1	*	projection in { Vector D-1 line-direction }
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Projection of Line A	*	Vector DA	+4	+2	+2	+2	+2	+4	*	{ Vector A-1 } minus { Vector DQ }
  Projection of Line B	*	Vector DB	-2	+2	-4	-4	+2	-2	*	{ Vector B-2 } minus { Vector DQ }
  Projection of Line C	*	Vector DC	-2	-4	+2	+2	-4	-2	*	{ Vector C-2 } minus { Vector DQ }
  Projection of Line E	*	Vector DE	-2	+2	+2	-4	+2	+4	*	{ Vector E-1 } minus { Vector DQ }
  Projection of Line F	*	Vector DF	+4	+2	-4	+2	+2	-2	*	{ Vector F-1 } minus { Vector DQ }

• We now makes these observations on the above table :
  • { Vector DA } and { Vector DB } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector DE } and { Vector DF } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector DC } intersects { Vector DA } / { Vector DB } / { Vector DE } / { Vector DF } at the angle of [ 120 degrees ] ;

  • { Vector DA } , { Vector DB } and { Vector DC } add up to [ zero ] ;

  • { Vector DC } , { Vector DE } and { Vector DF } add up to [ zero ] .

  As such :

  • { Vector DA } , { Vector DB } and { Vector DC } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

  • { Vector DC } , { Vector DE } and { Vector DF } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

• The 2 sets of { 4 Symmetric Lines in a 3-Dimension Vector Space } identified here are then :
  • { Line A / Line B / Line C / Line D } , and

  • { Line C / Line D / Line E / Line F } .

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Selecting { Line E } as the { Line-Of-Concern } :

• Let us now select { Line E } as our { Line-Of-Concern } for the time-being ,
  • using { Vector E-1 } as the primary vector for our analysis .

  We then bring in the 5 vectors intersecting { Vector E-1 } at the angle of [ 70.53 degrees ] ,

  • i.e. : { Vector A-1 } , { Vector B-2 } , { Vector C-1 } , { Vector D-1 } , and { Vector F-2 } , respectively ;
    • as shown in the table below .

• We then follow the same analystic procedure as in { Line A } above :
  • to arrive at the projections in the { 3-D sub-space } ,
    • as shown below .

  Line I.D.	* * *	Vector for the Line							* * *	Comments
  	*	Vector I.D.	Co-ordinates						*
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line E	*	Vector E-1	-3	+3	+3	-3	+3	+3	*	Line-Of-Concern
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line A	*	Vector A-1	+3	+3	+3	+3	+3	+3	*	intersects { Vector E-1 } at [ 70.53 degrees ]
  Line B	*	Vector B-2	-3	+3	-3	-3	+3	-3	*	intersects { Vector E-1 } at [ 70.53 degrees ]
  Line C	*	Vector C-1	+3	+3	-3	-3	+3	+3	*	intersects { Vector E-1 } at [ 70.53 degrees ]
  Line D	*	Vector D-1	-3	+3	+3	+3	+3	-3	*	intersects { Vector E-1 } at [ 70.53 degrees ]
  Line F	*	Vector F-2	-3	-3	+3	-3	-3	+3	*	intersects { Vector E-1 } at [ 70.53 degrees ]
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  1/3 x Line-Of-Concern	*	Vector EQ	-1	+1	+1	-1	+1	+1	*	projection in { Vector E-1 line-direction }
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Projection of Line A	*	Vector EA	+4	+2	+2	+4	+2	+2	*	{ Vector A-1 } minus { Vector EQ }
  Projection of Line B	*	Vector EB	-2	+2	-4	-2	+2	-4	*	{ Vector B-2 } minus { Vector EQ }
  Projection of Line C	*	Vector EC	+4	+2	-4	-2	+2	+2	*	{ Vector C-1 } minus { Vector EQ }
  Projection of Line D	*	Vector ED	-2	+2	+2	+4	+2	-4	*	{ Vector D-1 } minus { Vector EQ }
  Projection of Line F	*	Vector EF	-2	-4	+2	-2	-4	+2	*	{ Vector F-2 } minus { Vector EQ }

• We now makes these observations on the above table :
  • { Vector EA } and { Vector EB } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector EC } and { Vector ED } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector EF } intersects { Vector EA } / { Vector EB } / { Vector EC } / { Vector ED } at the angle of [ 120 degrees ] ;

  • { Vector EA } , { Vector EB } and { Vector EF } add up to [ zero ] ;

  • { Vector EC } , { Vector ED } and { Vector EF } add up to [ zero ] .

  As such :

  • { Vector EA } , { Vector EB } and { Vector EF } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

  • { Vector EC } , { Vector ED } and { Vector EF } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

• The 2 sets of { 4 Symmetric Lines in a 3-Dimension Vector Space } identified here are then :
  • { Line A / Line B / Line E / Line F } , and

  • { Line C / Line D / Line E / Line F } .

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Selecting { Line F } as the { Line-Of-Concern } :

• Let us now select { Line F } as our { Line-Of-Concern } for the time-being ,
  • using { Vector F-1 } as the primary vector for our analysis .

  We then bring in the 5 vectors intersecting { Vector F-1 } at the angle of [ 70.53 degrees ] ,

  • i.e. : { Vector A-1 } , { Vector B-2 } , { Vector C-1 } , { Vector D-1 } , and { Vector E-2 } , respectively ;
    • as shown in the table below .

