Curve of Intersection in two Three-Dimensional EQs

In summary, the curve of intersection of the surfaces x^2 + y^2 + z^2 = 4 and x + y + z = 1 is a circle with center at (1/3, 1/3, 1/3) and radius of 3.
  • #1
Reflected
13
0
Determine the curve of intersection of the surfaces x^2 + y^2 + z^2 = 4 and x + y + z = 1. The curve should be in parametric form.

With this problem, I'm really not sure what direction to go in. I had thought about using the quadratic formula in some manner, but really unclear. Any advice would be great! Thanks!
 
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  • #2
First, think about it. Do you know what kind of surfaces each of these are? If so, do you see how they can intersect?
 
  • #3
I'm actually not too sure; I want to graph them in Mathematica, but I don't know the command.

edit: Ok, I graphed them in Mathematica as follows

first = ContourPlot3D[{x^2 + y^2 + z^2 - 4}, {x, -10, 10}, {y, -10,
10}, {z, -10, 10}];
second = ContourPlot3D[{(x + y + z - 1)}, {x, -10, 10}, {y, -10,
10}, {z, -10, 10}];
Show[first, second]

The first function is simply a sphere while the second is just a plane intersecting it.
I'm still unsure how to find the curve of intersection, although I can now visually see it.
 
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  • #4
The next step you should work on is parametrizing the plane. Can you do this?
 
  • #5
Mathdope said:
The next step you should work on is parametrizing the plane. Can you do this?

By parametrizing the plane, are you referring to the second equation? If so, can't you just do something like:

x = u
y = v
z = 1 - u - v
 
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  • #6
Yep. Now how about the sphere?
 
  • #7
Since its a sphere of radius 2 it should be:
x = 2sinΦcosΘ
y = 2sinΦsinΘ
z = 2cosΦ
 
  • #8
Now the tricky part. You need to eliminate a parameter so that your curve of intersection expresses the set of points of the intersection with one parameter (this way it's a curve rather than a surface).
 
  • #9
I'm really not sure how to attempt doing this: I tried messing around in Mathematica by trying to somehow equate the two parameterizations together such as...

x = 2 Sin Cos[v],
y = 2 Sin Sin[v]
z = 1 - 2 SinCos[v] - 2 Sin Sin[v]

Graphing this in Mathematica gives a slanted circular plane and not the curve that we're looking for.
 
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  • #10
Above you aren't using the spherical definition for z.

Edit: Hmmm, this is harder than I thought, hope I didn't bring you to a dead end. I'll keep working on it.

Edit 2: Somehow you need to combine the two equations together to get something like U^2 + V^2 = constant^2, at which point you then declare U = (constant) cos theta and V = (constant) sin theta, but how to get U and V here is not something I am able to discover yet. Anybody else have a clue here?
 
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  • #11
Mathdope said:
Above you aren't using the spherical definition for z.

Edit: Hmmm, this is harder than I thought, hope I didn't bring you to a dead end. I'll keep working on it.

Ok, thanks for your help!
 
  • #13
Ai. I'm also trying to solve this problem. The examples they show only have 2 variables for the Planes instead of three variables, so their methods work easily.
 
  • #14
Reflected, are you by any chance in my class? lol
 
  • #15
Frankly, I am concerned that Reflected seems more concerned with learning to use Mathematica than with learning mathetmatics! If you are expected to be able to do problems like this one, you certainly should not need Mathematica to realize that x2+ y2+ z2= 4 is a sphere with center at the origin and radius 2, or that x+ y+ z= 1 is a plane through (0, 0, 1), (0, 1, 0), and (1, 0, 0). Tools are good, crutches are not.

The intersection of the two is a circle with center at (1/3, 1/3, 1/3) (that is the point on x+ y+ z= 1 closest to (0,0,0), the center of the sphere). The distance from (0, 0, 0) to that point is [itex]\sqrt{(1/3)^2+ (1/3)^2+ (1/3)^2}= \sqrt{3}/3[/itex]. From the Pythagorean theorem then, the radius of the circle of intersection is [itex]\sqrt{4- 1/3}= \sqrt{33}/3[/itex].

It's not really necessary to know all that to find the parameterization of the circle. Here's how I would do it.

From the plane equation, z= 1- x- y. Putting that into the equation of the sphere, [itex]x^2+ y^2+ (1- x- y)^2= x^2+ y^2+ 1- 2x- 2y+ x^2+ y^2+ 2xy= 2x^2+ 2y^2+ 2xy- 2x- 2y= 3[/itex]. Dividing through by 2, [itex]x^2+ y^2+ xy- x-y= 3/2[/itex]. We can eliminate the "xy" term by making the substitutions u= x+ y, v= x- y so that x= (1/2)u+ (1/2)v, y= (1/2)u- (1/2)v and the equation becomes [itex](3/4)u^2- u+ (1/4)v^2= 3[/itex], eliminating the "mixed" term. completing the square in "u" part, we have [itex](u- 2/3)^2+ (1/3)v^2= 40/9[itex] or [itex](9/40)(u- 2/3)^2+ (3/40)v^2= 1[/itex]. An obvious parameterization for that is [itex]u= (3/2\sqrt{10})cos(\theta)+ 2/3[/itex], v= (\sqrt{3}/2\sqrt{10})sin(\theta)[/itex]. Now, work backwards using u= x+ y, v= x- y to get x and y in terms of [itex]\theta[/itex] and then use z= 1- x- y to get z in terms of [itex]\theta[/itex].

