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Three Dimensional Geometry

Question
CBSEENMA12033278

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.

Solution
The line passes through (5, 1, 6) and (3, 4, 1)
∴   the equation of line is
              fraction numerator straight x minus 5 over denominator 3 minus 5 end fraction space equals space fraction numerator straight y minus 1 over denominator 4 minus 1 end fraction space equals fraction numerator straight z minus 6 over denominator 1 minus 6 end fraction
or          fraction numerator straight x minus 5 over denominator negative 2 end fraction space equals space fraction numerator straight y minus 1 over denominator 3 end fraction space equals space fraction numerator straight z minus 6 over denominator negative 5 end fraction
or        fraction numerator straight x minus 5 over denominator 2 end fraction space equals space fraction numerator straight y minus 1 over denominator negative 3 end fraction space equals space fraction numerator straight z minus 6 over denominator 5 end fraction                                           ...(1)
It meets yz-plane where x = 0
  ∴     putting x = 0 in (1), we get,
fraction numerator 0 minus 5 over denominator 2 end fraction space equals space fraction numerator straight y minus 1 over denominator negative 3 end fraction space equals space fraction numerator straight z minus 6 over denominator 5 end fraction
therefore space space space space space space space space space space minus 5 over 2 space equals space fraction numerator straight y minus 1 over denominator negative 3 end fraction space equals space fraction numerator straight z minus 6 over denominator 5 end fraction
therefore space space space space space space straight y minus 1 space equals space 15 over 2 comma space space straight z minus 6 space equals space minus 25 over 2
therefore space space space space space space space space space space space space space straight y space equals space 17 over 2 comma space space straight z space equals space minus 13 over 2
therefore space space space space point space is space open parentheses 0 comma space 17 over 2 comma space minus 13 over 2 close parentheses.