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Vector inteception

  1. Jun 15, 2010 #1
    1. The problem statement, all variables and given/known data
    where does the line z=<1,2,-3>+t(2,-3,-2) intersect the plane 2x+3y+z=12

    i got the answer as 3,-1,-5



    2. Relevant equations
    i follow the steps listed here:
    https://www.physicsforums.com/showthread.php?t=277585


    3. The attempt at a solution

    i got the answer as 3,-1,-5

    can someone please answer if the answer is correct or not, if not please explain what i did wrong.
    many thanks
     
  2. jcsd
  3. Jun 15, 2010 #2
    Not quite....what value did you get for t?

    It appears that you've used a value of t being 1, unfortunately it isn't. Your working should look something like this;

    x for the line = (1+2t)
    y for the line = (2-3t)
    z for the line = (-3-2t)

    Then inserting for t in the plane as in replacing 2x+3y+z=12 with the values above and solving for t, having done that putting it back into the line equation to get;

    (1,2,-3) + t(2, -3,-2)
     
  4. Jun 15, 2010 #3
    hi thx for the prompt reply

    these are my working out:
    x=1+2t
    y=2-3t
    z=-3-2t

    then i sub it in

    2(1+2t)+3(2-3t)+(-3-2t)-12=0
    i got t=-7/7=/-1
     
  5. Jun 15, 2010 #4
    If a≠0, t=-b/a

    So the point where the line intersects the plane is:

    LaTeX Code: M(x_1 - \\frac{b}{a}a_1 , y_1 - \\frac{b}{a}a_2 , z_1 - \\frac{b}{a}a_3)

    so i switch:
    1-(-1)(2),(2)-(-1)(-3),(-3)-(-1)(-2)
    which i got
    3,-1,-5
     
  6. Jun 15, 2010 #5
    so is the answer -1,5,-5?
    anyone?
     
  7. Jun 15, 2010 #6
    Right apart from the last one, be careful with signs.
     
  8. Jun 15, 2010 #7
    so its' -1,5,-1?
    thx chewy
     
  9. Jun 15, 2010 #8
    That's right, no problem.
     
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