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#26 2014-07-16 01:59:59

Agnishom
Real Member
From: The Complex Plane
Registered: 2011-01-29
Posts: 15,932
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Re: PerpendicularQ

A friend.


'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'
'Humanity is still kept intact. It remains within.' -Alokananda

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#27 2014-07-16 02:04:26

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,624

Re: PerpendicularQ

I will look at it today but do not expect too much.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#28 2014-08-17 03:03:56

Agnishom
Real Member
From: The Complex Plane
Registered: 2011-01-29
Posts: 15,932
Website

Re: PerpendicularQ

Solution

pZGPc1A.jpg
Let us call the curves c1, c2 and c3

It is evident that:

c1 = (cos t, sin t, 0)
c2 = (cos t, 0, sin t)
c3 = (0, sin t, cos t)

[I am considering the circles to be centered at the origin and of radius 1. This is just to make the algebra simple]

You wanted me to prove that c1 intersects c2 making an angle of pi/2 and c2 intersects c3 making an angle of pi/2 and c3 intersects c1 making an angle of pi/2

Consider the intersection of c1 and c2. At this point, c1(t) = c2(t)
Or, (cos t, sin t, 0) = (cos t, 0, sin t)
Solving, you get t = 0

Now, c1(0) = c2(0) = (1, 0, 0)

Differentiating the curves,
c1' = (-sin t, cos t, 0)
c2' = (-sin t, 0, cos t)

At t=0, c1'(0) = (0, 1, 0) and c2'(0) = (0, 0, 1)

Let x1 be the tangent of c1 at the intersection point
We have, x1= c1(0) + t(c1'(0)) = (1, 0, 0) + t(0, 1, 0) = (1, t, 0)

Let x2 be the tangent of c2 at the intersection point
We have, x2= c2(0) + t(c2'(0)) = (1, 0, 0) + t(0, 0, 1) = (1, 0, t)

By the angle between c1 and c2, we obviously mean the angle between x1 and x2.
Let that angle be  θ

x1 . x2 = (1, t, 0) . (1, 0, t) = 1*1 + t*0 + t*0 = 1

|x1| = Sqrt[1+t^2]
|x2| = Sqrt[1+t^2]

At the intersection point, t=0. So,


Similarly, you can go on proving the same thing for the two other intersection points (but that has been left to the reader as an exercise tongue)

QED


'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'
'Humanity is still kept intact. It remains within.' -Alokananda

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