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Math Help - Smallest distance between a line and a plane

  1. #1
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    Exclamation Smallest distance between a line and a plane

    I have a plane: 1 x + -5 y + 5 z = -7

    and a line: (x,y,z) = (0,0,0) + t(-1,-3,2)


    and i need to find thesmallest distance between the two.



    My work:

    Plane : normal = (1,-5,5) & point P = (-7,0,0)
    Line : direction vector = (-1,-3,2) & point Q = (0,0,0)

    Need to find vector PQ = (7,0,0)
    The distance = ||Projection of PQ onto normal n ||

    ( ( (7,0,0)(1,-5,5) ) / ||(1,-5,5)|| ) * (1,-5,5)

    = (7/51) * sqrt51 or 7/sqrt51




    This is wrong, and after looking online and through my textbook, i think itís because Iím not getting the shortest distance, but rather the distance to P
    I canít get a hold of my prof and the help room at my school thought it looked fine but it isnít right

    Iíve searched online for the past three days and i just KNOW this type of question will be on a test
    I need to know how to do this, please help
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  2. #2
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    Hello, williamb!

    Is this a trick question?


    Given the plane: . x -5y + 5z \:=\: -7

    and the line: . (x,y,z) \:= \:(0,0,0) + t(\text{-}1,\text{-}3,2)

    find the smallest distance between the two.

    The normal vector of the plane is: . \vec{n} \:=\:\langle 1,\text{-}5,5\rangle

    The direction vector of the line is: . \vec{v} \:=\:\langle \text{-}1, \text{-}3, 2\rangle

    Since \vec{n}\cdot\vec{v}\,\neq\,0, the plane and line are not parallel.


    Therefore, they intersect. .Their minimum distance is 0.

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  3. #3
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    little confused?

    I'm a little confused i was taught that if a dot b equals zero they are perpendicular.
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  4. #4
    Super Member Matt Westwood's Avatar
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    If the line is perpendicular to the normal then it is parallel to the plane. If not then not.
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