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Sep 22 2015 09:18pm
problem is solved and done.
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Sep 23 2015 07:13pm
Quote (FamilyGuyViewer @ Sep 22 2015 10:05am)
its gonna come out to be x=0,y=0,z=0

which obviously doesnt fit the constraint


It was a rounding issue from the professor that caused a problem meant to have a free variable not to have it, right?
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Sep 23 2015 10:48pm
Quote (Amaston @ Sep 23 2015 08:13pm)
It was a rounding issue from the professor that caused a problem meant to have a free variable not to have it, right?


wat do u mean
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Sep 24 2015 06:13pm
Quote (FamilyGuyViewer @ Sep 20 2015 11:12am)
yea they are approximate solution cause he rounded it lol


^A solution doesn't exist as it is written, as we've shown. If the coefficients were slightly different (probably not rounded vs rounded) in the initial 3 linear equations, there would've been a free variable (REF form would have a row of zeros), so there would be infinitely many solutions to this system (i.e. the 3 planes would have met on an entire line instead of just a single point). Given this, that line would've hit the sphere at two places and exact solutions satisfying the sphere constraint would have existed.

This post was edited by Amaston on Sep 24 2015 06:15pm
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