Introduction to Vector calculations (Calculus 3)

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Homework Help Overview

The discussion revolves around finding nonzero scalars a, b, c such that the equation au + b(u - v) + c(u + v) = 0 holds for every pair of vectors u and v. This is framed as a conceptual exercise rather than a traditional homework question.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the equation by attempting to rearrange it and express it in terms of vector components. There are discussions about setting up systems of equations based on the vector components and the challenges faced in isolating the scalars.

Discussion Status

Some participants have offered insights into the relationships between the scalars, suggesting that a = -2c and b = c could be valid solutions. Others reflect on the nature of the problem, indicating a need to adjust expectations regarding its complexity.

Contextual Notes

There is an acknowledgment that the scalars a, b, and c are not unique, and the discussion includes considerations about the arbitrary nature of their values as long as the relationships hold true. Participants also note the challenge of working with multiple unknowns in the equation.

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!
 

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