Express one vector as a linear combination of others

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The discussion revolves around demonstrating that the set of vectors {(1,2,3), (3,2,1), (5,5,5)} is linearly dependent by expressing one vector as a linear combination of the others. The user systematically sets up equations based on the assumption of linear dependence and derives relationships among the coefficients a, b, and c. After some confusion, they realize that by choosing a specific value for a, they can express b and c accordingly, ultimately confirming the linear dependence. The calculations show that substituting these values back into the original equation yields a valid identity. The user concludes that they have successfully shown the linear dependence of the vectors.
Saladsamurai
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Homework Statement



Show that the set of vectors is linearly dependent (LD) by expressing one vector as a linear combination (LC) of the others:

{(1,2,3), (3,2,1), (5,5,5)}


The Attempt at a Solution



I would like to do this systematically (without guess and check). So I assumed that if the set is LD, then there exists some values of a,b,c not all zero such that:

a(1,2,3) + b(3,2,1) + c(5,5,5) = 0

hence,

a + 3b +5c = 0
2a + 2b +5c = 0
3a + 1b +5c = 0

Now I should be able to solve for 2 of the parameters (assuming only one of the equations is not independent, else I an solve for 1 ...)) a,b,c in terms of the third which would remain arbitrary.

Eliminating c from the first and second equations, we find that a = b .

Eliminating b, from the second and third equations we find that c = -4a/5 .

Now, if I plug these values back into anyone of the equations, I simply get the identity. What am I missing here? Am I messing something up? Or are they ALL multiples of each other?

Thanks!
 
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I'm confused. Haven't you done what you wanted to? You seem to have found everything in terms of a.
 
hgfalling said:
I'm confused. Haven't you done what you wanted to? You seem to have found everything in terms of a.

Oh. Maybe I have :redface: So here, a is my arbitrary parameter. Say, I let a = 1, then b = 1 and c = -4/5. Plugging these into the original equation, yields:

(1,2,3) + (3,2,1) - (4,4,4) = 0 which is indeed true.

I'm a little slow tonight. :smile:
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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