Introduction to Vector calculations (Calculus 3)

Pinedas42
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Homework Statement


Find nonzero scalars a, b, c, such that au+b(u-v)+c(u+v)=0 for every pair of vectors u and v

This isn't a homework question, more of a conceptual exercise, but I want to understand it thoroughly.

The Attempt at a Solution


I've gone to u(a+b+c) + v(c-b)=0
then I couldn't quite figure where to go next. There is so many unknowns at once it's a little disorienting where to start first.

Then I figured to try splitting it into the vector pairs,
a(u1,u2)+b(u1-v1, u2-v2)+c(u1+v1, u2+v2)=0

but I am still stumped as to where to go from here. It seems like it's painfully simple but I'm not quite seeing it.

EDIT1: I've attempted putting the vectors in a system of equations;

u1a+u1b+u1c-v1b+v1c=0
u2a+u2b+u2c-v2b+v2c=0

but once again I hit a dead end and only get the scalars equaling zero.
 
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Pinedas42 said:

Homework Statement


Find nonzero scalars a, b, c, such that au+b(u-v)+c(u+v)=0 for every pair of vectors u and v

This isn't a homework question, more of a conceptual exercise, but I want to understand it thoroughly.

The Attempt at a Solution


I've gone to u(a+b+c) + v(c-b)=0
then I couldn't quite figure where to go next. There is so many unknowns at once it's a little disorienting where to start first.

Then I figured to try splitting it into the vector pairs,
a(u1,u2)+b(u1-v1, u2-v2)+c(u1+v1, u2+v2)=0

but I am still stumped as to where to go from here. It seems like it's painfully simple but I'm not quite seeing it.

EDIT1: I've attempted putting the vectors in a system of equations;

u1a+u1b+u1c-v1b+v1c=0
u2a+u2b+u2c-v2b+v2c=0

but once again I hit a dead end and only get the scalars equaling zero.
au+b(u-v)+c(u+v)=0 has to be true for any pair of vectors u & v.

u(a+b+c) + v(c-b)=0 is equivalent to au+b(u-v)+c(u+v)=0. So you can't count on anything from u(a+b+c) cancelling anything from v(c-b). Therefore, each of those has to be zero:
u(a+b+c) = 0

v(c-b) = 0​
The solution for a, b, and c, isn't unique, but you can find a solution that does the job.
 
So what I get is

a=-2c
and
b=c

the actual numbers inside will be arbitrary? just as long as a=-2c and b=c above are true?
 
Pinedas42 said:
So what I get is

a=-2c
and
b=c

the actual numbers inside will be arbitrary? just as long as a=-2c and b=c above are true?
That looks good.
 
Thanks, I think I just looked deeper than what the question actually pertained to. I need to learn to drop expectations I suppose. :P

Thanks again!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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