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Three Dimensional Geometry

Question
CBSEENMA12033508

Show that the lines fraction numerator straight x minus 5 over denominator 4 end fraction space equals space fraction numerator straight y minus 7 over denominator 4 end fraction space equals space fraction numerator straight z plus 3 over denominator negative 5 end fraction and fraction numerator straight x minus 8 over denominator 7 end fraction space equals space fraction numerator straight y minus 4 over denominator 1 end fraction space equals space fraction numerator straight z minus 5 over denominator 3 end fraction intersect each other. 

Solution

The equations of lines are
                       fraction numerator straight x minus 5 over denominator 4 end fraction space equals space fraction numerator straight y minus 7 over denominator 4 end fraction space equals space fraction numerator straight z plus 3 over denominator negative 5 end fraction                    ...(1)
and                fraction numerator straight x minus 8 over denominator 7 end fraction space equals space fraction numerator straight y minus 4 over denominator 1 end fraction space equals space fraction numerator straight z minus 5 over denominator 3 end fraction                    ...(2)
Any point on line (1) is (4r + 5, 4r + 7, - 5r -3), It lies on line (2)
if   fraction numerator 4 straight r plus 5 minus 8 over denominator 7 end fraction space equals space fraction numerator 4 straight r plus 7 minus 4 over denominator 1 end fraction space equals fraction numerator negative 5 straight r minus 3 minus 5 over denominator 3 end fraction
i.e, if  fraction numerator 4 straight r minus 3 over denominator 7 end fraction space equals space fraction numerator 4 straight r plus 3 over denominator 1 end fraction space equals space fraction numerator negative 5 straight r minus 8 over denominator 3 end fraction                              ...(3)
From the first and second members,
fraction numerator 4 straight r minus 3 over denominator 7 end fraction space equals space fraction numerator 4 straight r plus 3 over denominator 1 end fraction space space space space space space space or space space 28 straight r plus 21 space equals 4 straight r minus 3
therefore space space space space space space 24 straight r space equals space minus 24 space space space space space space space rightwards double arrow space space space straight r space equals space minus 1
Substituting this value of r in (3), we get,
                 fraction numerator negative 4 minus 3 over denominator 7 end fraction space equals space fraction numerator negative 4 plus 3 over denominator 1 end fraction space equals space fraction numerator 5 minus 8 over denominator 3 end fraction
or  negative 1 space equals space minus 1 space equals space minus 1 comma space space space which space is space true
∴ the lines intersect at (– 4 + 5, –4 + 7, 5 – 3) i.e. (1,3,2).