The proof of mass vector.
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From: Bilge on 20 Apr 2006 21:59 Ka-In Yen: >According to mathematician's opinion: >"Again there are two unknowns, |V| and u, in the equation so there are >infinitely many answers. Therefore cross division is also undefined." > ---- http://www.mcasco.com/qa_vdq.html > >That's not true. Sure it is. There are infinitely many ways to define the vectors A and B such that the cross product is the same. >Stupid mathematicians hinder the development of 3D vector algebra. No, stupid people simply don't bother to stop and think before declaring that thousands of mathematicians are stupid. >An area vector is A<i:3>m^2, and its length is l<j:3>m. >where <i:3> and <j:3> are unit vectors and m is meter. >We can divide the area vector by the length vector, and >we get the height(vector) of rectangle. The cross product of two vectors is a pseudovector. >A<i:3>m^2 / l<j:3>m >=(A/l) <i:3>x<j:3> m (x is cross product) >=(A*sin(theta)/l) <k:3> m (<k:3>=(<i:3>x<j:3>)/sin(theta)) >where theta is the angle between <i:3> and <j:3>. ><k:3> is a unit vector and perpendicular to <i:3> and <j:3>. I'm not about to sort out your silly notation. > >As soon as you accept Clifford's method, you will realize that >LXB is vector by vector division. Do yourself a favor. Purchase a copy of ``Clifford Algebras and Spinors,'' Perti Lounesto.
From: Ka-In Yen on 21 Apr 2006 20:08 Bilge wrote: > Ka-In Yen: > >A<i:3>m^2 / l<j:3>m > >=(A/l) <i:3>x<j:3> m (x is cross product) > >=(A*sin(theta)/l) <k:3> m (<k:3>=(<i:3>x<j:3>)/sin(theta)) > >where theta is the angle between <i:3> and <j:3>. > ><k:3> is a unit vector and perpendicular to <i:3> and <j:3>. > > I'm not about to sort out your silly notation. I have define the notation in my paper. If you do not read my paper, how can you comment on my paper?
From: Bilge on 22 Apr 2006 13:04 Ka-In Yen: > >Bilge wrote: >> Ka-In Yen: >> >A<i:3>m^2 / l<j:3>m >> >=(A/l) <i:3>x<j:3> m (x is cross product) >> >=(A*sin(theta)/l) <k:3> m (<k:3>=(<i:3>x<j:3>)/sin(theta)) >> >where theta is the angle between <i:3> and <j:3>. >> ><k:3> is a unit vector and perpendicular to <i:3> and <j:3>. >> >> I'm not about to sort out your silly notation. > >I have define the notation in my paper. If you do not read >my paper, how can you comment on my paper? I'm commenting on the article you posted. This is a newsgroup, not a journal.
From: Ka-In Yen on 23 Apr 2006 22:12 Bilge wrote: > Ka-In Yen: > > >According to mathematician's opinion: > >"Again there are two unknowns, |V| and u, in the equation so there are > >infinitely many answers. Therefore cross division is also undefined." > > ---- http://www.mcasco.com/qa_vdq.html > > > >That's not true. > > Sure it is. There are infinitely many ways to define the vectors > A and B such that the cross product is the same. Although there are infinite solutions, but we are not forbidden to find a specific solution. > > > >As soon as you accept Clifford's method, you will realize that > >LXB is vector by vector division. > > Do yourself a favor. Purchase a copy of ``Clifford Algebras and > Spinors,'' Perti Lounesto. Inverse of a vector is widely accepted by mathematicians and physicists. A^(-1) = 1/A = A/A^2 (where A is a vector) Please refer to: http://en.wikipedia.org/wiki/Geometric_algebra
From: Bilge on 24 Apr 2006 21:21
Ka-In Yen: > >Bilge wrote: >> Ka-In Yen: >> >> >According to mathematician's opinion: >> >"Again there are two unknowns, |V| and u, in the equation so there are >> >infinitely many answers. Therefore cross division is also undefined." >> > ---- http://www.mcasco.com/qa_vdq.html >> > >> >That's not true. >> >> Sure it is. There are infinitely many ways to define the vectors >> A and B such that the cross product is the same. > >Although there are infinite solutions, but we are not forbidden >to find a specific solution. What's your point? >> > >> >As soon as you accept Clifford's method, you will realize that >> >LXB is vector by vector division. >> >> Do yourself a favor. Purchase a copy of ``Clifford Algebras and >> Spinors,'' Perti Lounesto. > >Inverse of a vector is widely accepted by mathematicians and >physicists. You said ``division by a vector.'' Don't change the subject. >A^(-1) = 1/A = A/A^2 (where A is a vector) Note that the denominator is a scalar. |