# Introduction to Vector calculations (Calculus 3)

1. Jan 21, 2012

### Pinedas42

1. The problem statement, all variables and given/known data
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.

3. 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.

Last edited: Jan 21, 2012
2. Jan 21, 2012

### SammyS

Staff Emeritus
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.

3. Jan 21, 2012

### Pinedas42

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?

4. Jan 21, 2012

### SammyS

Staff Emeritus
That looks good.

5. Jan 21, 2012

### Pinedas42

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!