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Math Help - Another Vector Parametric Problem

  1. #1
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    Another Vector Parametric Problem

    How do I find equations of the planes for two similar yet different problems such as:

    Find an equation of the plane through the point and parallel to the given plane:

    (-5, 9, -4)
    x + y + z + 4 = 0

    and

    Find an equation of the plane that contains the line and parallel to the given plane

    x = 2 + 2t, y = t, z = 6 - t
    2x + 4y + 8z = 17
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  2. #2
    Member Abu-Khalil's Avatar
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    For the first one, you must find the normal vector of the plane.

    Observe that x+y+z+4=0\Rightarrow (x,y,z+4)\cdot (1,1,1)=0\Rightarrow \left((x,y,z)+(0,0,4)\right)\cdot(1,1,1)=0. So the plane looked for is ((x,y,z)+(-5,9,-4))\cdot(1,1,1)=0\Rightarrow x+y+z=0.

    For the second one, observe that the line can be written as \displaystyle\frac{x-2}{2}=\frac{y}{1}=\frac{z+6}{-1} so the directional vector is (2,1,-1). From Linear Algebra, you know that the plane given belongs to (17/2,0,0)+<(-2,1,0),(-4,0,1)>. Also, observe that (2,1,-1)=(-2,1,0)-(-4,0,1) what means the directional vector can be contained in a parallel plane to the given one. Finally, if you choose a positional vector (so the line can really be contained and not just be parallel to it) you get \Pi: \ (-2,0,6)+<(-2,1,0),(-4,0,1)>.
    Last edited by Abu-Khalil; February 5th 2009 at 11:37 PM.
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