Intersection of a sphere and plane

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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!
 

Answers and Replies

  • #2
tiny-tim
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hi yy205001! :smile:
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
LCKurtz
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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
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And I am thinking is there anything do with the normal vector of the plane.
 
  • #5
tiny-tim
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what does that have to do with the parametric equations? :confused:
 
  • #6
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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
tiny-tim
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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
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I got it!
tiny-tim, LCKurtz, thank you so much for the help!
 

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