Circle in plane parameterization

In summary, the homework statement is to parameterize a circle with radius 2, centered at 1,2,3 and lies on the plane x+y+z=6. The Attempt at a Solution says that you can solve the problem by using a rotation matrix that rotates the normal vector of the flat circle, say (0,0,1) into the normal vector of the plane (1,1,1), and then applies it to your parameterized circle. However, you may not know how to do this yet. If you want to stick with vectors, you can think about if it is possible to construct two orthogonal unit vectors that are parallel with the plane. If so, you can combine those two vectors
  • #1
quietrain
655
2

Homework Statement


parameterize the following

a circle with radius 2 , centered at 1,2,3 and lies on the plane x+y+z=6


The Attempt at a Solution



ok i think i know how to get radius 2, centered 1,2,3
namely, r(t) = (1,2,3) + (2cos(t),2sin(t),t)

but how do i fit into the plane?

thanks!
 
Physics news on Phys.org
  • #2
Perhaps you could find a rotation matrix that rotates the normal vector of the flat circle, say (0,0,1) into the normal vector of the plane (1,1,1), and then apply it to your parameterized circle?

By the way, you wrote r(t) as a spiral, but I assume you meant it to be a circle.
 
  • #3
Filip Larsen said:
Perhaps you could find a rotation matrix that rotates the normal vector of the flat circle, say (0,0,1) into the normal vector of the plane (1,1,1), and then apply it to your parameterized circle?

By the way, you wrote r(t) as a spiral, but I assume you meant it to be a circle.

oh my z component should be 0 instead of t

but anyway,as for the rotation matrix, is it (cosx cosy z) ?

so that if i do (cosx cosy z) (0 0 1) i get (1 1 1)?

but i have no idea how to apply into the parameterized expression
 
  • #4
If you don't know what a rotation matrix is (yet) then there is no need to start worrying about those now. You can solve it without them.

If you want to stick with vectors, then perhaps you can think about if it is possible to construct two orthogonal unit vectors that are parallel with the plane. If so, perhaps you can combine those two vectors with a radius and some trigonometry to make a circle in the plane?
 
  • #5
erm i don't know if this is correct

(0 0 1)(0)
(0 0 1)(0)
(0 0 1)(1)
= (1 1 1) column

but i am free to play with the 6 zeros to the left... somehow i think its wrong?
 
  • #6
Filip Larsen said:
If you don't know what a rotation matrix is (yet) then there is no need to start worrying about those now. You can solve it without them.

If you want to stick with vectors, then perhaps you can think about if it is possible to construct two orthogonal unit vectors that are parallel with the plane. If so, perhaps you can combine those two vectors with a radius and some trigonometry to make a circle in the plane?

ok, 2 vectors parallel to plane are 0,1,-1 and 0,-1,1

but how do i continue from here :(
 
  • #7
I would use a completely different method.

The desired circle is the intersection of the plane x+ y+ z= 6 and the sphere with center at (1, 2, 3) (which does lie on the plane) and radius 2.

Parametric equations for the sphere are
[itex]x= 1+ 2cos(\theta)sin(\phi)[/itex]
[itex]y= 2+ 2sin(\theta)sin(\phi)[/itex]
[itex]z= 3+ 2cos(\phi)[/itex].

Its intersection with the plane requires that
[tex]x+ y+ z= 1+ 2+ 3+ 2(sin(\theta)+ cos(\theta))sin(\phi)+ 2 cos(\phi)= 6[/tex]
which reduces to
[tex]tan(\phi)= -\frac{1}{cos(\theta)+ sin(\theta)}[/tex]
So that we can write [itex]sin(\phi)[/itex] and [itex]cos(\phi)[/itex] in terms of sine and cosine of [itex]\theta[/itex].
(Think of a right triangle with "opposite side" -1 and "near side" [itex]-(cos\theta)+ sin(\theta))[/itex] and find the hypotenuse using the Pythagorean theorem.)

That gives parametric equations for the circle in terms of the parameter [itex]\theta[/itex].
 
  • #8
The two vectors (0,1,-1) and (0,-1,1), while indeed both parallel to the plane, are not orthogonal to each other (their inner product is not zero).

But since you have one vector, say a = (0,1,-1), which is parallel to the plane and thus orthogonal to the normal vector n=(1,1,1) you can easily make a third vector b that is orthogonal to both a and n (hint: its another kind of "product").

Assuming you have two orthogonal vectors you can make those into unit vectors by dividing them by their length, so you get two orthogonal unit vectors parallel to the plane. Let's call the a and b again.

