123.
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](/application/zrc/images/qvar/MAEN12064664.png)
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.](/application/zrc/images/qvar/MAEN12064664-1.png)
       Â
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The equations of the two lines are          Â
![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](/application/zrc/images/qvar/MAEN12064664-2.png)
          ...(1)
and       Â
![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](/application/zrc/images/qvar/MAEN12064664-3.png)
        ...(2)
Comparing these equations withÂ
![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](/application/zrc/images/qvar/MAEN12064664-4.png)
 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](/application/zrc/images/qvar/MAEN12064664-5.png)
Let S be the point on the line (1) with position vectorÂ
 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](/application/zrc/images/qvar/MAEN12064664-8.png)
![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](/application/zrc/images/qvar/MAEN12064664-9.png)
Now,Â
![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](/application/zrc/images/qvar/MAEN12064664-10.png)
        Â
![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](/application/zrc/images/qvar/MAEN12064664-11.png)
![therefore](/application/zrc/images/qvar/MAEN12064664-12.png)
     Â
![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](/application/zrc/images/qvar/MAEN12064664-13.png)
                Â
![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](/application/zrc/images/qvar/MAEN12064664-14.png)
![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](/application/zrc/images/qvar/MAEN12064664-15.png)
LetÂ
![PQ with rightwards arrow on top](/application/zrc/images/qvar/MAEN12064664-16.png)
 be the S.D. vector between given lines. Therefore, it is parallel toÂ
![stack straight b subscript 1 with rightwards arrow on top space cross times space stack straight b subscript 2 with rightwards arrow on top](/application/zrc/images/qvar/MAEN12064664-17.png)
.
IfÂ
![straight n with rightwards arrow on top](/application/zrc/images/qvar/MAEN12064664-18.png)
 is a vector alongÂ
![PQ with rightwards arrow on top comma](/application/zrc/images/qvar/MAEN12064664-19.png)
 then
       Â
![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](/application/zrc/images/qvar/MAEN12064664-20.png)
Now. S.D. = Projection ofÂ
![ST with rightwards arrow on top space on space PQ with rightwards arrow on top](/application/zrc/images/qvar/MAEN12064664-21.png)
 = Projection ofÂ
![ST with rightwards arrow on top space on space straight n with rightwards arrow on top](/application/zrc/images/qvar/MAEN12064664-22.png)
        Â
![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](/application/zrc/images/qvar/MAEN12064664-23.png)
         Â
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