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Circle in plane parameterization

  • Thread starter quietrain
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  • #1
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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!
 

Answers and Replies

  • #2
Filip Larsen
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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
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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
Filip Larsen
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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
654
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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
654
2
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
HallsofIvy
Science Advisor
Homework Helper
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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
Filip Larsen
Gold Member
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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. Lets 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
654
2
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. Lets 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 realise 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
654
2
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
654
2
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
Filip Larsen
Gold Member
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183
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 realise 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
654
2
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
HallsofIvy
Science Advisor
Homework Helper
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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
654
2
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?
 

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