Surface defined by vector product

[bigevil]

Homework Statement

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]

Homework Equations

Nothing really.

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]?

 

[tiny-tim]

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]
…
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]?

Hi bigevil!

(have a theta: θ )

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.

Just write out the defining equation … r.u = |r| times …? and then bung that into the original equation.

 

[bigevil]

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?

 

[tiny-tim]

Hi bigevil!

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?

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

hmm … I hadn't noticed before, that you'd said that in the middle of your first post:

As the values of m are continuous, the surface so generated is a cone with the axis parallel to u.

… I'd stopped when I saw your solutions for m = ±1 were wrong.

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'.

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.

 

[HallsofIvy]

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.

 

[bigevil]

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].