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Planes in 3 space

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A plane is perpendicular to the line given by x=3+6t, Y=7+4t, and z=7-9t. What are the components of the normal to the plane

    2. Relevant equations



    3. The attempt at a solution
    I don't understand what the question is asking me all I have figured out that the normal should be parallel to the perpendicular line, but I can't find the normal from the parametric.
     
  2. jcsd
  3. Oct 1, 2012 #2

    Dick

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    You can find the tangent vector to the parametric, right? What is it? Then sure, that should be parallel to the normal. Do you want a unit normal vector?
     
  4. Oct 1, 2012 #3
    The only thing I don't understand is finding the tangent vector, all my book and teacher have told me is how to find the vector orthogonal to the plane.
     
  5. Oct 1, 2012 #4

    Dick

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    The tangent vector to the parametric curve (x(t),y(t),z(t)) is the derivative with respect to t of that. It's (x'(t),y'(t),z'(t)). What is that?
     
  6. Oct 1, 2012 #5
    So 6,4,-9?
    So the answer would be <6,4,-9>
     
  7. Oct 1, 2012 #6

    Dick

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    That's a normal vector to your plane alright. Any multiple of that is also a normal, yes? Do they want any specific one? That's why I asked if you want a 'unit normal'.
     
  8. Oct 1, 2012 #7
    I guess that should do I just don't understand the conceptual behind it.
     
  9. Oct 1, 2012 #8

    Dick

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    When you said, "I have figured out that the normal should be parallel to the perpendicular line" then that pretty much sums it up. Finding a vector tangent to the perpendicular line will then give you a normal. Just as you did.
     
  10. Oct 1, 2012 #9
    Thank you very much it actually makes sense now.
     
  11. Oct 2, 2012 #10

    HallsofIvy

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    Part of your problem is shown here:
    A plane doesn't have "the tangent vector". Every line in the plane gives a tangent vector. A plane is completely determined by a single point and a vector normal to the plane.
     
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