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Help finding the directed vector of a plane

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data
    If the line l is given by the equations 2x-y+z=0, x+z-1=0, and if M is the point (1,3,-2), fine a Cartesian equation of the plane,

    (a) passing through M and l

    I don't get how I can find the directed vector of the plane

    2. Relevant equations

    This is a past paper question so the answer is given as x-2y-z+3=0
    However I don't know how to get there.

    3. The attempt at a solution

    I have tried to get the solution assuming (2,-1,1) to be the directed vector and hence putting it into the equation,

    2(x-1)-1(y-3)+(z+2)=0

    However I get the wrong result so help regarding how to get the directed vector would help
     
  2. jcsd
  3. Feb 6, 2014 #2

    maajdl

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    Gold Member

    Why do you assume a directed vector of (2,-1,1), while you have no idea of what it actually is?
    Therefore, why don't you instead assume that the directed vector is (a,b,c),
    and try to find out what a and b and c might be.
    (note: the length of the directed vector is arbitrary: only it direction matters)
     
  4. Feb 6, 2014 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    First, I must admit that I have no idea what you mean by "the directed vector of the plane". I would think of a vector determining or determined by a plane to be a normal or perpendicular vector to the plane. But there exist an infinite number of such vectors of different lengths. And certainly not "directed" because such normal vectors could be in either of two directions.

    I would do this: subtracting the second equation, x+ z= 1, from the first, 2x- y+ z= 0 eliminates z and gives x- y= -1 or y= x+ 1. Taking x= t, we get the parametric equations for the line x= t, y= t+ 1, z= -t+ 1. That line has direction vector <1, 1, -1>, the coefficients of the parameter. (Or: taking x= 0, from x+ z= 1, z= 1 and then 2x- y+ z= 0 becomes 2(0)- y+ 1= 0 so y= 1; one point on the line is (0, 1, 1). Taking x= 1, from x+ z= 1, z= 0 and then 2x- y+ z= 0 becomes 2- y+ 0= 0 so y= 2; another point on the line is (1, 2, 0). The vector from one point to the other is <1- 0, 2- 1, 0- 1>= <1, 1, -1> again.). A vector from the point (0, 1, 1), on the line, to the given point, M= (1, 3, -2), is <1- 0, 3- 1, -2- 1>= <1, 2, -3>. The cross product of those two vectors, both lying in the plane, will give a vector perpendicular to the plane (which is what I presume you mean by the "directed vector of the plane").
     
    Last edited: Feb 6, 2014
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