Now back to the circle. You described a circle in the x-y plane by making combination of the x unit vector (1,0,0) and y unit vector (0,1,0) with sin and cos adding an offset and called it r. Can you make a similar construction with vector a and b instead of the x and y unit vector?
 
  • #9
Filip Larsen said:
The two vectors (0,1,-1) and (0,-1,1), while indeed both parallel to the plane, are not orthogonal to each other (their inner product is not zero).

But since you have one vector, say a = (0,1,-1), which is parallel to the plane and thus orthogonal to the normal vector n=(1,1,1) you can easily make a third vector b that is orthogonal to both a and n (hint: its another kind of "product").

Assuming you have two orthogonal vectors you can make those into unit vectors by dividing them by their length, so you get two orthogonal unit vectors parallel to the plane. Let's call the a and b again.

Now back to the circle. You described a circle in the x-y plane by making combination of the x unit vector (1,0,0) and y unit vector (0,1,0) with sin and cos adding an offset and called it r. Can you make a similar construction with vector a and b instead of the x and y unit vector?

ok i get another vector using cross product (2,-1,-1)

so normalizing them i get

a= 1/sqrt6 (2,-1,-1) and b = 1/sqrt2 (0,1,-1)

so a = rcost
b = rsint? , t being angle between a and b

so my parameterization at 1,2,3 and radius 2 will be

x = 1+ 4/√ 6 cost + 2(0)sint ( excluding 1, the first term gives me x component of a, 2nd term gives me x component of b right?)
y = 2 - 2/√ 6 cost + 2/√ 2 sint
z = 3 - 2/√ 6cost - 2/√ 2 sint

is this right? but i realize that since i only take into account the normal vector 1,1,1 of the plane, then the circle i create can occur in any plane that have 1,1,1 as its normal vector ? but the value 6 on the RHS of x+y+z = 6 tells me that the plane i want cuts the x,y,z axis at 6 each.

so how do i account for this stuff?
 
  • #10
HallsofIvy said:
I would use a completely different method.

The desired circle is the intersection of the plane x+ y+ z= 6 and the sphere with center at (1, 2, 3) (which does lie on the plane) and radius 2.

Parametric equations for the sphere are
[itex]x= 1+ 2cos(\theta)sin(\phi)[/itex]
[itex]y= 2+ 2sin(\theta)sin(\phi)[/itex]
[itex]z= 3+ 2cos(\phi)[/itex].

Its intersection with the plane requires that
[tex]x+ y+ z= 1+ 2+ 3+ 2(sin(\theta)+ cos(\theta))sin(\phi)+ 2 cos(\phi)= 6[/tex]
which reduces to
[tex]tan(\phi)= -\frac{1}{cos(\theta)+ sin(\theta)}[/tex]
So that we can write [itex]sin(\phi)[/itex] and [itex]cos(\phi)[/itex] in terms of sine and cosine of [itex]\theta[/itex].
(Think of a right triangle with "opposite side" -1 and "near side" [itex]-(cos\theta)+ sin(\theta))[/itex] and find the hypotenuse using the Pythagorean theorem.)

That gives parametric equations for the circle in terms of the parameter [itex]\theta[/itex].

let t = theta
so my hypoteneuse will be sqrt ( 1 + cos2t + 2costsint + sin2t)
= sqrt ( 2 + sin2t)

so sin(phi) = opposite / hypo
= -1 / sqrt ( 2+2sin2t)

cos(phi) = adj / hypo
= cost+sint / sqrt(2+2sin2t)

so i sub sin(phi) and cos(phi) back into the spherical parameterized equations of the sphere and i get the circle's parameterization?

so
x= 1 + 2 cost ( -1 / sqrt ( 2+2sin2t))
y = 2 +2 sint ( -1 / sqrt ( 2+2sin2t))
z = 3 + 2 (cost+sint / sqrt(2+2sin2t))

is this right?
 
  • #11
btw, is

x= 1 + 2 cost ( -1 / sqrt ( 2+2sin2t))
y = 2 +2 sint ( -1 / sqrt ( 2+2sin2t))
z = 3 + 2 (cost+sint / sqrt(2+2sin2t))

equals to

x = 1+ 4/√ 6 cost + 2(0)sint
y = 2 - 2/√ 6 cost + 2/√ 2 sint
z = 3 - 2/√ 6cost - 2/√ 2 sint

?
 
