1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Where does the line intersect the plane?

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Where does the line through (1, 0, 1) and (4, -2, 2) intersect the plane x+y+z=6 ?


    2. Relevant equations
    eqn of plane: a(x-x0)+b(y-y0)+c(z-z0)=0
    ??
    ?? anything else idk..


    3. The attempt at a solution
    I started off by thinking that I should use the given points for the line to make a vector. lets call ... AB? so that would be <3, -2, 1>. but i dont really know if/why i should start with that. thats about as far as i got
     
  2. jcsd
  3. Sep 15, 2011 #2

    Mark44

    Staff: Mentor

    Use the two points to find the parametric equations of the line, and then solve for the parameter value of the point at which the line and plane intersect
     
  4. Sep 15, 2011 #3
    How do I know what a,b, and c are for the parametric eqns?
     
  5. Sep 15, 2011 #4

    Mark44

    Staff: Mentor

    You know two points on the line - call them A(1, 0, 1) and B(4, -2, 2). Let P(x, y, z) be an arbitrary point on the line. AB and AP have the same direction, so AP = t*AB, for some parameter t. Plug in the given numbers, and that gives you your parametric equations.
     
  6. Sep 15, 2011 #5
    yea i did that right before you answered. in a similar/slightly diffferent way. but thanks. and so.... how do you solve the parameter value? i have the normal vector <1,1,1> for the plane..
     
  7. Sep 15, 2011 #6

    Mark44

    Staff: Mentor

    The parametric equations for your line should look something like this:
    x = at + d
    y = bt + e
    z = ct + f
    where a, b, c, d, e, and f are known numbers.

    The equation of your plane is x + y + z = 6. Substitute for x, y, and z in the plane, using the parametric equations, and you'll have an equation that involves only the parameter t. Solve for t, and then substitute it back into your parametric equations to find the point of intersection.

    The plane's normal doesn't enter into things here.
     
  8. Sep 15, 2011 #7
    i got (7, -4, 3) is that right ? i think so
     
  9. Sep 15, 2011 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Does it lie on the line? Does it satisfy the equation of the plane?
     
  10. Sep 15, 2011 #9
    it satisfies the eqn of the plane but how do you know for the line?
     
  11. Sep 15, 2011 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Note that the vector from (1, 0, 1) and (4, -2, 2) is <4-1, -2- 0, 2- 1>= <3, -2, 1> while the vector from (1, 0, 1) to (7, -4, 3) is <7- 1, -4- 0, 3- 1>= <-6, -4, 2> which is just 2 times <3, -2, 1>. Since one vector is a multiple of the other they are parallel and define the same line through (1, 0, 1).
     
  12. Sep 15, 2011 #11
    Sooo that means it's right !
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook