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Three Dimensional Geometry

Question
CBSEENMA12033506

Show that the lines:
fraction numerator straight x minus 1 over denominator 2 end fraction space equals fraction numerator straight y minus 2 over denominator 3 end fraction space equals space fraction numerator straight z minus 3 over denominator 4 end fraction space and space fraction numerator straight x minus 4 over denominator 5 end fraction space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals straight z intersect. Find the point of intersection also. 

Solution
The equations of lines are
                   fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y minus 2 over denominator 3 end fraction space equals fraction numerator straight z minus 3 over denominator 4 end fraction                                             ...(1)
and           fraction numerator straight x minus 4 over denominator 5 end fraction equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 0 over denominator 1 end fraction                                                ...(2)
Any point on line (1) is (2r + 1, 3r + 2, 4r + 3)
It lies on the line (2), if  fraction numerator 2 straight r plus 1 minus 4 over denominator 5 end fraction space equals fraction numerator 3 straight r plus 2 minus 1 over denominator 2 end fraction space equals fraction numerator 4 straight r plus 3 over denominator 1 end fraction
i.e.,  if   fraction numerator 2 straight r minus 3 over denominator 5 end fraction space equals space fraction numerator 3 straight r plus 1 over denominator 2 end fraction space equals space fraction numerator 4 straight r plus 3 over denominator 1 end fraction                                          ...(3)
Taking fraction numerator 2 straight r minus 3 over denominator 5 end fraction space equals space fraction numerator 3 straight r plus 1 over denominator 2 end fraction comma space space we space get comma
                  15 straight r plus 5 space equals space 4 straight r minus 6 comma space space space space space space space space space space space space space space space space space space space therefore space space space space 11 straight r space equals space minus 11 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 2 minus 3 over denominator 5 end fraction space equals space fraction numerator negative 3 plus 1 over denominator 2 end fraction space equals space fraction numerator negative 4 plus 3 over denominator 1 end fraction
or               negative 1 space equals space minus 1 space space equals space minus 1 comma space space space which space is space true. space
∴  the lines intersect and the point of intersection is
(–2 + 1, –3 + 2, –4 + 3), i.e. (–1, –1, –1).