Describe the surface created by the equations.

  • #1
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.
 

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  • #2
Dick
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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.
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
HallsofIvy
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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.
What do you mean by "n= a" if n is "a Cartesian unit vector" and a is a scalar?

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.
 
  • #4
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
Dick
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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. :(
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
Ray Vickson
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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.
Since ##\vec{n}## is a unit vector, the equation
[tex] \vec{r} \cdot \vec{n} = |\vec{r}| a [/tex]
becomes very special if ##a = 1##, and even more special if ##a > 1##.
 
  • #7
a < 1 for this problem, though the professor did say a special case arises for a > 1 :)
 
  • #8
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
Dick
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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?!
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
Ray Vickson
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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?!
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.
 

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