Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Introduction to Vector calculations (Calculus 3)

  1. Jan 21, 2012 #1
    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;


    but once again I hit a dead end and only get the scalars equaling zero.
    Last edited: Jan 21, 2012
  2. jcsd
  3. Jan 21, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
  4. Jan 21, 2012 #3
    So what I get is


    the actual numbers inside will be arbitrary? just as long as a=-2c and b=c above are true?
  5. Jan 21, 2012 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That looks good.
  6. Jan 21, 2012 #5
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook