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Math Help - 3D: 2 planes intersecting

  1. #1
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    3D: 2 planes intersecting

    I believe I've figured out a and b, but I don't understand c


    Consider the points A(1,2,0), B(1,0,1), C(0,1,-1) and D(1,0,-1)

    a) Find the equation of the plane passing through the points A, B, C. Is the point D on this plane?
    b)Find a parametric equation for the line passing through D which is perpendicular (the line) to the plane of the triangle ABC.
    c) Find the equation of the plane passing through A and D, which is perpendicular to the plane of the triangle ABC. Note: Two planes are perpendicular if their normal vectors are perpendicular.

    a)equation of the plane ABC is -3x+y+2z=-1 and D is not on the plane
    \begin{gathered}<br />
  \overrightarrow u  =  < 0,2, - 1 >  \hfill \\<br />
  \overrightarrow v  =  <  - 1,1, - 2 >  \hfill \\ <br />
\end{gathered} <br />
    \begin{gathered}<br />
  \overrightarrow n  =  - 3i + j + 2k \hfill \\<br />
  \overrightarrow r  - \overrightarrow {{r_o}}  =  < x - 0,y - 1,z - ( - 1) >  \hfill \\ <br />
\end{gathered}<br />
    \begin{gathered}<br />
  \overrightarrow r  - \overrightarrow {{r_o}} \overrightarrow n  = 0 \hfill \\<br />
   < x - 0,y - 1,z - ( - 1) >  <  - 3,1,2 >  = 0 \hfill \\ <br />
\end{gathered} <br />
    \begin{gathered}<br />
   - 3x + 3 + y + 2z - 2 = 0 \hfill \\<br />
   - 3x + y + 2x =  - 1 \hfill \\ <br />
\end{gathered} <br />
    b)equation of line perpendicular to plane:
    \begin{gathered}<br />
  \overrightarrow {{r_o}}  =  < 1,0, - 1 >  \hfill \\<br />
  \overrightarrow v  =  <  - 1,1, - 2 >  \hfill \\ <br />
\end{gathered} <br />
    \begin{gathered}<br />
  \overrightarrow r  = \overrightarrow {{r_o}}  + t\overrightarrow v  \hfill \\<br />
   < x,y,z >  =  < 1,0, - 1 >  + t <  - 1,1, - 2 >  \hfill \\ <br />
\end{gathered} <br />
    x=1-t
    y=t
    z=-1-2t

    and like I said, I have no idea what to do for c
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  2. #2
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    On B, I'm confused on why you are using <-1, 1, -2> as your direction vector. If the line is through D and perpindicular to the plane of the triangle ABC, why not just use the normal to the plane as your direction vector? Perhaps I'm not seeing something in your reasoning.

    For question C, we need two vectors (create the normal) and a point. You have one vector in the direction vector of whatever line you made in B. You also have the vector AD (or DA).
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  3. #3
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    The line you found in part(b) is perpendicular to the plane of the triangle ABC at every point of the plane (I'll call it the ABC-plane). That is, you can pick any point you want in the ABC-plane and the line from part(b) that goes through this point is perpendicular to the ABC-plane.

    You're looking for a plane through A and D (I'm going to call it the AD-plane) that is also perpendicular to the ABC-plane. Since both planes go through A, find the particular expression of the line from part(b) that goes through A.

    Here's the key: That line (perpendicular to the ABC-plane that goes through the point A) lives inside of the AD-plane (which is also perpendicular to the ABC-plane and goes through A)!

    This means you can pick any other point on that line. That will give you a third point in the plane. Judging from the work you've posted, I'll bet you can take it from here if you haven't already
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  4. #4
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    I'm not sure why I used <-1,1,-2> but thank you for pointing that out...

    for part c the direction vector, do you mean n=<-3,1,2>?
    and does that work because its perpendicular?
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  5. #5
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    It does indeed work. Remember that vectors aren't like points that lie in the plane. The direction vector is a "representation vector" in the sense that its just the difference (or addition, or whatever) of two vectors that lie in a plane. Any plane parallel to the plane in question, will have points in each of them that can make similar representation vectors - even if they do not use the same points. Thus, we can use any vector parallel to the plane we want to create as our direction vector.

    Of course, this does not work for points, so we must use the points that actually lay IN the plane. So you'll use one of those points (A or D) to make your plane, but first use both of them to create ANOTHER vector parallel to the plane you wish to create, and cross it with the vector you've already got: instant normal. From here its simply a matter of using the definition of the plane (which you used above) to create your plane equation.
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