Three Dimensional Geometry

Find the shortest distance between the lines: and - Wired Faculty

Question

Find the shortest distance between the lines:
$straight r with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 15 space straight j with hat on top space plus space 9 space straight k with hat on top space plus space straight lambda left parenthesis 2 space straight i with hat on top space minus space 7 space straight j with hat on top space plus space 5 space straight k with hat on top right parenthesis$
and $straight r with rightwards arrow on top space equals space left parenthesis 2 space straight mu minus 1 right parenthesis space straight i with hat on top space plus space left parenthesis 1 plus straight mu right parenthesis space straight j with hat on top space plus space left parenthesis 9 minus 3 space straight mu right parenthesis space straight k with hat on top$

Solution

The equations of two lines are
$straight r with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 15 space straight j with hat on top space plus space 9 space straight k with hat on top space plus space straight lambda left parenthesis 2 space straight i with hat on top space minus space 7 space straight j with hat on top space plus space 5 space straight k with hat on top right parenthesis$
and $straight r with rightwards arrow on top space equals space left parenthesis 2 straight mu minus 1 right parenthesis thin space straight i with hat on top space plus space left parenthesis 1 plus straight mu right parenthesis space straight j with hat on top space plus space left parenthesis 9 minus 3 space straight mu right parenthesis space straight k with hat on top$
or    $straight r with rightwards arrow on top space equals space minus straight i with hat on top space plus space straight j with hat on top space plus space 9 space straight k with hat on top space plus space straight mu open parentheses 2 space straight i with hat on top space plus space straight j with hat on top space minus space 3 space straight k with hat on top close parentheses$
Comparing these equations with
$straight r with rightwards arrow on top space equals space stack straight a subscript 1 with rightwards arrow on top space plus space straight lambda space stack straight b subscript 1 with rightwards arrow on top space space and space space space straight r with rightwards arrow on top space space equals space stack straight a subscript 2 with rightwards arrow on top space plus space straight mu space stack straight b subscript 2 with rightwards arrow on top comma space we space get comma$
$stack straight a subscript 1 with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 15 space straight j with hat on top space plus space 9 space straight k with hat on top comma space space stack straight b subscript 1 with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 7 space straight j with hat on top space plus space space 5 space straight k with hat on top$
$stack straight a subscript 2 with rightwards arrow on top space space minus space stack straight a subscript 1 with rightwards arrow on top space equals space minus 4 space straight i with hat on top space plus space 16 space straight j with hat on top$
Now, $stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top space equals space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 cell negative 7 end cell 5 row 2 1 cell negative 3 end cell end table close vertical bar space equals space open vertical bar table row cell negative 7 end cell 5 row 1 cell negative 3 end cell end table close vertical bar straight i with hat on top space minus space open vertical bar table row 2 5 row 2 cell negative 3 end cell end table close vertical bar straight j with hat on top space plus space open vertical bar table row 2 cell negative 7 end cell row 2 1 end table close vertical bar straight k with hat on top$
   $equals space left parenthesis 21 minus 5 right parenthesis space straight i with hat on top space minus space left parenthesis negative 6 minus 10 right parenthesis space straight j with hat on top space plus space left parenthesis 2 plus 14 right parenthesis space straight k with hat on top equals space 16 space straight i with hat on top space plus space 16 space straight j with hat on top space plus space 16 space straight k with hat on top space equals space 16 space open parentheses straight i with hat on top plus space straight j with hat on top space plus space straight k with hat on top close parentheses$
$therefore space space space space space space space open vertical bar stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top close vertical bar space equals space 16 square root of 1 plus 1 plus 1 end root space equals space 16 square root of 3$
S.D. = $open vertical bar fraction numerator left parenthesis stack straight a subscript 2 with rightwards arrow on top minus stack straight a subscript 1 with rightwards arrow on top right parenthesis thin space left parenthesis stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top right parenthesis over denominator open vertical bar stack straight b subscript 1 with rightwards arrow on top cross times stack straight b subscript 2 with rightwards arrow on top close vertical bar end fraction close vertical bar space equals space space open vertical bar fraction numerator left parenthesis negative 4 space straight i with hat on top space plus space 1 space 6 space straight j with hat on top right parenthesis. space 16 space left parenthesis straight i with hat on top space plus straight j with hat on top space plus space straight k with hat on top right parenthesis over denominator 16 square root of 3 end fraction close vertical bar$
   $equals space fraction numerator 1 over denominator square root of 3 end fraction open square brackets left parenthesis negative 4 right parenthesis space left parenthesis 1 right parenthesis space plus space left parenthesis 16 right parenthesis thin space left parenthesis 1 right parenthesis space plus space open parentheses 0 close parentheses space left parenthesis 1 right parenthesis close square brackets space equals space fraction numerator 1 over denominator square root of 3 end fraction left parenthesis negative 4 plus 16 plus 0 right parenthesis equals space fraction numerator 12 over denominator square root of 3 end fraction space equals 4 square root of 3$

For Daily Free Study Material Join wiredfaculty