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Find the intersection of a particle and a plane in R3

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    The starting position of a particle in R3 is (1,1,1) and it's traveling with constant velocity (2,-1,1). Where does it hit the plane {(x,y,z)|x - 2y + z = 4}. And find the angle between the path of the particle and the plane.

    2. Relevant equations


    3. The attempt at a solution
     
    Last edited: Feb 3, 2010
  2. jcsd
  3. Feb 3, 2010 #2

    Mark44

    Staff: Mentor

    Are there any relevant equations (e.g., equation of a line in space, parametric equation of a plane in space, angle between two vectors)?

    What attempt have you made?
     
  4. Feb 3, 2010 #3
    yeah i am having trouble figuring out how to find the equation for the path of the particle. From there I think it should be pretty easy, just setting them equal and finding their intersection. and then for the angle between them i can just use the dot product, but with what vectors?
     
  5. Feb 3, 2010 #4

    Mark44

    Staff: Mentor

    I'm trying to get you to fill in the information you neglected to enter in the first post.
     
  6. Feb 3, 2010 #5
    one can express a line by a pair of linear equations

    ax+by+cz+d=0 and a'x+b'y+c'z+d'=0

    such that bc'-cb', ca'-ac', and ab'-ba' are not all zero. The line thus defined is parallel to the vector

    (bc'-cb', ca'-ac', ab'-ba').

    Line through (x0,y0,z0) parallel to the vector (a,b,c):

    (x-x0)/a = (y-y0)/b = (z-z0)/c

    Now, the angle between two vectors can be obtained by using the dot product:

    angle = arccos ((a . b)/|a||b|)



    as for my attempt, i am having trouble just getting off the ground.
     
  7. Feb 3, 2010 #6

    Mark44

    Staff: Mentor

    Well, that's a start. One that I had in mind was how do you write the equation of a line in space, give a point P(x0, y0, z0) on the line and a vector v = <A, B, C> with the same direction as the line?

    This is directly relevant to your problem, whereas what you wrote about the intersection of two planes isn't.
     
  8. Feb 3, 2010 #7
    So, the equation for a line in space is: r = r0 +tv = (x0,y0,z0) + t<a,b,c>

    but once i have that, how do i solve for their intersection? and the angle between them?

    For their intersection, do i just solve the system of equations:

    x = x0 + ta
    y = y0 + tb
    z = z0 + tc
    x - 2y + z = 4

    but what about t?


    and for the the angle between them, can i just use the vector form of the line, and a vector parallel to the plane and then use the dot product?
     
    Last edited: Feb 3, 2010
  9. Feb 3, 2010 #8

    Mark44

    Staff: Mentor

    Now this is something you should have included in the Relevant equations part in your first post.

    Put the data of your problem into the equation r = (x0,y0,z0) + t<a,b,c>

    Every point on your plane satisfies
    x = 2y -z + 4
    y = y
    z = z

    If the line and the plane intersect, their x coordinates, y coordinates, and z coordinates have to be equal.
     
  10. Feb 3, 2010 #9
    Ok, so i get the equation for the line:

    (x,y,z) = (1 + 2t, 1 - t, 1 + t)

    x = 1 + 2t
    y = 1 - t
    z = 1 + t

    So, using the equations for our plane:

    x = 2y - z + 4
    y = y
    z = z

    we get that:

    y = 1 - t
    z = 1 + t
    x = 1 + 2t = 2(1 - t) - (1 + t) + 4
    t = 4/5

    So, is their intersection point (13/5, 1/5, 9/5)?
     
  11. Feb 3, 2010 #10

    Mark44

    Staff: Mentor

    Check and see. Does this point satisfy your equations for the line and the plane? If so, you're golden.
     
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