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Writing parameters in terms of given vectors

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Four 3-vectors a, b, c, and d are related by the equation
    ax + by + cz = d;
    where x, y, and z are real parameters. Using a suitable combination of scalar and vector
    products, findd x, y, and z in terms of the vectors

    2. Relevant equations

    3. The attempt at a solution

    So I tried to eliminate one vector at a time by doing the cross product of d and b etc so:

    d^b=(a^b)x + (c^b)z and
    d^c=(a^c)x + (b^c)y the other term cancelling as b^b=0


    And substituting into the equation in the question everything seems to cancel out and I get:

    xa=d, not sure if my approach is just entirely wrong. Any help would be greatly appreciated.
  2. jcsd
  3. May 27, 2013 #2


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    Gold Member

    I find your choice of symbols and your notation very confusing and hard to read. Your choice to label a,b,c as vectors and x,y,z as real constants doesn't help, when common notation would be the other way. Neither does your notation for cross product help. So I'm going to change your notation and give you a hint that might help. Let's say you are given vectors ##\vec u,\, \vec v,\, \vec w,\, \vec z## and real numbers ##a,\, b,\, c## such that$$
    a\vec u + b\vec v + c \vec w =\vec z$$Now consider what would happen if you dot ##\vec u \times \vec v## into both sides. See if that helps you solve for one of the constants.
    Last edited: May 27, 2013
  4. May 28, 2013 #3
    Hi, the notation used for defining vectors and parameters is just the one in the question and I didn't want to confuse the vector x with a cross product, so I used the other notation for cross product, nevertheless I accept that it was a bit confusing, apologies.

    Ok I see this totally, makes perfect sense. Obviously 2 out of 3 of the terms would be reduced to zero with the triple scalar product leaving only the one without repeated vectors. I tried using a matrix method in the end but this was a bit useless without actually knowing what the vectors are. Thanks, I think this has totally solved it for me but I'll just give it a go and come back later. Thank you very much.
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