I\'m not sure how they\'re getting the final step here. Isee where the \'s and \

Question

I'm not sure how they're getting the final step here. Isee where the 's and 's come from, but the not thee's... for a rigid body m at (a,0,0) 2m at (0,a,a) and 3m at((0,a,-a). We find that 1=10ma^2 and plug it in to get       0=0       y=z=0 2=7ma^2 and plug it in to get       x=0       y=z 3=5ma^2 and plug it in to get        x=0        y= - z Then the conclusion is e1=(1,0,0)                                   e2=1/2(0,1,1)                                   e3=1/2(0,1,-1) It's the conclusion and don't follow. thanks. I'm not sure how they're getting the final step here. Isee where the 's and 's come from, but the not thee's... for a rigid body m at (a,0,0) 2m at (0,a,a) and 3m at((0,a,-a). We find that 1=10ma^2 and plug it in to get       0=0       y=z=0 2=7ma^2 and plug it in to get       x=0       y=z 3=5ma^2 and plug it in to get        x=0        y= - z Then the conclusion is e1=(1,0,0)                                   e2=1/2(0,1,1)                                   e3=1/2(0,1,-1) It's the conclusion and don't follow. thanks.

Explanation / Answer

okay, i'm responding to my own q, because i think ifigured it out...this might not be right! it looks like the system is symmetrical about the x-axis, sothe we don't change in the first vector, where y and z are 0. now look at the value (0,1,1) . it goes over one up one the =45degrees, and thecos(45)=2/2=1/2, thus giving the angle by which wemust rotate this vector, so we multiply 1/2(0,1,1) to gete2. then e3 is found similarly.

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