Three Dimensional Geometry

Show that the lines are coplanar. - Wired Faculty

Question

Show that the lines $fraction numerator straight x plus 3 over denominator negative 3 end fraction space equals space fraction numerator straight y minus 1 over denominator 1 end fraction space equals space fraction numerator straight z minus 5 over denominator 5 end fraction space and space fraction numerator straight x plus 1 over denominator negative 1 end fraction space equals fraction numerator straight y minus 2 over denominator 2 end fraction space equals fraction numerator straight z minus 5 over denominator 5 end fraction$ are coplanar.

Solution

The given lines are

fraction numerator straight x plus 3 over denominator negative 3 end fraction space equals fraction numerator straight y minus 1 over denominator 1 end fraction space equals fraction numerator straight z minus 5 over denominator 5 end fraction space and space fraction numerator straight x plus 1 over denominator negative 1 end fraction space equals space fraction numerator straight y minus 2 over denominator 2 end fraction space equals space fraction numerator straight z minus 5 over denominator 5 end fraction

These equations can be written as

fraction numerator straight x minus left parenthesis negative 3 right parenthesis over denominator negative 3 end fraction space space equals fraction numerator straight y minus 1 over denominator 1 end fraction space equals fraction numerator straight z minus 5 over denominator 5 end fraction space and space fraction numerator straight x minus left parenthesis negative 1 right parenthesis over denominator negative 1 end fraction space equals space fraction numerator straight y minus 2 over denominator 2 end fraction space equals space fraction numerator straight z minus 5 over denominator 5 end fraction

∴ x₁ = –3, y₁= 1, z₁ = 5, a₁ = –3, b₁= 1, c₁ = 5
x₂ = –1, y₂ = 2, z₂ = 5, a₂ = –1 b₂ = 2, c₂ = 5
The given lines will be coplanar
if

open vertical bar table row cell straight x subscript 1 minus straight x subscript 1 end cell cell straight y subscript 2 minus straight y subscript 1 end cell cell straight z subscript 2 minus straight z subscript 1 end cell row cell straight a subscript 1 end cell cell straight b subscript 1 end cell cell straight c subscript 1 end cell row cell straight a subscript 2 end cell cell straight b subscript 2 end cell cell straight c subscript 2 end cell end table close vertical bar space equals space 0

i.e., if

open vertical bar table row 2 1 0 row cell negative 3 end cell 1 5 row cell negative 1 end cell 2 5 end table close vertical bar space equals space 0

i.e. if

2 space open vertical bar table row 1 5 row 2 5 end table close vertical bar minus 1 open vertical bar table row cell negative 3 end cell 5 row cell negative 1 end cell 5 end table close vertical bar space plus space 0 space open vertical bar table row cell negative 3 end cell 1 row cell negative 1 end cell 2 end table close vertical bar space equals space 0

i.e. if

2 space left parenthesis 5 minus 10 right parenthesis minus 1 space left parenthesis negative 15 plus 5 right parenthesis plus 0 space left parenthesis negative 6 plus 1 right parenthesis space equals space 0

i.e. if –10+10 + 0 = 0
i.e. if 0 = 0, which is true
∴ given lines are coplanar.