# Surface defined by vector product

1. Nov 10, 2008

### bigevil

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.

$${\bf r} \cdot {\bf u} = m |{\bf r}| , -1 \le {\bf m} \le 1$$

2. Relevant equations

Nothing really.

3. The attempt at a solution

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

For m = 1, $${\bf r} \cdot {\bf u} = |{\bf r}|$$. The equation describes a plane (whose normal vector is u) distance $$|{\bf r}|$$ 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 $$|{\bf r}|$$ 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

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

2. Relevant equations

3. The attempt at a solution

2. Nov 10, 2008

### tiny-tim

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.

3. Nov 10, 2008

### 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 $${\bf \hat{r}} \cdot {\bf u} = m$$ 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 $$arccos m$$. 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?

4. Nov 10, 2008

### tiny-tim

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

5. Nov 10, 2008

### HallsofIvy

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

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

I am puzzled by your saying that it is a cone along the z-axis. That would imply that $\bf{u}$ is a vector pointing along the z-axis and I don't believe you said that.

6. Nov 10, 2008

### 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, $$\hat{r} \cdot \hat{u} = m, -1 \le m \le 1$$.