Modulo any arithmetic errors that should do it.
 
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  • #16
I don't understand how you parametrizted the final equation after all of the substitution and completing the square.
 
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  • #17
Not sure if you need still this or not but what I did was set z = 1 - x - y
plug it in x^2+y^2+z^2=4 and get this long equation, simplified it and got a xy term. then I tried to put equation so that it's quadratic ax^2+bx+c = 0 then used quadratic formula and go x = f(y); set y = t so x = f(t) , plug everything back in z = 1- x -y and got the parametric function for z. Boy, this problem was driving me nuts and now it's over. LOL
 
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  • #18
Reflected, this should work. I got my equation already and ParametricPlot3D'ed in Mathematica and showed me this beautiful (T_T) circle around the sphere and on the plane. Try it. =)
 
  • #19
dragon4chen said:
Not sure if you need still this or not but what I did was set z = 1 - x - y
plug it in x^2+y^2+z^2=4 and get this long equation, simplified it and got a xy term. then I tried to put equation so that it's quadratic ax^2+bx+c = 0 then used quadratic formula and go x = f(y); set y = t so x = f(t) , plug everything back in z = 1- x -y and got the parametric function for z. Boy, this problem was driving me nuts and now it's over. LOL

yes, i tried doing that as well. How did you set that expression into the quadratic formula?
You have 2x^2 + 2xy - 2x + 2y^2 -2y = 3
so how does that go into the quad form?
 
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  • #20
erm i set it so that
2y^2 +(2x-2)y+(2x^2-2x-3) .. do you see the a b and c now?
OH.. one thing.. you in my class tho? Lol i got the exact same problem. Ohlone. Bradshaw?
Hope this helps you a lot.
a = 2
b = 2x-2
c = 2x^2-2x-3
 
  • #21
Hmm okay.. so after you plug everything into the quad formula you get something like
[itex] (-2x + 2 +- (\sqrt{(-12x^2+8x+28)}))/4 [/itex]
What would you do at this point? and yes, I am in your class apparently
 
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  • #22
at that point you should have y = f(x) .. put x = t so y = f(x) then plug 'em in for z .. and BAM. you got your parametric equation for the circle
 
  • #23
Hmm... I'm still a bit confused.
By y = f(x) do you mean [itex] y = -2x + 2 +- (\sqrt{(-12x^2+8x+28)})/4 [/itex]
and then just substitute all of those X's for T's?
 
  • #24
Yep... That should do the trick
 
  • #25
Oh. try to reduce it tho. you can factor out 4 out of squareroot and 2 comes out then cancel out the 2s by the 4
 
  • #26
So after factoring it you get
[itex] x=-x + 1 ± \sqrt{-3x^2+2x+7} / 2 [/itex]
and your saying to set it so..
[itex] x=-t + 1 ± \sqrt{-3t^2+2t+7} / 2 [/itex]

y = t

z=1-(-t + 1 ± \sqrt{-3t^2+2x+7} / 2) - t

Also, what do u do with the ± sign?

edit: nevermind, i figured it out, thanks for your help.
 
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  • #27
What kind of sadistic prof assigned this?
 
  • #28
The one who loves math. lol.. great prof though.
 

1. What is a curve of intersection in two three-dimensional equations?

A curve of intersection is the set of points where two three-dimensional equations intersect in three-dimensional space. It represents the points that satisfy both equations simultaneously.

2. How many curves of intersection can two three-dimensional equations have?

Depending on the equations, there can be zero, one, or an infinite number of curves of intersection. It is possible for two equations to have no points of intersection in three-dimensional space, or they can have one point of intersection that creates a single curve, or they can have an infinite number of points that create a continuous curve.

3. What is the significance of a curve of intersection in solving equations?

A curve of intersection helps us find the points that satisfy both equations simultaneously. This can be useful in solving systems of equations, where we are looking for the values of variables that satisfy multiple equations at the same time.

4. Can a curve of intersection exist in two dimensions?

No, a curve of intersection only exists in three-dimensional space. In two dimensions, two equations will intersect at a point, creating a single point of intersection rather than a curve.

5. How can we visualize a curve of intersection in three-dimensional space?

A curve of intersection can be visualized as the intersection of two planes in three-dimensional space. It can also be represented graphically as a curve on a three-dimensional coordinate system.

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