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p6 Vectors Shortest distance (question26) watch

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    Page 141 qu.26 P6

    Show shortest distance between line with equation
    (x+4)/3=(y-3)/2=(z+6)/5 and line with equations:x-2y-z=0 and x-10y-3z=-7 is a half(root14)


    I get through the working but get the wrong answer it seems. Can't find mistake in my working...
    Please help.
    All the best
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    Oh by the way i got for Line 1: r= -4i+3j-6k+Lamda(3i+2j+5k)
    FOr line 2: -3.5i+3.5k+myoo(-2i+j-4k) lol

    If you find which of these is wrong if any, then it's easier to find the mistake.

    Shortest distance is
    (a-c).bxd / l bxd l all in modulus sign.

    WHere r1= a+lamda(b)
    r2=c+myoo(d)
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    We have:
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \frac{x+4}{3}=\frac{y-3}{2}=\frac{z+6}{5}=\lambda\\

    x=3\lambda-4,y=2\lambda+3,z=5\lambda-6\\

    r_{1}=-4i+3j-6k+\lambda [3i+2j+5k]\\



    x-2y-z=0,x-10y-3z=-7\\

    x-2y-z=0,-x+10y+3z=7\\

    8y+2z=7\\

    z=\frac{7-8y}{2}\\

    3x-6y-3z=0,-x+10y+3z=7\\

    2x+4y=7\\

    x=\frac{7-4y}{2}\\

    y=\frac{2z-7}{-8}=\frac{2x-7}{-4}=t\\

    x=\frac{-4t+7}{2},y=t,z=\frac{-8t+7}{2}\\

    x=3.5-2t,y=t,z=3.5-4t\\

    r_{2}=3.5i+0j+3.5k+t[-2i+j-4k]\\

    \textit{Shortest distance} = \left| \frac{(a-c).b\times d}{|b\times d|} \right|\\

    a-c = -7.5i+3j-9.5k.\\

    b\times d=

    \left|

    \begin{array} {c c c}

    i & j & k\\

    3 & 2 & 5\\

    -2 & 1 & -4

    \end{array}

    \right|

    =i(-13)-j(-2)+k(7)\\

    |b\times d|=\sqrt{222}\\

    \textit{shortest distance} = \frac{(-7.5i+3j-9.5k).(-13i+2j+7k)}{\sqrt{222}}\\

    =\frac{37}{\sqrt{222}\\

    =\frac{37\sqrt{222}}{222}\\

    =\frac{1}{6}\sqrt{222}

    The book gives the answer as \frac{1}{2}\sqrt{14}. Anyone else?
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    how do you in GENERAL find the shortest distance between two skew lines? (time is running out!!!)
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    (Original post by lesser weevil)
    how do you in GENERAL find the shortest distance between two skew lines? (time is running out!!!)
    The distance between r=a+tb and r=c+\lambda d is given by \left| \frac{(a-c).(b\times d)}{|b\times d|} \right|
    If you're that way inclined there's a nice fat proof in your P6 book but it's not really all that elegant :rolleyes:
    Can you spot my mistake?
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    no i cant. in the state i am at the moment i find it hard enough to remember stuff without mistakes let alone finding other people's mistakes...
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    54/root222 Is the answer i get... SO is the book wrong? And it seems i'm wrong somewhere for getting 54 and not 37?
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    (Original post by sjgarson)
    54/root222 Is the answer i get... SO is the book wrong? And it seems i'm wrong somewhere for getting 54 and not 37?
    I think it was one of the signs in the second line where mine differed from yours. The book has been wrong in the past.
    Would anyone else care to check this?
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    (Original post by Gaz031)
    I think it was one of the signs in the second line where mine differed from yours. The book has been wrong in the past.
    Would anyone else care to check this?
    If I understood it, I would :/
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    (Original post by Gaz031)
    I think it was one of the signs in the second line where mine differed from yours. The book has been wrong in the past.
    Would anyone else care to check this?
    Looking through your working I got the same
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    (Original post by lgs98jonee)
    If I understood it, I would :/
    He's finding equations for both lines in the form r=a+λb, and r=c+µd.
    Then using the shortest distance relationship to find the shortest distance.

    And I couldn't see any mistake either.
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    i have x= -2myoo-3.5 by accident... Should be a +3.5 there... Mistake in simple old BODMAS as usual...

    I still think book is wrong.
 
 
 
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