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Homework Help: Help on systems of eqtns.

  1. Jan 25, 2008 #1
    Need some help to solve this problem.I have tried using a system of equation(matrices) but hasnt worked out.

    Find nonzero scalars a,b and c such that aA+b(A-B)+c(A+B)=0 for every pair of vectors A and B.

    Thanks for the help.
  2. jcsd
  3. Jan 25, 2008 #2


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    Staff: Mentor

    What do you mean by "vectors"? If A and B are vectors, and they are orthogonal to each other, I don't believe there is a general solution.

    EDIT -- oops, yeah there still is a set of solutions. Just distribute the terms, and gather the terms multiplying A and those multiplying B. Assume that A and B are indeed orthogonal. What can you say about those two sets of terms....?
  4. Jan 25, 2008 #3


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    Science Advisor

    For every pair of vectors A and B? Do you mean one set of numbers a, b, c such that that is true for all vectors? Obviously that is impossible.

    If it were true for all A, B, it must be true for any A and B= 0. Then you must have (a+ b+ c)A= 0 for any A, that is a+ b+ c= 0.

    On the other hand, if A= B, you have (a+ 2c)A= 0 so a+ 2c= 0. If A= -B, you have (a+ 2b)A= 0 so a+ 2b= 0. The only numbers that satisfy those three equations are a= b= c= 0.

    If, however, you mean find a, b, c so that aA+ b(A- B)+ c(A+ B)= 0 for specific A, B, then you need to have (a+ b+ c)A+ (a- b+ c)B. There will be an infinite number of values of a, b, c such that that is true for any given A and B.
  5. Jan 25, 2008 #4
    Hmm, the book says one possible answer is a=-2,b=c=1
  6. Jan 25, 2008 #5
    What is the exact statement of the problem from your book? Does it says anything about the vectors A,B, independent or something else?
  7. Jan 25, 2008 #6
    what i posted is the EXACT statement, its from a vector analysis course.
  8. Jan 25, 2008 #7
    Ok, then! If A,B are arbitrary vectors I would write

    << exact solution edited out by berkeman, but too late to keep the OP from seeing it >>

    yielding the book's solution, but I assumed that A,B are arbitrary vectors.
    Last edited by a moderator: Jan 28, 2008
  9. Jan 25, 2008 #8
    I hope nobody succeeded seeing my previous post, which I deleted quickly. I cannot believe how frozen my brains were.....
  10. Jan 25, 2008 #9
    Ahh!cant believe this easy question gave me problems..so that answer is one of many possible answers just like the book says!..thnx for the help!
  11. Jan 28, 2008 #10


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    Staff: Mentor

    Only the Mentors can see it. We're wispering about it now.... :bugeye:
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