$u{\times}v$ is orthogonal to both $u$ and $v$.
I tried to visualise the above, but couldn't do it. If $u{\times}v$ is orthogonal to both $u$ and $v$, then provided that the angle between $u$ and $v$ $\theta \neq 0$, then the triangle formed between $u$, $v$ and $u{\times}v$ would have a total internal angle value greater than $180^o$.
On
Try this: if $u$ is a vector on the ground, pointing North, and $v$ is a vector on the ground, pointing in some other direction, say southwest, from the same point, then $u\times v$ would be a vector pointing straight up to the zenith of the sky. Thus, it is perpendicular to both $u$ and $v$.
Does that visualization help?
Hold you finger's like this:
Suppose you want to do $B \times I$. The angle between $B$ and $I$ does not have to be 90. It is that the vector that the cross product produces, is $90$ to the plane that B and I span. Notice that $F$ and $B$ are $90$ degrees apart, and same for $F$ and $I$.
Now consider this:
These are three dimensional planes, that you two vectors that you are crossing span. When you cross the two vectors, it produces a vector in another plane that is perpendicular ($90$ degrees) to the other two. The other two need not be perpendicular too. It's just that the result that you get is.