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Vectors: Please check my work

  1. Feb 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Please check my work for the following problem:

    Find the point(s) where the curve r = <t,t2,-3t> intersects the plane 2x-y+z=-2.



    2. The attempt at a solution

    t + t2 -3t = -2

    (t-2)(t+0) = -2

    t=0 and t=-2

    plugging those values in yields: r(0) = <0,0,0>

    r(-2) = <-2,4,6>

    Is that what the problem is asking for?
     
  2. jcsd
  3. Feb 4, 2014 #2

    Dick

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    Check your answers. <0,0,0> is not on the plane. Why? (Hint: you solved the wrong quadratic - check it).
     
  4. Feb 4, 2014 #3
    I see that <0,0,0> is not on the plane however I'm not sure what other quadratic I could solve.
     
  5. Feb 4, 2014 #4

    Dick

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    There's a wrong sign and a coefficient mistake in your quadratic. It's just an error. There's nothing wrong with your procedure Just fix it.
     
    Last edited: Feb 4, 2014
  6. Feb 4, 2014 #5
    I honestly can't see the mistake all I did was match the signs with the given equations.
     
  7. Feb 4, 2014 #6

    Dick

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    Then you are doing something strange. You have 2x-y+z=(-2). x=t, y=t^2 and z=(-3t). If I put those values of x, y and z into the plane equation, I get 2t-t^2-3t=(-2). Why didn't you? What did you do???
     
  8. Feb 4, 2014 #7
    Oh I see, I misunderstood the process. The values of x, y, z go into the plane equation.

    I was using the components of r which is where I got positive t^2.

    Thank you.
     
  9. Feb 4, 2014 #8

    Dick

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    You're welcome. I still don't see quite what you did, but that's ok. Just don't do it again.
     
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