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Linear Algebra Problem - Finding The Equation Of A Plane

  1. Dec 5, 2011 #1
    Not sure if I'm in the right section, but here it goes:


    1. The problem statement, all variables and given/known data

    Find the equation of the plane which contains:
    i) the intersection of -3x-2y-2z=9 and -2x-3y-2z=9
    ii) the point (-2,-3,-4)
    Write the equation in the form Ax+By+Cz=D


    2. Relevant equations

    Cross Product, perhaps triple product?

    3. The attempt at a solution

    I've solved for the direction vector where the two planes intersect:

    This vector I solved for was [-2,-2,5]

    At this point I'm stuck.... I don't know what my next move is. If anyone can help that would be greatly appreciated.
     
  2. jcsd
  3. Dec 5, 2011 #2

    Mark44

    Staff: Mentor

    Find either the vector equation of the line of intersection or its parametric form. From either of these equations, find two points on the line.

    With those two points and the given third point, you should be able to find a vector that is normal to the plane. Use that vector and any one of the three points to find the equation of the plane.

    I find it helpful for these kinds of problems to draw a sketch in three dimensions. The sketch doesn't have to be very accurate, but it gives you a better idea of what you're looking for.
     
  4. Dec 5, 2011 #3

    THANK YOU FOR THE HELP!

    I understand that we need the equation of the line, but how do I do this when I only have a directional vector?

    Lastly, how may one find these two points?
     
  5. Dec 5, 2011 #4
    How did you obtain the vector?

    You have two planes, and in [itex]\Re[/itex][itex]^{3}[/itex] their intersection will be a line. I just did this question by subtracting one of the equations from a multiple of the other, and eliminating x. This then leaves an equation in terms of y and z (let's say ay +bz = c).

    From here, you can choose y or z freely to be any value in [itex]\Re[/itex][itex]^{3}[/itex], and out of habit I then make the substitution [itex]\lambda[/itex] for either x,y, or z. In this case I said y=[itex]\lambda[/itex], and using the equation of ay + bz = c, you can define z:

    y=[itex]\lambda[/itex]
    a[itex]\lambda[/itex] + bz = c
    z= (c - a[itex]\lambda[/itex]) / a

    Then you substitute these values of y and z into one of the original equations, and then solve for x.

    You will then have three parametric equations:

    x= g + h[itex]\lambda[/itex]
    y= i + j[itex]\lambda[/itex]
    z= k + l[itex]\lambda[/itex]

    Or you could write this in the form:

    (g, i, k) + [itex]\lambda[/itex] (h, k, l)


    You then have the equation of a line and a point. From here you can work out the equation of the plane, remembering that any additional line drawn from the point you know (-2, -3, -4) to the equation of the line of intersection you found above [ (g, i, k) + [itex]\lambda[/itex] (h, k, l) ] will also lie in the plane.
     
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