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Intersection of line through a plane

  1. Sep 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Where does the line through (−2, 3, 2) and (3, 5, −1) intersect the plane x + y − 2z = 6?

    2. Relevant equations


    3. The attempt at a solution
    i used r = r0 + tv

    the vector between the 2 given points is <5,2,-3>

    r = (-2,3,2) + t<5,2-3>

    x = -2 + 5t y=3+2t z=2 - 3t

    plugging these into the plane given: -2+5t + 3+2t -2(2-3t)= 6

    solving for t , t should equal 9/13 if my algebra is correct.

    plugging into the parametric equations , the point of intersection should be (19/13 , 57/13, -1/13)

    can someone check all this?
     
  2. jcsd
  3. Sep 18, 2015 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is correct. Nice work.:oldcool:
     
  4. Sep 18, 2015 #3

    Mark44

    Staff: Mentor

    It's not that hard to check it for yourself, and it's a good habit to get into. You can do this by verifying that the point you found is on both the line and on the plane.

    To verify that the point is on the line, find the value of t so that 19/13 = -2 + 5t, 57/13 = 3 + 2t, and -1/13 = 2 - 3t. The same value of t should work in all three equations.

    To verify that the same point is on the plane, confirm that 19/13 + 57/13 - 2(-1/13) = 6.
     
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