1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Surface defined by vector product
  1. Vector product (Replies: 4)

  2. Vectors product proof (Replies: 3)