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Describe the surface created by the equations.

  1. Sep 3, 2013 #1
    1. Position vector r DOT n(hat) = a, where n(hat) is a unit vector and a is a scalar.

    2. Position vector r DOT n(hat) = a times (magnitude of) r.

    I know the first question describes a plane of n = a, if n is a Cartesian unit vector. However I don't know if I should worry about other unit vectors. The second question I would think the plane just changes by mag of r, but that seems too simple. Thanks for any help.
     
  2. jcsd
  3. Sep 4, 2013 #2

    Dick

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    Rotate the vectors so that ##\hat n## becomes ##\hat x## (the x unit vector). Now write the equations out in coordinates where r=(x,y,z). So your first equation is x=a. So sure, it's a plane (possibly rotated). The second becomes ##x=a \sqrt{x^2+y^2+z^2}##. What kind of surface is that?
     
  4. Sep 4, 2013 #3

    HallsofIvy

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    What do you mean by "n= a" if n is "a Cartesian unit vector" and a is a scalar?

     
  5. Sep 4, 2013 #4
    I am not sure what I mean haha. I know that if n is a unit vector, then it will be n = a. I am not sure how it is if it is not a Cartesian unit vector. Thank you Dick! That was very clear and exactly what I needed. z = a√(x^2+y^2+z^2) brings back memories of multivariable calculus, but I am not exactly sure how to proceed. :(
     
  6. Sep 4, 2013 #5

    Dick

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    Go back and review conic sections. The form of the graph will depend on whether a is less than, greater than or equal to 1.
     
  7. Sep 4, 2013 #6

    Ray Vickson

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    Since ##\vec{n}## is a unit vector, the equation
    [tex] \vec{r} \cdot \vec{n} = |\vec{r}| a [/tex]
    becomes very special if ##a = 1##, and even more special if ##a > 1##.
     
  8. Sep 4, 2013 #7
    a < 1 for this problem, though the professor did say a special case arises for a > 1 :)
     
  9. Sep 4, 2013 #8
    I got to: z^2 = (a^2(x^2 + y^2)) / (1 - a^2), at which point I can just call it a cone based upon the standard equation of a cone. Right?!
     
  10. Sep 4, 2013 #9

    Dick

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    Pretty close, it's part of a cone. Remember when you square something that not all solutions of the final equation are solutions of the equation you started with.
     
  11. Sep 4, 2013 #10

    Ray Vickson

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    It is easier to just note that ##\vec{r}\cdot \vec{n} = r \cos(\theta), ## where ##r = |\vec{r}|## and ##\theta## is the angle between ##\vec{r}## and ##\vec{n}##. So, the equation says
    ##r \cos(\theta) = a r,## hence either ##r = 0## or ##\cos(\theta) = a.## For ##a \in (-1,1),## this says that ##\theta =## constant, so describes a cone.
     
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