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Math Help - Zero Vector Question

  1. #1
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    Zero Vector Question

    Hmm...starting to do Vector spaces and such, and i'm having a bit of trouble!

    Why does:
     R^3: x + y + z = 0; x - y + 2z = 8 ?
    Not have a Zero vector?

    Surely all vectors have Zero vectors as you can just add (0, 0, 0) ?

    Thankyou
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  2. #2
    MHF Contributor matheagle's Avatar
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    I'm not sure what you are asking.
    These are two planes, are you asking about the intersection of the two?
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    Quote Originally Posted by matheagle View Post
    I'm not sure what you are asking.
    These are two planes, are you asking about the intersection of the two?
    Apologies. Thankyou for the reply.

    the question was: Does such form a space?

    The answer was it did not because there was no 'Zero vector' - And im unsure as to why there is no Zero vector. But there is a zero vector when the equation is equal to 0 instead of 8.

    Thanks
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  4. #4
    MHF Contributor matheagle's Avatar
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    I'm still not sure I follow.
    I found the cross product of the two normal vectors and I found a point of intersection, by letting z=0.
    The line of intesection is
    x=3t+4, y=-t-4, z=-2t which does not pass through the origin.
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  5. #5
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    Quote Originally Posted by matheagle View Post
    I'm still not sure I follow.
    I found the cross product of the two normal vectors and I found a point of intersection, by letting z=0.
    The line of intesection is
    x=3t+4, y=-t-4, z=-2t which does not pass through the origin.
    Ah, so as it does not pass through the Origin, it has no Zero vector?
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  6. #6
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    I believe you are asking about the subset of R^3, that is all vectors of the form <x, y, z>, such that x+ y+ z= 0 and such that x- y+ 2z= 8. That is, the set of all <x, y, z> that satisfy both equations. That is, in fact, the intersection of the sets of vectors that satisfy each equations separately.

    That obviously does not include the 0 vector, <0, 0, 0>, because it does not satisfy the second equation: 0 - 0+ 2(0)= 0, not 8. You could also argue that it is not a subspace because it does not "closed under scalar multiplication". If <x, y, z> satisfies x- y+ 2z= 8 and n is any number other than 1, then <nx, ny, nz> satisfies nx- ny+ 2(nz)= n(x- y+ 2z)= 8n, not 8. One could also argue that it is not closed under addition. If <x, y, z> and <a, b, c> are two vectors not satisfying x- y+ 2z= 8 and a- b+ 2c= 8, then their sum, <x+ a, y+ b, z+ c> satisifies (x+ a)- (y+ b)+ 2(z+ c)= (x- y+ 2z)+ (a- b+ 2c)= 8+ 8= 16, not 8.
    Last edited by HallsofIvy; November 19th 2010 at 03:03 AM.
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