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Math Help - Vectors - shortest distance between two 3d lines

  1. #1
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    Vectors - shortest distance between two 3d lines

    'Show that the shortest distance between the line with equations

    (x + 4)/3 = (y - 3)/2 = (z + 6)/5

    and the line with equations x - 2y - z = 0, x - 10y - 3z = -7
    is 1/2 sqrt 14.'

    I rewrote the equations in vector form:

    -4i + 3j - 6k + s(3i + 2j + 5k)
    7i/2 + 7k/2 + t(-2i + J - 4k)

    I found the cross product to be: -13i + 2j + 7k
    (-15i/2 + 3j - 19k/2 ).(-13i + 2j + 7k) = 97.5 + 6 - 66.5 = 37
    |-13i + 2j + 7k| = sqrt 222
    so my answer is 37/sqrt 222.
    Can someone please tell me where I'm going wrong?
    Thanks.
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  2. #2
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    Re: Vectors - shortest distance between two 3d lines

    Quote Originally Posted by jimi View Post
    'Show that the shortest distance between the line with equations

    (x + 4)/3 = (y - 3)/2 = (z + 6)/5

    and the line with equations x - 2y - z = 0, x - 10y - 3z = -7
    is 1/2 sqrt 14.'
    If \ell _1: P+t\overrightarrow D ~\&~\ell _2: Q+t\overrightarrow E are two skew lines
    then the distance between them is d({\ell _1},{\ell _2}) = \frac{{\left| {\overrightarrow {PQ}  \cdot (\overrightarrow D  \times \overrightarrow E )} \right|}}{{\left\| {\overrightarrow D  \times \overrightarrow E } \right\|}}
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  3. #3
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    Re: Vectors - shortest distance between two 3d lines

    Plato

    Yes, that's the formula I'm trying to use, but I don't agree with their answer. Can you tell me where I'm going wrong?
    Thanks.
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