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Find an equation of plane that contains the line

  1. Jan 25, 2012 #1
    I have looked at previous post that were on similar topics in hopes of not having to create a new post but I seem to be stuck still.

    1. The problem statement, all variables and given/known data
    Find an equation of the Plane that contains the line y=-2x+3 in the plane z=5, and goes through the point (2,-7,23)


    2. Relevant equations
    General equation of plane:
    f(x,y,z)=mx+ny+c



    3. The attempt at a solution
    First I set up the equation of a plane using slopes from the equation of the line and values for the point it should pass through

    23=-2(2)+n(-7)+3

    then solved for y-slope

    24=-7n

    n=24/7

    this gives me the function:

    z=-2x-(24y/7)+3

    I'm not sure my procedure was correct at all and I can't seem to figure out how to check this. If i plug in the values , I get the right z point, but this only tells me that I did my algebra correct (I hope :-!). I know I'm probably missing a major concept probably but some I cannot seem to make any progress on this problem.

    Thanks for the help
     
    Last edited: Jan 25, 2012
  2. jcsd
  3. Jan 25, 2012 #2

    Dick

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    I don't think that's a very good expression for the general equation of a plane. Use ax+by+cz=k, with a,b,c,k all constants. Written in that form (a,b,c) is a normal vector to the plane. You can find a normal vector to the plane by finding two different vectors that are parallel to the plane and taking their cross product. Have you ever done it that way?
     
  4. Jan 25, 2012 #3

    lanedance

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    I find the easiest way to do it is to use the fact that any vector that lies in the plane will be normal to the plane normal (by definition)

    If n is normal to the plane and if you have a point in the plane q=(u,v,w), now any and take any other point in the plane p=(x,y,z), the vector formed by the difference (p-q) between the points will be parallel to the plane.

    The difference between every point outside the plane and p will give a vector that is not parallel to te hplane

    the points in the plane p are exactly the ones given by
    [tex](p-q) \bullet n = 0 [/tex]
     
  5. Jan 25, 2012 #4
    Yeah, I agree that vectors would be the best way to do this. Unfortunately we are supposed to use the "non vector" method. Although I am comfortable with vectors for the most part the class has just started teaching them. I think I see where I was wrong. I believe the equation that i need to use is:

    z=m(x-x_0) + n(y-y_0) + z_0 that would give...

    z = m(x-2) + n(y+7) + 23

    I'm still not sure how to check the work... If I plug in 5 for z, should i get the equation of the line after solving for y?
     
    Last edited: Jan 25, 2012
  6. Jan 25, 2012 #5

    Dick

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    Ok, I think that form will work. But you still need to find m and n. Plug in a couple of points on the line y=-2x+3 in the plane z=5 and try to solve for m and n.
     
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