# Intersection of a sphere and plane

1. May 15, 2013

### yy205001

1. The problem statement, all variables and given/known data
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)

2. Relevant equations

3. 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!

2. May 15, 2013

### tiny-tim

hi yy205001!
why don't you just bung the three parameter formulas into the two original equations, and show that they work?
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 !

3. May 15, 2013

### LCKurtz

That would show the parametric curve lies on the circle. But is it the whole circle? Looks like there is more to do.

4. May 15, 2013

### yy205001

And I am thinking is there anything do with the normal vector of the plane.

5. May 16, 2013

### tiny-tim

what does that have to do with the parametric equations?

6. May 16, 2013

### yy205001

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. May 16, 2013

### tiny-tim

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

(and, as LCKurtz says, you'll also have to prove that that parametrisation gives the whole intersection)

8. May 16, 2013

### yy205001

I got it!
tiny-tim, LCKurtz, thank you so much for the help!