On { N+1 Symmetric Lines } in a { N-Dimension Vector Space }

by Frank C. Fung ( 1st published in March, 2016. )

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Section VII - { Symmetric Lines in 4-D } vs. { Symmetric Lines in 3-D }

Summary for the Section :

• In this Section , we shall do a demonstration on the relation :
  • between :
    • the { 5 Symmetric Lines } in a { 4-Dimension Vector Space }

    and

    • the { 4 Symmetric Lines } in a { 3-Dimension Vector Space } ;

  in preparation of our statement on :

  • the Hierarchy of Symmetric Lines across { Spaces with different Number of Dimensions } ,

    in the next Section .

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Recalling the { 5 Symmetric Lines } in a { 4-Dimension Space } :

• Let us recall that , for the { 4-Dimension Vector Space } ,
  • we first set up a { 5-Dimension Vector Space } defined by the 5 { unit vectors } :
    • { 1, 0, 0, 0, 0 }
    • { 0, 1, 0, 0, 0 }
    • { 0, 0, 1, 0, 0 }
    • { 0, 0, 0, 1, 0 }
    • { 0, 0, 0, 0, 1 } .

  We then created a { 4-Dimension Sub-Space } containing :

  • all vectors orthogonal to the Vector { 1, 1, 1, 1, 1 } .

• We then created the following 5 pairs of vectors :
  • 1st pair :
    • { Vector A1 } = { -1, -1, -1, -1, +4 }
    • { Vector A2 } = { +1, +1, +1, +1, -4 }

  • 2nd pair :
    • { Vector B1 } = { -1, -1, -1, +4, -1 }
    • { Vector B2 } = { +1, +1, +1, -4, +1 }

  • 3rd pair :
    • { Vector C1 } = { -1, -1, +4, -1, -1 }
    • { Vector C2 } = { +1, +1, -4, +1, +1 }

  • 4th pair :
    • { Vector D1 } = { -1, +4, -1, -1, -1 }
    • { Vector D2 } = { +1, -4, +1, +1, +1 }

  • 5th pair :
    • { Vector E1 } = { +4, -1, -1, -1, -1 }
    • { Vector E2 } = { -4, +1, +1, +1, +1 }

• And we note , here again , that :
  • First ,
    • the size of the above 10 vectors are all-the-same , with the size here being given by :
      • Size of Vector = { Square-Root of [ 20 ] } .

  • Secondly :
    • the 5 Vectors { Vector A1 / B1 / C1 / D1 / E1 } do intersect one-another at the angle of [ 104.48 degrees ] ,

      i.e. :

      • arccos( -1/4 ) = 104.48 degrees

        and

      • cosine of [ 104.48 degrees ] = -1/4 .

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Spelling-Out the Relation :

• Let us now take any one of { 5 Symmetric Lines } in the { 4-Dimension Space }
  • and mark it off as our [ Line-of-Concern ] for the time-being .

  And we take note here that :

  • all vectors orthogonal to the [ Line-of-Concern ] would then lie in a { 3-Dimension Sub-Space } .

• Let us now determine as to whether :
  • the projection of the [ 4 remaining lines ] onto this { 3-Dimension Sub-Space } is indeed symmetric .

  And in fact it is ,

  • as per our demonstration below .

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Doing the Demonstration :

• For our demonstration , let us first set up the quantity [ Q ] :
  • [ Q ] = { square-root of [ 20 ] } ,

    with [ Q ] here being the size of { Vectors A1 / A2 / B1 / B2 / C1 / C2 / D1 / D2 / E1 / E2 }

    • noting here again that all 10 vectors are of-the-same-size .

• Let us now select the [ A1-A2 direction ] as our { Line-of-Concern } ,
  • notwithstanding that :
    • had we selected any one of the other 4 Symmetric Lines as our { Line-of-Concern } ,
      • the general flow of our Demonstration here would not have changed .

  We then set up a unit-vector corresponding to { Vector A1 } , naming it { Unit Vector UA } ,

  i.e. :

  • { Unit Vector UA } = { -1/Q, -1/Q, -1/Q, -1/Q, +4/Q }

    corresponding to :

    { Vector A1 } = { -1, -1, -1, -1, +4 }

• Let us now establish :
  • the component of { Vector B1 } in the [ A1-A2 direction ] , naming it { Vector FB } ,

    i.e. :

    • { Vector FB } = [ size of Vector B1 ] x [ cosine of 104.48 degrees ] x { Unit Vector UA }

  • the component of { Vector C1 } in the [ A1-A2 direction ] , naming it { Vector FC } ,

    i.e.:

    • { Vector FC } = [ size of Vector C1 ] x [ cosine of 104.48 degrees ] x { Unit Vector UA }

  • the component of { Vector D1 } in the [ A1-A2 direction ] , naming it { Vector FD } ,

    i.e. :

    • { Vector FD } = [ size of Vector D1 ] x [ cosine of 104.48 degrees ] x { Unit Vector UA }

  • the component of { Vector E1 } in the [ A1-A2 direction ] , naming it { Vector FE } ,

    i.e. :

    • { Vector FE } = [ size of Vector E1 ] x [ cosine of 104.48 degrees ] x { Unit Vector UA }

• Again we note here that :
  • the size of { Vectors B1 / C1 / D1 / E1 } is [ Q ] , being the { square-root of [ 20 ] } , and

  • { cosine of 104.48 degrees } is equal to [ -1/4 ] .

