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Three Dimensional Geometry

Question
CBSEENMA12033341

Find the shortest distance between the lines:
               straight r with rightwards arrow space on top equals space straight i with hat on top space plus space 2 space straight j with hat on top space plus space space 3 space straight k with hat on top space plus space straight lambda space open parentheses 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top close parentheses
and       straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space 4 space straight j with hat on top space plus space 5 space straight k with hat on top space plus space straight mu space open parentheses 3 space straight i with hat on top space plus space 4 space straight j with hat on top space plus space 5 space straight k with hat on top close parentheses.
               

      

Solution
The equations of the two lines are
                    straight r with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top space plus space straight lambda space open parentheses 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top close parentheses                   ...(1)
and              straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space 4 space straight j with hat on top space plus space 5 space straight k with hat on top space plus space straight mu space open parentheses 3 space straight i with hat on top space plus space 4 space straight j with hat on top space plus space 5 space straight k 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 space space and space 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 we get
stack straight a subscript 1 with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top comma space space space space space space stack straight b subscript 1 with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top
stack straight a subscript 2 with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space 4 space straight j with hat on top space plus space 5 space straight k with hat on top comma space stack straight b subscript 1 with rightwards arrow on top space equals space 3 space straight i with hat on top space plus space 4 space straight j with hat on top space plus space 5 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 space so space 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 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
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 2 3 4 row 3 4 5 end table close vertical bar space equals left parenthesis 15 minus 16 right parenthesis space straight i with hat on top space minus space left parenthesis 10 minus 12 right parenthesis space straight j with hat on top space plus space left parenthesis 8 minus 9 right parenthesis space straight k with hat on top
                 equals space minus straight i with hat on top space plus space 2 space straight j with hat on top space minus space straight k with hat on top
therefore          open vertical bar stack straight b subscript 1 with rightwards arrow on top cross times space stack straight b subscript 2 with rightwards arrow on top close vertical bar space equals space square root of left parenthesis negative 1 right parenthesis squared plus left parenthesis 2 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared end root space equals space square root of 6
                                equals space minus space straight i with hat on top space plus space 2 space straight j with hat on top space minus 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 left parenthesis negative 1 right parenthesis squared plus left parenthesis 2 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared end root space equals square root of 6
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 space stack straight b subscript 2 with rightwards arrow on top.
If straight n with rightwards arrow on top is a 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 square root of 6 end fraction left parenthesis negative straight i with hat on top space plus space 2 space straight j with hat on top space minus 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 = Projection of 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 straight n with rightwards arrow on top space equals space open parentheses 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 close parentheses. space fraction numerator 1 over denominator square root of 6 end fraction space open parentheses negative straight i with hat on top space plus space 2 space straight j with hat on top space minus space straight k with hat on top close parentheses
                  equals space fraction numerator 1 over denominator square root of 6 end fraction left parenthesis negative 1 plus 4 minus 2 right parenthesis space equals space fraction numerator 1 over denominator square root of 6 end fraction

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.