Parameterise the circle that results from intersecting the unit sphere in centered at the origin with the plane

I tried

however the last equation is an ellipse????

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- January 24th 2011, 12:24 PMFGT12Parameterising a certain circle
Parameterise the circle that results from intersecting the unit sphere in centered at the origin with the plane

I tried

however the last equation is an ellipse???? - January 24th 2011, 12:43 PMahaok
- January 24th 2011, 07:23 PMxxp9
You result is only the projection of the circle to the xy plane, which is really an ellipse.

- January 25th 2011, 07:36 AMFGT12
so how would you parameterise the circle in

- January 25th 2011, 09:57 AMMazerakham
It's definitely going to be a circle, because really: how could a plane intersect a sphere and make an ellipse! You're right that it's absurd.

Have you tried drawing a sketch? Try visualizing the picture with the y axis pointing directly out of the paper, and that may give you some insight. I solved the problem by first coming up with an equation based on the sketch, then proving that it was right. So, like I said. Start with the picture. - January 25th 2011, 11:06 AMHallsofIvy
ahaok is wrong, a sphere cut by a plane gives a circle, not an ellipse. xxp9 is right- a circle, projected onto a plane at an angle to a line perpendicular to the circle gives an ellipse. That is why you get the equation of an ellipse- you are projecting the circle onto the xy-plane, not parameterizing it.

Here, your sphere is the unit sphere, and the plane is given by x= z. The "standard" parameterization of the unit sphere can be derived from spherical coordinates:

by taking the radius variable, equal to 1:

The fact that z= x means that we must have so that .

So one way of getting the parameterization of the circle is to replace by , reducing from two parameters to one:

Of course, . And, we can use to say that

and so have

. - January 25th 2011, 11:34 AMTheEmptySet
The vector form for the plane is

This gives that Using the equation of the sphere gives

So a parametric representation of the circle is

Since the plus or minus is really messy it can be eliminated using trigonometry.

If we define this gives

Using this we get

Attachment 20599

http://www.mathhelpforum.com/math-he...2&d=1295983968

Edit: too slow