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    Express the vector (a,b,c,) as a linear combination of v1, v2 and v3 with coefficients in terms of the contstant a,b and c
    v1= (1,-1,0)
    v2= (1,0,-1)
    v3= (1,1,1)

    absolutely no idea where to start , never been taught this and it isn't covered in any textbook i have. looked at the answer in the back for a clue and it says:
    (1/3)(a-2b+c)v1 + (1/3)(a+b-2c)v2 + (1/3)(a+b+c)v3

    which has confused me even more as i can't see how i'd get to that at all

    help?
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    if it were me doing this question, I would first try and create the vectors (1,0,0) (0,1,0) and (0,0,1) from v1 v2 and v3, then the answer should be obvious.

    so (v1+v2+v3)/3=(1,0,0) so then a(v1+v2+v3)/3=(a,0,0)

    Doing a same thing for the others yields the result. I hope this was clear.
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    hot damn that worked!
    thanks a tonne
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    (Original post by Misiak)
    Express the vector (a,b,c,) as a linear combination of v1, v2 and v3 with coefficients in terms of the contstant a,b and c
    v1= (1,-1,0)
    v2= (1,0,-1)
    v3= (1,1,1)

    absolutely no idea where to start , never been taught this and it isn't covered in any textbook i have. looked at the answer in the back for a clue and it says:
    (1/3)(a-2b+c)v1 + (1/3)(a+b-2c)v2 + (1/3)(a+b+c)v3

    which has confused me even more as i can't see how i'd get to that at all

    help?
    The linear combination of v1,v2,v3 is
    \displaystyle \lambda_1 \vec v_1 +\lambda_2 \vec v_2 +\lambda_3 \vec v_3 = \vec v
    where \displaystyle \lambda_1, \lambda_2, \lambda_3 \in R and v1,v2,v3 are independent linearly.
    This vector equation means three scalar equations (for x, y, and z coordinates)
    \lambda_1\cdot 1 +\lambda_2 \cdot 1+\lambda_3 \cdot 1=a
    \lambda_1\cdot (-1) +\lambda_3 \cdot 1=b
    \lambda_2 \cdot (-1)+\lambda_3 \cdot 1=c
    Solve simultaneously for \lambda_i
 
 
 
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