# proving some vector is not a linear combination by dimension of subspace

So I have a set of 2 vectors, $$v1,v2$$ which are linearly independent and the dimension of the subspace they belong to is 3 ($$dim(W) =3)$$. How can I prove there is some vector $$W$$ that is not a linear combination of $$v1,v2$$?

edit: my attempt so far, I assumed that some vector $$W$$ is in $$span(v1,v2)$$

I show some vector in the subspace, $$W$$ can be written as a linear combination of $$v1,v2$$

then i showed $$v1 + v2 -W = 0$$ and hence $$W,v1,v2$$ and linearly dependent which contradicts the original assumption that $$W$$ is in $$span(v1,v2)$$

Suppose for all $$v \in W$$, the vectors $$v_1,v_2,v$$ are linearly dependent. Then we can write $$v$$ as a linear combination of $$v_1,v_2$$ (why?). But that means $$v_1,v_2$$ span $$W$$. Hence $$v_1,v_2$$ is a basis and hence dimension of $$W$$ is 2 (contradiction).