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Math Help - Lines in a Plane

  1. #1
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    Lines in a Plane

    The distance between the skew lines r=OP + t d_1 and r= OQ + s d_2 is |Proj(PQ onto n)| or |PQ . n|/|n| wherer n=d_1 x d_2.

    Find the distance between the lines x=6, y=-4-t, z=t, and x = -2s, y=5, z=3+s.

    Bolded = vectors (arrow on top)

    My answer is as follows:

    x=6
    y=-4-t
    z=t

    therefore in parametric eqtn form
    (x-6)/0
    (y+4)/-1
    z/1
    d_1 = (0,-1,1)

    x/-2
    (y-5)/0
    (z-3)/1
    d_2=(-2,0,1)

    PQ=d=(-2,1,0)

    (-2,1,0) . (-1,0,-2)
    -------------------
    sqrt5

    final answer
    1/sqrt5

    is this correct??
    please let me know.
    thanks!
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  2. #2
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    Does anyone know how to do this question=?
    Last edited by skeske1234; August 30th 2009 at 09:12 AM.
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  3. #3
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    ..?
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  4. #4
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    DO NOT BUMP. It is rude.

    Suppose that + tD\;\& \;\ell _2 :q + tE" alt="\ell _1 + tD\;\& \;\ell _2 :q + tE" /> are two skew lines.
    The distance between them is given by: D\left( {\ell _1 ,\ell _2 } \right) = \frac{{\left| {\overrightarrow {pq}  \cdot \left( {D \times E} \right)} \right|}}{{\left\| {\left( {D \times E} \right)} \right\|}}
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  5. #5
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    ok, sorry about that Plato

    From my work, i think it looks like I have vector PQ wrong only. Is that correct? Here let me try that again:

    l1 P (6,-4,0)
    l2 P(0, 5,3)
    so, PQ would be
    (-6,9,-3)

    D=(-6,9,-3) . (-1,0,-2)
    ----------------------
    sqrt5
    =6+5
    -----
    sqrt5
    =11/sqrt5

    is this correct now? thanks for your help by the way
    Last edited by skeske1234; August 30th 2009 at 10:42 AM.
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  6. #6
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    From the given: 6,-4,0) + t\left\langle {0, - 1,1} \right\rangle \;\& \;\ell _2 0,5,3) + t\left\langle { - 2,0,1} \right\rangle " alt="\ell _1 6,-4,0) + t\left\langle {0, - 1,1} \right\rangle \;\& \;\ell _2 0,5,3) + t\left\langle { - 2,0,1} \right\rangle " />.

    So, p = (6,-4,0)\;,\,D = \left\langle {0, - 1,1} \right\rangle \;,\,q = (0,5,3)\;\& \,E = \left\langle { - 2,0,1} \right\rangle .

    What you have above is incorrect: \overrightarrow {pq}  = \left\langle { - 6,9,3} \right\rangle .

    Now all you need is \left( {D \times E} \right) = ?

    EDIT: Correct
    Last edited by Plato; August 30th 2009 at 11:58 AM.
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  7. #7
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    how did you get
    p(6,4,0)?

    Given:
    y=-4-t

    (x,y,z)=(6,-4,0)+t(0,-1,1)
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  8. #8
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    It was a typo.
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  9. #9
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    Ok, then I get

    D= (-6,9,3) . (-1,-2,-2)
    ----------------------
    3

    therefore

    6 units..

    (hmm :S turns out same as another q i had, i think)
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  10. #10
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    Quote Originally Posted by skeske1234 View Post
    Ok, then I get

    D= (-6,9,3) . (-1,-2,-2)
    ----------------------
    3

    therefore

    6 units..
    Yes that is correct.
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