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Math Help - Equation of a plane which parallel to a line

  1. #1
    moheelave4
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    Exclamation Equation of a plane which parallel to a line

    Please help me with this question(URGENT):

    Find the equation of the plane through the line
    r=(1,-3,4)+t(2,1,1)
    and parallel to the line
    r=s(1,2,3)


    help me please...
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  2. #2
    Super Member Rebesques's Avatar
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    Call it (p).


    The line r=(1,-3,4)+t(2,1,1) is parallel to the vector (2,1,1). Since (p) contains r, the vector (2,1,1) is parallel to (p) also.

    Now (p) is also parallel to r=s(1,2,3), so it is parallel to (1,2,3).



    ------To find its parametric equation, we use the fact that planes are generated by two linearly independent parallel vectors:

    r=a(2,1,1)+b(1,2,3), for a,b real.


    -----To find its equation in (x,y,z), we need a vector perpendicular to (p). For a perpendicular vector, we find the cross product of the two vectors parallel to (p).

    So, a vector normal to the surface is (2,1,1)x(1,2,3)=(1,-5,1).

    A point on (p) can be found by assigning a random value for t in the equation of the line r=(1,-3,4)+t(2,1,1), say t=0: r=(1,-3,4).

    If (x,y,z) is a point (p), then (x,y,z)-(1,-3,4) is a vector parallel to the plane, and must be normal to (1,-5,1). We now have the equation of the plane: (1,-5,1) \cdot(x-1, y+3,z-4)=0 or (x-1)-5(y+3)+(z-4)=0.
    Last edited by Rebesques; August 19th 2007 at 05:05 AM. Reason: keyboard alzheimer
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by moheelave4 View Post
    Please help me with this question(URGENT):

    Find the equation of the plane through the line
    r=(1,-3,4)+t(2,1,1)
    and parallel to the line
    r=s(1,2,3)

    ...
    Hello,

    call the plane p. Since p contains the given line the point A(1, -3, 4) belongs to the plane too. Now you need two linear independent vectors to span(?) the plane. These vectors are known to you. If \vec x is the position vector of/to every point of the plane the equation of the plane is:
    <br />
p: \vec x = \left(\begin{array}{c}1 \\-3 \\ 4 \end{array}  \right) + r \cdot \left(\begin{array}{c} 2 \\1 \\ 1 \end{array}  \right) + s \cdot \left(\begin{array}{c} 1 \\ 2\\ 3 \end{array}  \right)~, ~r,s~\in~\mathbb{R}
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  4. #4
    Super Member Rebesques's Avatar
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    Oops, for the parametric, do as Earboth did and don't forget to add the point ()
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  5. #5
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    Quote Originally Posted by earboth View Post
    Hello,

    call the plane p. Since p contains the given line the point A(1, -3, 4) belongs to the plane too. Now you need two linear independent vectors to span(?) the plane. These vectors are known to you. If \vec x is the position vector of/to every point of the plane the equation of the plane is:
    <br />
p: \vec x = \left(\begin{array}{c}1 \\-3 \\ 4 \end{array}  \right) + r \cdot \left(\begin{array}{c} 2 \\1 \\ 1 \end{array}  \right) + s \cdot \left(\begin{array}{c} 1 \\ 2\\ 3 \end{array}  \right)~, ~r,s~\in~\mathbb{R}
    Is this correct? I think when the question says the plane passes through the line, the line intersects the plane at one point, but the whole line is not contained in the plane... help?
    Last edited by Arivaz; March 25th 2008 at 03:32 PM.
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