• We then follow the same analystic procedure as in { Line A } above :
  • to arrive at the projections in the { 3-D sub-space } ,
    • as shown below .

  Line I.D.	* * *	Vector for the Line							* * *	Comments
  	*	Vector I.D.	Co-ordinates						*
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line F	*	Vector F-1	+3	+3	-3	+3	+3	-3	*	Line-Of-Concern
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Line A	*	Vector A-1	+3	+3	+3	+3	+3	+3	*	intersects { Vector F-1 } at [ 70.53 degrees ]
  Line B	*	Vector B-2	-3	+3	-3	-3	+3	-3	*	intersects { Vector F-1 } at [ 70.53 degrees ]
  Line C	*	Vector C-1	+3	+3	-3	-3	+3	+3	*	intersects { Vector F-1 } at [ 70.53 degrees ]
  Line D	*	Vector D-1	-3	+3	+3	+3	+3	-3	*	intersects { Vector F-1 } at [ 70.53 degrees ]
  Line E	*	Vector E-2	+3	-3	-3	+3	-3	-3	*	intersects { Vector F-1 } at [ 70.53 degrees ]
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  1/3 x Line-Of-Concern	*	Vector FQ	+1	+1	-1	+1	+1	-1	*	projection in { Vector F-1 line-direction }
  ********************	*	***********	***	***	***	***	***	***	*	**********************************
  Projection of Line A	*	Vector FA	+2	+2	+4	+2	+2	+4	*	{ Vector A-1 } minus { Vector FQ }
  Projection of Line B	*	Vector FB	-4	+2	-2	-4	+2	-2	*	{ Vector B-2 } minus { Vector FQ }
  Projection of Line C	*	Vector FC	+2	+2	-2	-4	+2	+4	*	{ Vector C-1 } minus { Vector FQ }
  Projection of Line D	*	Vector FD	-4	+2	+4	+2	+2	-2	*	{ Vector D-1 } minus { Vector FQ }
  Projection of Line E	*	Vector FE	+2	-4	-2	+2	-4	-2	*	{ Vector E-2 } minus { Vector FQ }

• We now makes these observations on the above table :
  • { Vector FA } and { Vector FB } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector FC } and { Vector FD } intersects one-another at an angle of [ 120 degrees ] ;

  • { Vector FE } intersects { Vector FA } / { Vector FB } / { Vector FC } / { Vector FD } at the angle of [ 120 degrees ] ;

  • { Vector FA } , { Vector FB } and { Vector FE } add up to [ zero ] ;

  • { Vector FC } , { Vector FD } and { Vector FE } add up to [ zero ] .

  As such :

  • { Vector FA } , { Vector FB } and { Vector FE } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

  • { Vector FC } , { Vector FD } and { Vector FE } are 3 vectors lying in a [ 2-Dimension Plane ] intersecting one-another at [ 120 degrees ] ;

• The 2 sets of { 4 Symmetric Lines in a 3-Dimension Vector Space } identified here are then :
  • { Line A / Line B / Line E / Line F } , and

  • { Line C / Line D / Line E / Line F } .

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Summarizing our Findings :

• For the { 6 Symmetric Lines in 4-Dimensions } , we were able to identify only :
  • three (3) sets of { 4 Lines in a 3-Dimension Vector Space intersecting one-another at an angle of [ 70.53 degrees ] } .

  And the 3 sets are :

  • { Line A / Line B / Line C / Line D } ,

  • { Line A / Line B / Line E / Line F } , and

  • { Line C / Line D / Line E / Line F } .

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The Number of Choices :

• Based on the above findings :
  • there is only a [ choice of 3 alternatives ]
    • when we go from the { 6 Symmetric Lines in 4-D } to the { 4 Symmetric Lines in 3-D } situation .

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A Special Catch on playing domino's :

• Let us recall that the { 8 nodal-points } of the cube can be split into 2 sets of { 4 nodal-points } ,
  • with each { set of 4 nodal-points } being in the formation of a tetrahedron .

• Based on our analysis above , let us now identify the 2 tetrahedra associated with each set of 4 Symmetric Lines :
  • for { Lines A / B / C / D } , the 2 tetrahedra are :
    • { Vectors A2 / B1 / C1 / D1 } and { Vectors A1 / B2 / C2 / D2 } .

  • for { Lines A / B / E / F } , the 2 tetrahedra are :
    • { Vectors A2 / B1 / E1 / F1 } and { Vectors A1 / B2 / E2 / F2 } .

  • for { Lines E / F / E / F } , the 2 tetrahedra are :
    • { Vectors C1 / D1 / E2 / F2 } and { Vectors C2 / D2 / E1 / F1 } .

• Thus , when playing domino's with the 6 tetrahedron-pieces above ,
  • the circular-loop sequence would be :
    • A2 / B1 / C1 / D1 ,
    • C1 / D1 / E2 / F2 ,
    • E2 / F2 / A1 / B2 ,
    • A1 / B2 / C2 / D2 ,
    • C2 / D2 / E1 / F1 ,
    • E1 / F1 / A2 / B1 .

  Mobius Strip ?

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What's Next :

• We shall take a quick look at Orthogonal Lines within the { 6-D Quasi-Cube } framework , next ;
  • with interesting results .

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go to the next Section : Section XXVIII - Orthogonal Lines arising from the { 6-D Quasi-Cube }

go to the last Section : Section XXVI - Identifying { 6 Symmetric Lines in 4-D }

return to the Symmetry Through-put HomePage

Original dated 2017-7-23