I am trying to determine the dimension of the intersection of a Koch curve and a Cantor set, as illustrated below.
The Koch curve has dimension $\frac{\ln4}{\ln 3}$ and the Cantor set has dimension $\frac{\ln2}{\ln 3}$. Since both fractals intersect transversally in $\mathbb{R}^2$, the dimension of the intersection equals $$ \frac{\ln4}{\ln 3}+\frac{\ln2}{\ln 3}-2<0, $$ a contradiction. Am I missing something? Or does this mean both fractals typically do not intersect ? It is true that in the drawing above, they do not, but it is easy to see that by lowering the Cantor set, it is plausible that they would.
The formula that you state is simply not true in general. We can show this by example, by illustrating specific intersections of this type with easily computed different values for the dimension:
I think the general question is quite a bit more difficult. In theorem 8.1 of the first edition of Falconer's Fractal Geometry, we find that for the Hausdorff dimension of Borel sets in $\mathbb R^n$:
$$\dim(E\cap (F+x)) \leq \max\{0,\dim(E\times F)-n\},$$ for almost all $x\in\mathbb R^n$. This is a much weaker statement than your assertion. Theorem 8.2 states a related result for the reverse inequality under more technical assumptions.