Question

Find the shortest distance between the lines whose vector equations are
$straight r with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space straight lambda left parenthesis 2 space straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top right parenthesis$
and $straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top space plus space straight mu space left parenthesis 3 straight i with hat on top space minus space 5 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis$

Solution

The equations of two lines are
   $straight r with rightwards arrow on top equals space straight i with hat on top space plus space straight j with hat on top space plus space straight lambda left parenthesis 2 straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top right parenthesis$ ...(1)
and    $straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top space plus space straight mu space left parenthesis 3 straight i with hat on top space minus space 5 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis$ ...(2)
Comparing these equations with
   $stack straight r subscript 1 with rightwards arrow on top space equals space stack straight a subscript 1 with rightwards arrow on top plus space straight lambda space stack straight b subscript 1 with rightwards arrow on top space space space and space space stack straight r subscript 2 with rightwards arrow on top space equals space stack straight a subscript 2 with rightwards arrow on top space plus space straight lambda space stack straight b subscript 2 with rightwards arrow on top space space space we space get comma stack straight a subscript 1 with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top comma space space space stack straight b subscript 1 with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top stack straight a subscript 2 with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top comma space space stack straight b subscript 2 with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 5 space straight j with hat on top space plus space 2 space straight k with hat on top$
Let S be point 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 $stack straight a subscript 2 with rightwards arrow on top$ so that
$ST with rightwards arrow on top space equals space stack straight a subscript 2 with rightwards arrow on top space minus space stack straight a subscript 1 with rightwards arrow on top space equals space straight i with hat on top space minus space straight k with hat on top$
Now, $stack straight b subscript 1 with rightwards arrow on top space cross times space stack straight b subscript 2 with rightwards arrow on top space equals space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 cell negative 1 end cell 1 row 3 cell negative 5 end cell 2 end table close vertical bar space equals space straight i with hat on top space left parenthesis negative 2 plus 5 right parenthesis space minus space straight j with hat on top space left parenthesis 4 minus 3 right parenthesis space plus space straight k with hat on top space left parenthesis negative 10 plus 3 right parenthesis space equals space 3 space straight i with hat on top space minus space straight j with hat on top space minus space 7 space straight k with hat on top$
$therefore$    $open vertical bar stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top close vertical bar space equals space square root of 9 plus 1 plus 49 end root space equals space square root of 59$
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 unit vector along $PQ with rightwards arrow on top$ , then
   $straight n with rightwards arrow on top space equals space fraction numerator stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top over denominator open vertical bar stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator 1 over denominator square root of 58 end fraction space left parenthesis 3 space straight i with hat on top space minus space straight j with hat on top space minus space 7 space stack straight k right parenthesis with hat on top$
Now. S.D. = Projection of $ST with rightwards arrow space on top space on space PQ with rightwards arrow on top space equals space Projection space of space ST with rightwards arrow on top space on space straight n with rightwards arrow on top space equals space ST with rightwards arrow on top. space straight n with rightwards arrow on top$
   $equals space left parenthesis straight i with hat on top space minus space straight k with hat on top right parenthesis. space fraction numerator 1 over denominator square root of 59 end fraction left parenthesis 3 space straight i with hat on top space minus space straight j with hat on top space minus space 7 space straight k with hat on top right parenthesis space equals space fraction numerator 1 over denominator square root of 59 end fraction left parenthesis 3 minus 0 plus 7 right parenthesis space equals space fraction numerator 10 over denominator square root of 59 end fraction$

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