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Line perpendicular to a plane

  1. Oct 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Line perpendicular to a plane
    Plane: z(x,y)= 0.882531*x-0.346494*y+0.383108
    P1(0.5;0.1;Z)
    P2(0.4;0.15;Z)
    2. Relevant equations


    3. The attempt at a solution

    z(0.5,0.1) = 0.7897
    z(0.4,0.15) = 0.68414

    vector n=ai+bj+ck

    vector P2P1 = (0.5-0.4)i + (0.1-0.15)j +(0.7897-0.68414)k

    n*(P2P1) = 0.1a - 0.005b + 0.10556c
    c= -0.947329(a-0.5*b)

    parametric

    x(t,u) = t+0.5
    y(t,u)=t+0.1
    z(t,u)=-0.947329(t-0.5*t)

    when I put this on the graph, its not perpendicular at all... :( help
     
  2. jcsd
  3. Oct 16, 2015 #2

    Mark44

    Staff: Mentor

    For the first one, I get .7897241, and for the second, I get .6841463
    Your 2nd coordinate is wrong -- it should be -.05, not -.005 .
    Also, your plane can be written as 0.882531*x - 0.346494*y - z = .383108. A normal to this plane is <0.882531, -0.346494, -1> .

     
    Last edited: Oct 16, 2015
  4. Oct 16, 2015 #3

    Mark44

    Staff: Mentor

    BTW, what terrible numbers in this problem! I started by using a calculator, but then gave up and made a small Excel spreadsheet.
     
  5. Oct 16, 2015 #4
    When you have a plane of equation ##ax+by+cz+d = 0##, you know that the perpendicular line is directed by ##\vec u = (a,b,c)##.
    So the equation of your line passing through ##P_1## is ##{\cal D} = \{ P_1 + t \vec u, t\in\mathbb{R}\} ##
     
    Last edited: Oct 16, 2015
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