1. ## Zero Vector Question

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

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

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

Thankyou

2. I'm not sure what you are asking.
These are two planes, are you asking about the intersection of the two?

3. Originally Posted by matheagle
I'm not sure what you are asking.
These are two planes, are you asking about the intersection of the two?

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

4. 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.

5. Originally Posted by matheagle
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?

6. I believe you are asking about the subset of $\displaystyle R^3$, 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.