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Homework Help: Surface defined by vector product

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Identify the following surface. I'm not sure what the answer is, but I believe I am on the right track, appreciate if anyone can check this for me.

    [tex]{\bf r} \cdot {\bf u} = m |{\bf r}| , -1 \le {\bf m} \le 1 [/tex]

    2. Relevant equations

    Nothing really.

    3. The attempt at a solution

    Split cases. For m=0, [tex]{\bf r} \cdot {\bf u} = {\bf 0} [/tex], the locus of which is the origin.

    For m = 1, [tex]{\bf r} \cdot {\bf u} = |{\bf r}| [/tex]. The equation describes a plane (whose normal vector is u) distance [tex]|{\bf r}|[/tex] from the origin. For m = -1, the equation describes a plane an equal distance from the origin and in the opposite direction. The locus described is any vector whose projection on the plane is [tex]|{\bf r}|[/tex] from the origin, forming the locus of a circle.

    I'm already about halfway there I think. As the values of m are continuous, the surface so generated is a cone with the axis parallel to u. But I also need to define the semi-parallel angle.

    4. More attempt

    [tex]{\bf \hat{r}} \cdot {\bf u} = m[/tex], and [tex]{\bf \hat r} = 1[/tex] Therefore, [tex]cos\theta = m[/tex]?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 10, 2008 #2


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    Hi bigevil! :smile:

    (have a theta: θ :wink:)

    Your original attempt is completely wrong … but I think you've worked that out.

    Your "More attempt" is on the right lines - cosθ is definitely the important factor. :smile:

    Just write out the defining equation … r.u = |r| times …? and then bung that into the original equation. :wink:
  4. Nov 10, 2008 #3
    Oh dear, that's bad. So my answer is right but the working is completely wrong? I'm even more lost than I thought now....

    Or maybe what I should have done is to describe the vector product as [tex]
    {\bf \hat{r}} \cdot {\bf u} = m
    [/tex] all along, and save myself all the trouble. I sort of know what is going on, I just don't know how to put it down in 'proper working'.

    I forgot to state my (attempted) answer, which is a cone (hourglass) along the z-axis, passing thru the origin (when m=0), with semi-parallel angle of [tex]arccos m[/tex]. Did I get it right? From what I gather, the answer is wrong, but the semi-parallel angle is correct. Or is my answer correct and the reasoning absolutely wrong?
  5. Nov 10, 2008 #4


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    Hi bigevil! :smile:
    Yes, if |u| = 1, a half-cone (not a full one) of semi-angle arccos(m) is right! :biggrin:

    hmm … I hadn't noticed before, that you'd said that in the middle of your first post:
    … I'd stopped when I saw your solutions for m = ±1 were wrong. :redface:
    In 'proper working', you need to start with the definition, and then put that definition in the equation:

    r.u = |r| cosθ = m|r|,

    so, dividing by |r|,

    cosθ = m. :smile:
  6. Nov 10, 2008 #5


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    Fior one thing you should do a better job of defining your terms! Is [itex]\bf{r}[/itex] the position vector? Is [itex]\bf{u}[/itex] a given constant vector? If so then "[itex]\bf{r}\cdot\bf{u}= 0[/itex] does not say that [itex]\bf{r}[/itex] is the zero vector: it says only that [itex]\bf{r}[/itex] is perpendicular to [itex]\bf{u}[/itex], and gives a plane.

    You can write [itex]\bf{r}\cdot\bf{u}= |u||r|cos(\theta)[/itex] where [itex]\theta[/itex] is the angle between [itex]\br{r}[/itex] and [itex]\bf{u}[/itex] (and so depends on [itex]\bf{r}[/itex]). Then the equation becomes [itex]|u||r|cos(\theta)= m|r|[/itex] so that [itex]|u|cos(\theta)= m[/itex] that says that [itex]\theta[/itex] is in fact a constant: the angle between [itex]\bf{u}[/itex] and [itex]\bf{r}[/itex] is constant and so the surface is a cone with axis along the direction of [itex]\bf{u}[/itex].

    I am puzzled by your saying that it is a cone along the z-axis. That would imply that [itex]\bf{u}[/itex] is a vector pointing along the z-axis and I don't believe you said that.
  7. Nov 10, 2008 #6
    Sorry guys, I'm not a very subtle mathematician. I was doing a few other problems where I simply substituted for Cartesian coordinates, which is where all the things about the z-axis came from. This too isn't very mathematically elegant. So sorry, Tim and Halls.

    Yes Halls, r is the position vector, u is a fixed unit vector, (I defined the "origin" as the point where the vectors r and u converge) which clears things up here, [tex]\hat{r} \cdot \hat{u} = m, -1 \le m \le 1[/tex].
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