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Three Dimensional Geometry

Question
CBSEENMA12033349

Determine whether the following pair of lines intersect:
                                fraction numerator straight x minus 1 over denominator 2 end fraction space equals fraction numerator straight y plus 1 over denominator 3 end fraction space equals straight z

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


Solution
The equations of given lines are:
                          fraction numerator straight x minus 1 over denominator 2 end fraction space equals fraction numerator straight y plus 1 over denominator 3 end fraction space equals straight z comma space space fraction numerator straight x plus 1 over denominator 5 end fraction space equals space fraction numerator straight y minus 2 over denominator 1 end fraction semicolon space space straight z space equals space 2
or                   fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y plus 1 over denominator 3 end fraction space space equals fraction numerator straight z minus 0 over denominator 1 end fraction
and       fraction numerator straight x plus 1 over denominator 5 end fraction space equals space fraction numerator straight y minus 2 over denominator 1 end fraction space equals space fraction numerator straight z minus 2 over denominator 0 end fraction       
 
The equations of two lines are
                   straight r with rightwards arrow on top space equals space straight i with hat on top space minus space straight j with hat on top space plus space straight lambda open parentheses 2 straight i with hat on top space plus space 3 space straight j with hat on top space plus space straight k with hat on top close parentheses                 ...(1)
and             straight r with rightwards arrow on top space equals space minus straight i with hat on top space plus space 2 space straight j with hat on top space plus space 2 space straight k with hat on top space plus space straight mu open parentheses 5 space straight i with hat on top space plus space straight j with hat on top close parentheses        ...(2)
Comparing these equations with
straight r with rightwards arrow on top space equals space stack straight a subscript 1 with rightwards arrow on top space plus space straight lambda space stack straight b subscript 1 with rightwards arrow on top space and space straight r with rightwards arrow on top space equals space stack straight a subscript 2 with rightwards arrow on top space plus space straight mu space stack straight b subscript 2 with rightwards arrow on top comma space space we space get comma
stack straight a subscript 2 with rightwards arrow on top space equals space minus straight i with hat on top space plus space 2 space straight j with hat on top space plus space 2 space straight k with hat on top comma space space stack straight b subscript 2 with rightwards arrow on top space equals space 5 straight i with hat on top space plus space straight j with hat on top
Let S be points on line (1) with position vector stack straight a subscript 1 with rightwards arrow on top and T be point on line (2) with position vector stack straight alpha subscript 2 with rightwards arrow on top so that
                                 ST with rightwards arrow on top space equals space stack straight a subscript 2 with rightwards arrow on top space minus space stack straight a subscript 1 with rightwards arrow on top space equals space minus 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 2 space straight k with hat on top
stack straight b subscript 1 with rightwards arrow on top space cross times space stack straight b subscript 2 with rightwards arrow on top space equals space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 3 1 row 5 1 0 end table close vertical bar space equals space left parenthesis 0 minus 1 right parenthesis space straight i with hat on top space minus space left parenthesis 0 minus 5 right parenthesis space straight j with hat on top space plus space left parenthesis 2 minus 15 right parenthesis space straight k with hat on top space equals space minus straight i with hat on top space plus space 5 space straight j with hat on top space minus space 13 space straight k with hat on top
therefore space space space space space open vertical bar stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top close vertical bar space equals space square root of 1 plus 25 plus 169 end root space equals space square root of 195
Let PQ with rightwards arrow on top be the S.D. vector between given lines. Therefore, it is parallel to stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top.
If straight n with rightwards arrow on top is a unit vector along PQ with rightwards arrow on top,  then
                   straight n with rightwards arrow on top space equals space fraction numerator stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top over denominator open vertical bar stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator 1 over denominator square root of 195 end fraction left parenthesis negative straight i with hat on top space plus space 5 space straight j with hat on top space minus space 13 space space straight k with hat on top right parenthesis
Now S.D. = Projection of ST with rightwards arrow on top space on space PQ with rightwards arrow on top space equals space Projection space of space ST with rightwards arrow on top space on space straight n with rightwards arrow on top space equals space ST with rightwards arrow on top space. space straight n with rightwards arrow on top
                equals space open parentheses negative 2 straight i with hat on top space plus space 3 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses. space fraction numerator 1 over denominator square root of 195 end fraction left parenthesis negative space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 13 space straight k with hat on top right parenthesis
space equals space fraction numerator 1 over denominator square root of 195 end fraction left parenthesis 2 plus 15 minus 26 right parenthesis space equals negative fraction numerator 9 over denominator square root of 195 end fraction space equals space fraction numerator 9 over denominator square root of 195 end fraction left parenthesis in space magnitude right parenthesis space not equal to 0
∴ given lines do not intersect.