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Math Help - Simple 3D Vector geometry question about planes and lines!

  1. #1
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    Simple 3D Vector geometry question about planes and lines!

    How would you find the vector equation of the line h, if told it lies in the plane:

    E: x plus y = 4,

    and it passes through Q(2,2,1) which is on the plane,

    and the line is also perpendicular to the line -->


    g = (1 plus )i plus (3-)j plus (3)k



    find line h.



    please help!!! thanks!!!
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  2. #2
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    Quote Originally Posted by Yehia View Post
    How would you find the vector equation of the line h, if told it lies in the plane:

    E: x plus y = 4, and it passes through Q(2,2,1) which is on the plane,

    and the line is also perpendicular to the line -->


    g = (1 plus )i plus (3-)j plus (3)k

    Find line h.
    1. Not sure in which part of Germany you live but where I live we would use this notation:

    E:x+y=4

    g:\vec x = \begin{pmatrix}1\\3\\3\end{pmatrix} + \mu \cdot \begin{pmatrix}1\\-1\\0\end{pmatrix}

    2. You are looking for a point P with the following properties:

    p\in E~\wedge~P\in g~\wedge~(PQ) \perp g

    That means the distance |\overline{PQ}| has a minimum.

    There is a simple formula which yields the perpendicular distance of a point to a straight line which is probably known to you. I'll show you a different approach (it is always useful to have an alternative at hand ) :

    3. Calculate the vector from Q to any point of g:

    \vec d =  \begin{pmatrix}1\\3\\3\end{pmatrix} + \mu \cdot \begin{pmatrix}1\\-1\\0\end{pmatrix} - \begin{pmatrix}2\\2\\1\end{pmatrix}

    |\vec d| = \sqrt{6-4\mu +2\mu ^2}

    Determine the minimum of \vec d by differentiating wrt \mu and set the numerator of the quotient equal to zero.
    You'll get \mu = 1 which produces the point P(2,2,3).

    4. Thus the line you are looking for is the line \overline{PQ}.
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