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Find the equation of the plane given point and line (3D)

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  1. Apr 19, 2017 #1
    ∴1. The problem statement, all variables and given/known data
    Find an equation for the plane that contains the point (1,-1,2) and the line x=t, y=t+1, z=-3+2t?
    2. Relevant equations



    3. The attempt at a solution

    I dont know what Im doing wrong but everytime I try to do these plane questions my amswer is always the opposite sign of the real answer.

    P1(7,0,1)
    P2(7,3,0)

    P1P2 (0,3,-1)
    n(7,-1,2)

    P1P2 crossproducted with n = (5,-7,-21)

    ∴ 5(x-7)-7y-21(z-1)=0
     
  2. jcsd
  3. Apr 19, 2017 #2

    gv3

    User Avatar

    Where are your P1 and P2 coming from?
    Are you crossing P1P2 vector with the normal vector?
     
  4. Apr 19, 2017 #3

    Mark44

    Staff: Mentor

    If you have a point ##P_0(x_0, y_0, z_0)## and a normal to the plane N = <A, B, C>, the equation of the plane is ##A(x - x_0) + B(y - y_0) + C(z - z_0) = 0##. This equation can be obtained by taking the dot product of N and a vector ##\vec{P_0P}##, where P(x, y, z) is an arbitrary point in the plane.

    You're given a point, so you have some of the information you need. To get a normal to the plane, find two vectors that lie in the plane. For the first vector, find two points on the line, and construct a displacement vector between the two points. Call this ##\vec u##. To get another vector, construct a displacement vector between any point on the line and the given point. Call this ##\vec u##.
    Take the cross product of ##\vec u## and ##\vec v##, which will give you a vector perpendicular to both ##\vec u## and ##\vec v## (and therefore be a normal to the plane).
     
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