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B Vector addition error

  1. Apr 6, 2016 #1
    i have two unitary vectors in space U and V , U and V are perpendicular .
    W is a vector that verifies W ^ V = U - W .
    the following resolution is incorrect , i wan't to understand why :

    we use (o,U,V,(U^V)) . components of U (1,0,0) , V(0,1,0) , W(a,b,c) where a , b and c are real numbers .
    components of W ^ V ( c , 0 , a) . and U - W ( 1-a , -b, -c)
    where is the error here ?
     
  2. jcsd
  3. Apr 6, 2016 #2

    stevendaryl

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    I'm not sure about your notation: ^ means vector cross-product? I've always used [itex]\times[/itex]

    If so, then you are on the right track. [itex]U, V, U \times V[/itex] can be used as an orthonormal basis. So we can write [itex]W[/itex] as a linear combination:

    [itex]W = a U + b V + c (U \times V)[/itex]

    Then [itex]W \times V = U - W[/itex] becomes:

    [itex] (a U + b V + c (U \times V)) \times V = U - a U - b V - c (U \times V)[/itex]

    Now, we use the rules:
    [itex]X \times X = 0[/itex]
    [itex](X \times Y) \times Z = (X \cdot Z) V - X (Y \cdot Z)[/itex]

    where [itex]\cdot[/itex] is the vector scalar product.

    Applying these rules gives us:
    [itex]a U \times V + 0 + c (U \cdot V) V - c U (V \cdot V) = U - a U - bV - c (U \times V)[/itex]

    This simplifies to:
    [itex]a U \times V - c U = (1-a)U - b V - c(U \times V)[/itex]

    So if you just pair up the corresponding orthogonal vectors, this gives three equations:
    1. [itex]a = -c[/itex]
    2. [itex]0 = -b[/itex]
    3. [itex]-c = (1-a)[/itex]
     
  4. Apr 6, 2016 #3
    thank you sir :)
     
  5. Apr 6, 2016 #4

    stevendaryl

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    Looking at what you wrote, I think that your problem is that

    [itex](a, b, c) \times (0,1,0) = (-c, 0, a)[/itex]
     
  6. Apr 6, 2016 #5
    yes , i missed the sign . and flipped the numbers by mistake .
     
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