# Describe the surface created by the equations.

1. Sep 3, 2013

### paulharrylemon

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. Sep 4, 2013

### Dick

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?

3. Sep 4, 2013

### HallsofIvy

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

4. Sep 4, 2013

### paulharrylemon

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. :(

5. Sep 4, 2013

### Dick

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.

6. Sep 4, 2013

### Ray Vickson

Since $\vec{n}$ is a unit vector, the equation
$$\vec{r} \cdot \vec{n} = |\vec{r}| a$$
becomes very special if $a = 1$, and even more special if $a > 1$.

7. Sep 4, 2013

### paulharrylemon

a < 1 for this problem, though the professor did say a special case arises for a > 1 :)

8. Sep 4, 2013

### paulharrylemon

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?!

9. Sep 4, 2013

### Dick

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.

10. Sep 4, 2013

### Ray Vickson

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.