Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**debjit625****Member**- Registered: 2012-07-23
- Posts: 101

Find the radius of the circular section of sphere

by the planeMy answer came out to

Did I did it correctly,I am bit confused over the question.

Thanks

Debjit Roy

___________________________________________________

The essence of mathematics lies in its freedom - Georg Cantor

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,979

hi debjit625

I get that too.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**debjit625****Member**- Registered: 2012-07-23
- Posts: 101

Thanks Bob

Debjit Roy

___________________________________________________

The essence of mathematics lies in its freedom - Georg Cantor

Offline

how do you do it?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,979

Give me a moment while I look up an url

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Okay

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,979

OK. Got it.

You could treat the two equations as simultaneous and find the equation of their intersection. Buit this is hard work. So here's a quicker method.

http://www.enm.bris.ac.uk/admin/courses … ov2009.pdf

Section called Point and a Plane. 3.37

n is the normal to the plane ie n = (3,2,-1) and take a as a = (0,0,0)

d = 5root14

once you have p you can use pythag on the radius of the sphere as the hypotenuse and find r, the radius of the circle.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**debjit625****Member**- Registered: 2012-07-23
- Posts: 101

Well I did it like this

By equation of the sphere its already given that the sphere is at origin (0,0,0).

Now from there we get radius of the sphere is

Now we have to find the distance 'd' of the plane from origin and that will be

Now use Pythagoras theorem to find the radius of the section by the plane

Good Luck

*Last edited by debjit625 (2014-01-23 03:56:16)*

Debjit Roy

___________________________________________________

The essence of mathematics lies in its freedom - Georg Cantor

Offline

Pages: **1**