mjhjmtu

I am trying to linearize a vector expression
$$frac(mathbfu+dmathbfu)times(mathbfv+dmathbfv)vert(mathbfu+dmathbfu)times(mathbfv+dmathbfv)vert$$
Here is my code

$Assumptions = (u | v | du | dv) ∈ Vectors[3, Reals];
Simplify[Series[
Cross[(u + ϵ*du), (v + ϵ*dv)]/
Sqrt[Dot[Cross[(u + ϵ*du), (v + ϵ*dv)], 
Cross[(u + ϵ*du), (v + ϵ*dv)]]], ϵ, 
0, 1]]

And the solution is:

Two things I don't understand. First, is 1 a vector? 1=1,1,1?
And another thing is the meaning of the space between two vectors, does this means a cross product between two vectors?

As mikado pointed out, this is a bug and the return value is complete nonsense. Don't invest any time in interpreting it. Better use the time and send a bug report to Wolfram Support. This appears to be a really old bug and they won't fix it without sufficient peer pressure.

In the meantime, you can go on with

expr = Cross[(u + ϵ*du), (v + ϵ*dv)]/
 Sqrt[Dot[Cross[(u + ϵ*du), (v + ϵ*dv)]]];
(expr /. ϵ -> 0) + (D[expr, ϵ] /. ϵ -> 0) ϵ

$fracepsilon (textdutimes v+utimes textdv)2 sqrtutimes
v+sqrtutimes v$

With regard to the second question: A space between two vectors is still interpreted as Times, hence as a component-wise multiplication, producing again a vector of same Length (if the two vectors have same Length and an error message otherwise).