  Consequently , upon further evaluations , we have :

  • { Vector FB } = { +1/4, +1/4, +1/4, +1/4, -1 }

  • { Vector FC } = { +1/4, +1/4, +1/4, +1/4, -1 }

  • { Vector FD } = { +1/4, +1/4, +1/4, +1/4, -1 }

  • { Vector FE } = { +1/4, +1/4, +1/4, +1/4, -1 }

• And with this in-hand , we can now proceed to set up :
  • the projections of the 4 { Vectors B1 / C1 / D1 / E1 } onto the { 3-Dimension Sub-space } .

  We then have these for the above-said projections :

  • Projection of { Vector B1 } onto the { 3-Dimension Sub-Space } , naming it { Vector PB } :
    • { Vector PB }= { Vector B1 } minus { Vector FB }

      yielding :

      { Vector PB } = { -5/4, -5/4, -5/4, +15/4, 0 }

  • Projection of { Vector C1 } onto the { 3-Dimension Sub-Space } , naming it { Vector PC } :
    • { Vector PC }= { Vector C1 } minus { Vector FC }

      yielding :

      { Vector PC } = { -5/4, -5/4, +15/4, -5/4, 0 }

  • Projection of { Vector D1 } onto the { 3-Dimension Sub-Space } , naming it { Vector PD } :
    • { Vector PD }= { Vector D1 } minus { Vector FD }

      yielding :

      { Vector PD } = { -5/4, +15/4, -5/4, -5/4, 0 }

  • Projection of { Vector E1 } onto the { 3-Dimension Sub-Space } , naming it { Vector PE } :
    • { Vector PE }= { Vector E1 } minus { Vector FE }

      yielding :

      { Vector PE } = { +15/4, -5/4, -5/4, -5/4, 0 }

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Summarizing the Results :

• Let us now bring-in , from above , the four (4) projected vectors in the { 3-Dimension Sub-Space } :
  • { Vector PB } = { -5/4, -5/4, -5/4, +15/4, 0 }
  • { Vector PC } = { -5/4, -5/4, +15/4, -5/4, 0 }
  • { Vector PD } = { -5/4, +15/4, -5/4, -5/4, 0 }
  • { Vector PE } = { +15/4, -5/4, -5/4, -5/4, 0 }

  We take note here :

  • First-of-all ,
    • the 4 Vectors above are orthogonal to the Vector { 1, 1, 1, 1, 1 } ;

    • the 4 Vectors above are orthogonal to the { Vector PA } being the vector { - 1, -1, -1, -1, +4 } ; and

    • the Vector { 1, 1, 1, 1, 1 } and the Vector { -1, -1, -1, -1, +4 } are orthogonal to one-another ;

      so that :

      • we can now confirm that the 4 Vectors above do lie in a { 3-Dimension Space } .

  • Secondly ,
    • the 4 Vectors do intersect one-another at an angle of [ 109.47 degrees ] ,

      i.e. :

      • arccos( -1/3 ) = 109.47 degrees

        and

      • cosine of [ 109.47 degrees ] = -1/3

• Consequently , we can now set up these 4 pairs of vectors :
  • 1st Pair :
    • { -5/4, -5,4, -5/4, +15/4, 0 } and { +5/4, +5/4, +5/4, -15/4, 0 }

  • 2nd Pair :
    • { -5/4, -5,4, +15/4, -5/4, 0 } and { +5/4, +5/4, -15/4, +5/4, 0 }

  • 3rd Pair :
    • { -5/4, +15,4, -5/4, -5/4, 0 } and { +5/4, -15/4, +5/4, +5/4, 0 }

  • 4th Pair :
    • { +15/4, -5,4, -5/4, -5/4, 0 } and { -15/4, +5/4, +5/4, +5/4, 0 }

  And joining the end-points of these 4 pairs of Vectors would then yield us :

  • the { 4 Symmetric Lines } in a { 3 Dimension Space } ;

    that we were looking for .

• And the angle of intersection here is :
  • the { 109.47 degrees / 70.53 degrees combination } ;

    i.e. :

    • 109.47 degrees + 70.53 degrees = 180 degrees

    noting here that when 2 non-orthogonal lines interesects :

    • we can cite either the [ acute angle ] or the [ obtuse angle ] as the [ angle of intersection ] .

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What's next :

• We are now ready to make a general statement, on :
  • the Hierarchy of the { N+1 Symmetric Lines System } .

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go to the next Section : Section VIII - The Hierarchy of the { N+1 Symmetric Lines System }

go to the last Section : Section VI - { Symmetric Lines } in { Higher-Dimensioned Spaces }

return to the { N+1 Symmetric Lines } HomePage

Original dated 2016-3-18