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Vectors and planes

  1. Jan 20, 2005 #1
    I need to find the equation of a plane that is parallel to the plane [itex]2x + 4y + 8z = 17 [/itex] and contains the line [itex] { x = 3 +2t , y=t , z=8-t } [/itex].

    Not quite sure how or where to begin. Something having to do with a perpendicular vector maybe?

    Also, in a different problem, I need to find the equation of a plane that contains the lines:
    [itex] \vec{r} = <1,1,0>+t<1,-1,2>[/itex]
    [itex] \vec{r} = <2,0,2>+s<-1,1,0>[/itex]

    Don't quite know where to begin this one either.
     
    Last edited: Jan 20, 2005
  2. jcsd
  3. Jan 20, 2005 #2
    Do you know how to extract the normal vector form the equation of a plane?

    Do you know how to construct the equation of a a plane from a normal vector and a point? Do you know how to find a point on a line? (Hint: [itex] \vec{n} \cdot (\vec{r} - \vec{r_o}) = 0 [/itex])

    Can you find the normal vector somehow (one of the products, maybe) from two vectors that you know are perpendicular to it?

    --J
     
  4. Jan 20, 2005 #3
    You can find one point, which belongs to plane. For example when t=0.

    And You have perpendicular vector from 2x+4y+8z=17 (2, 4, 8)

    So, if You have point P (x1, y1, z1) and perpendicular vector n=(A, B, C), then
    A(x-x1)+B(y-y1)+C(z-z1)=0 is your equation of plane
     
  5. Jan 20, 2005 #4

    HallsofIvy

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    For the second problem, you can read off two vectors that lie in the plane (a lot like "slope"). Once you know that, the cross product gives you the normal to the plane and it's easy to find a point in the plane. (Since the given lines are in the plane, taking any value of t in either equation gives a point on that line and so in the plane.)
     
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