Intersection of a sphere and plane

In summary, the circle that is the intersection of the plane x + y + z = 0 and the sphere x2 + y2 + z2 = 1 can be expressed as: x(t) = [cos(t)-sqrt(3)sin(t)]/sqrt(6), y(t) = [cos(t)+sqrt(3)sin(t)]/sqrt(6), z(t) = -[2cos(t)]/sqrt(6).
  • #1
yy205001
60
0

Homework Statement


Show that the circle that is the intersection of the plane x + y + z = 0 and the
sphere x2 + y2 + z2 = 1 can be expressed as:

x(t) = [cos(t)-sqrt(3)sin(t)]/sqrt(6)
y(t) = [cos(t)+sqrt(3)sin(t)]/sqrt(6)
z(t) = -[2cos(t)]/sqrt(6)


Homework Equations





The Attempt at a Solution


At first, I tried to let z=-x-y then sub it into the equation of the sphere:
x2+y2+(-x-y)2=1
x2+y2+x2+2xy+y2=1
2x2+2xy+2y2 = 1
And i stuck at here, I can't make it into a circle equation.

So, I changed another way.
I equated two equations, so it becomes:
x2 + y2 + z2 -1 = x+y+z
x2+ x + y2 + y + z2 +Z -1 =0
(x-1/2)2 + (y-1/2)2 + (z-1/2)2 = 7/4 *By completing the square

But isn't this is an equation of a sphere??
Shouldn't it just a circle?

Any help is appreciated!
 
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  • #2
hi yy205001! :smile:
yy205001 said:
Show that the circle that is the intersection of the plane x + y + z = 0 and the
sphere x2 + y2 + z2 = 1 can be expressed as:

x(t) = [cos(t)-sqrt(3)sin(t)]/sqrt(6)
y(t) = [cos(t)+sqrt(3)sin(t)]/sqrt(6)
z(t) = -[2cos(t)]/sqrt(6)

why don't you just bung the three parameter formulas into the two original equations, and show that they work? :confused:
I equated two equations, so it becomes:
x2 + y2 + z2 -1 = x+y+z

but that also gives you the intersections of x + y + z = k and the
sphere x2 + y2 + z2 = 1 + k, for all values of k ! :wink:
 
  • #3
tiny-tim said:
hi yy205001! :smile:


why don't you just bung the three parameter formulas into the two original equations, and show that they work? :confused:

That would show the parametric curve lies on the circle. But is it the whole circle? Looks like there is more to do.
 
  • #4
And I am thinking is there anything do with the normal vector of the plane.
 
  • #5
what does that have to do with the parametric equations? :confused:
 
  • #6
tiny-tim said:
what does that have to do with the parametric equations? :confused:

So, I just sub in those 3 parametrize equations into the equation for the plane and sphere, then show they satisfy x2+y2+z2=1 and x+y+z=0?
 
  • #7
yes

the question starts "show", so all you need to do is show that the answer is correct, you don't have to find the answer from scratch :smile:

(and, as LCKurtz says, you'll also have to prove that that parametrisation gives the whole intersection)
 
  • #8
I got it!
tiny-tim, LCKurtz, thank you so much for the help!
 

Related to Intersection of a sphere and plane

1. What is the intersection of a sphere and a plane?

The intersection of a sphere and a plane is the set of points where the sphere and the plane meet or cross each other.

2. How many points can there be in the intersection of a sphere and a plane?

The number of points in the intersection of a sphere and a plane can vary, but it can be anywhere from 0 to infinite.

3. Can the intersection of a sphere and a plane be a circle?

Yes, the intersection of a sphere and a plane can be a circle. This occurs when the plane passes through the center of the sphere.

4. What determines the shape of the intersection of a sphere and a plane?

The shape of the intersection of a sphere and a plane is determined by the relative position and orientation of the sphere and the plane. It can be a point, a circle, a line, or another shape depending on these factors.

5. How can the intersection of a sphere and a plane be calculated?

The intersection of a sphere and a plane can be calculated using mathematical equations and algorithms. This involves determining the distance between the center of the sphere and the plane, as well as the radius of the sphere and the orientation of the plane. Various software and programs can also be used for this calculation.

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