  • #12
quietrain said:
ok i get another vector using cross product (2,-1,-1)

so normalizing them i get

a= 1/sqrt6 (2,-1,-1) and b = 1/sqrt2 (0,1,-1)

so a = rcost
b = rsint? , t being angle between a and b

so my parameterization at 1,2,3 and radius 2 will be

x = 1+ 4/√ 6 cost + 2(0)sint ( excluding 1, the first term gives me x component of a, 2nd term gives me x component of b right?)
y = 2 - 2/√ 6 cost + 2/√ 2 sint
z = 3 - 2/√ 6cost - 2/√ 2 sint

is this right?

Looks correct to me. If you clean up the zero term in the expression for x I'd say you have a solution.

but i realize that since i only take into account the normal vector 1,1,1 of the plane, then the circle i create can occur in any plane that have 1,1,1 as its normal vector ? but the value 6 on the RHS of x+y+z = 6 tells me that the plane i want cuts the x,y,z axis at 6 each.

In your solution above you did the right thing and translated the circle (which originally has center in the origin) to (1,2,3). Since (1,2,3) is in the plane (1+2+3=6) the center is in the plane, and since the circle itself is parallel to the plane (because the plane is being spanned by any linear combination of the two vectors a and b) the circle must now also be in the plane.
 
  • #13
Filip Larsen said:
Looks correct to me. If you clean up the zero term in the expression for x I'd say you have a solution.



In your solution above you did the right thing and translated the circle (which originally has center in the origin) to (1,2,3). Since (1,2,3) is in the plane (1+2+3=6) the center is in the plane, and since the circle itself is parallel to the plane (because the plane is being spanned by any linear combination of the two vectors a and b) the circle must now also be in the plane.

oh is this coincidence that 1,2,3 happen to be 6? what if x,y,z = 7?
 
  • #14
quietrain said:
oh is this coincidence that 1,2,3 happen to be 6? what if x,y,z = 7?
No, it is not a coincidence. The problem asked for the circle with center (1, 2, 3) lieing in the plane x+ y+ z= 6. If x+ y+ z, for point (x, y, z) were NOT 6, then that point would not be in that plane and there would be no such circle.
 
  • #15
quietrain said:
let t = theta
so my hypoteneuse will be sqrt ( 1 + cos2t + 2costsint + sin2t)
= sqrt ( 2 + sin2t)

so sin(phi) = opposite / hypo
= -1 / sqrt ( 2+2sin2t)

cos(phi) = adj / hypo
= cost+sint / sqrt(2+2sin2t)

so i sub sin(phi) and cos(phi) back into the spherical parameterized equations of the sphere and i get the circle's parameterization?

so
x= 1 + 2 cost ( -1 / sqrt ( 2+2sin2t))
y = 2 +2 sint ( -1 / sqrt ( 2+2sin2t))
z = 3 + 2 (cost+sint / sqrt(2+2sin2t))

is this right?

erm so is this right?
 

1. What is a circle in plane parameterization?

A circle in plane parameterization is a way of representing a circle on a two-dimensional plane using mathematical equations. It involves defining a set of parameters that determine the position, size, and shape of the circle, and using those parameters to plot points on the plane.

2. How is a circle in plane parameterization different from other ways of representing a circle?

Unlike other methods of representing a circle, such as using geometric constructions or Cartesian coordinates, a circle in plane parameterization allows for more flexibility and precision in defining the circle's properties. It also makes it easier to manipulate the circle mathematically, which is useful in many scientific and engineering applications.

3. What are the basic parameters used in a circle in plane parameterization?

The most commonly used parameters are the center coordinates (x,y) and the radius of the circle. Other parameters may include the orientation of the circle, any eccentricity or distortion, and the equation used to define the circle.

4. How is a circle in plane parameterization used in real-world applications?

Circle in plane parameterization is used in a variety of fields, including mathematics, physics, engineering, and computer graphics. It is used to model circular objects and phenomena, such as planetary orbits, gears, and lenses. It is also used in computer graphics to render 2D and 3D graphics, as well as in control systems and signal processing.

5. Are there any limitations to using a circle in plane parameterization?

One limitation is that a circle in plane parameterization can only represent a perfect circle, meaning all points on the circle are equidistant from the center. It cannot accurately represent imperfect or irregular circles. Additionally, the choice of parameters and equations used can affect the precision and accuracy of the representation. It is important to carefully consider the application and choose appropriate parameters for the desired level of accuracy.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
21
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
878
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
841
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
517
  • Calculus and Beyond Homework Help
Replies
14
Views
655
Replies
8
Views
669
  • Calculus and Beyond Homework Help
Replies
3
Views
556
Back
Top