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Three Dimensional Geometry

Question
CBSEENMA12033338

Find the shortest distance between the lines
  straight r with rightwards arrow on top space equals space left parenthesis straight lambda minus 1 right parenthesis space straight i with hat on top space plus space left parenthesis straight lambda plus 1 right parenthesis space straight j with hat on top space minus space left parenthesis straight lambda plus 1 right parenthesis space straight k with hat on top
and straight r with rightwards arrow on top space equals space left parenthesis 1 minus straight mu right parenthesis space straight i with hat on top space plus space left parenthesis 2 straight mu minus 1 right parenthesis space straight j with hat on top space plus space left parenthesis straight mu space plus space 2 right parenthesis space straight k with hat on top
                                     
      



Solution
The equations of given lines are
          straight r with rightwards arrow on top space equals space left parenthesis straight lambda minus 1 right parenthesis space straight i with hat on top space plus space left parenthesis straight lambda plus 1 right parenthesis space straight j with hat on top space minus space left parenthesis straight lambda plus 1 right parenthesis space straight k with hat on top
and    straight r with rightwards arrow on top space equals space left parenthesis 1 minus straight mu right parenthesis space straight i with hat on top space plus space left parenthesis 2 straight mu minus 1 right parenthesis space straight j with hat on top space plus space left parenthesis straight mu space plus space 2 right parenthesis space straight k with hat on top.
or       straight r with rightwards arrow on top space equals space minus space straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top space space plus space straight lambda left parenthesis straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top right parenthesis                                 ...(1)
and    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 2 space straight k with hat on top space plus space space straight mu left parenthesis negative straight i with hat on top space plus space 2 space straight j with hat on top space plus space straight k with hat on top right parenthesis                            ...(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 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 stack straight b subscript 2 with rightwards arrow on top comma
we get,
       stack straight a subscript 1 with rightwards arrow on top space equals space minus straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top comma space space space stack straight b subscript 1 with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top
stack straight a subscript 2 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 2 space straight k with hat on top comma space space space stack straight b subscript 1 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 straight k with hat on top
         
Let S be the point on the line (1) with position vector stack straight a subscript 1 with rightwards arrow on top and T be the point on the line (2) with position vector stack straight a subscript 2 with rightwards arrow on top comma 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 2 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top
Now, 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 1 1 cell negative 1 end cell row cell negative 1 end cell 2 1 end table close vertical bar space equals space left parenthesis 1 plus 2 right parenthesis space straight i with hat on top space minus space left parenthesis 1 minus 1 right parenthesis space straight j with hat on top space plus space left parenthesis 2 plus 1 right parenthesis space straight k with hat on top space equals space 3 space straight i with hat on top space plus space space 3 space straight k with hat on top
therefore            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 9 plus 0 plus 9 end root space equals space square root of 18 space equals space 3 space square root of 2
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 space 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 comma then 
              straight n with rightwards arrow on top space equals space fraction numerator stack straight b subscript 1 with rightwards arrow on top space cross times space 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 3 square root of 2 end fraction space left parenthesis 3 space straight i with hat on top space plus space 3 space straight k with hat on top right parenthesis space equals space fraction numerator 1 over denominator square root of 2 end fraction left parenthesis straight i with hat on top space plus space straight k with hat on top right parenthesis
Now, S.D. = Projection space of space 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
                 equals space ST with rightwards arrow on top. space straight n with rightwards arrow on top space equals space open parentheses 2 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space plus space fraction numerator 1 over denominator square root of 2 end fraction left parenthesis straight i with hat on top space plus space straight k with hat on top right parenthesis
                 equals space fraction numerator 1 over denominator square root of 2 end fraction left parenthesis 2 plus 0 plus 3 right parenthesis space equals space fraction numerator 5 over denominator square root of 2 end fraction space equals space fraction numerator 5 square root of 2 over denominator 2 end fraction space units. space