Diffraction from fcc(111) surface Take the (111) surface of a fcc lattice:

If we have a wave incident on the surface moving in a direction with a projection parallel to the orange line, then in principle we should observe diffraction in the scattering plane, since there is a periodicity in the surface in this direction.
However, what we would find is that we do not observe a 1st order, though there is a second order peak. Mathematically this is a consequence of a smaller periodicity in the directions indicated by the red and green lines.
Physically though it would be nice to understand the absence of 1st order peak in terms of destructive interference. Do you see a way to explain these absences in terms of a cancellation of scattered waves?
 A: I happen to have a slide about something similar, so I'll adjust it and paste below.
If the red and green lines are the (acute) 2D lattice vectors for the 2D hexagonal lattice on the fcc(111) face, we can call those directions [1, 0] and [0, 1]. Then "orange" direction could be called [1, 1]. As far as I can tell from the two images below, in the 3D lattice those directions could be called either [1, -1, 0] & [-1, 0 1] or [1, -1, 0] & [1, 0 -1].
Either way, add them together to get the orange arrow and it's [0, -1, 1] or [2, -1, -1] and the rules for allowed diffraction are all odd or all even and the orange line does not satisfy that
Before the intuitive answer, let's look at the selection rules for diffraction We can trust them while we learn to understand them.
Bravais lattices           Allowed reflections           Forbidden reflections

Simple cubic               Any h, k, ℓ                   None

Body-centered cubic (BCC)  h + k + ℓ = even              h + k + ℓ = odd

Face-centered cubic (FCC)  h, k, ℓ all odd or all even   h, k, ℓ mixed odd and even

Diamond FCC                All odd, or all even with     h, k, ℓ mixed odd and even 
                           h + k + ℓ = 4n                or all even with h + k + ℓ ≠ 4n

Triangular lattice         ℓ even, h + 2k ≠ 3n           h + 2k = 3n for odd ℓ

For the orange-arrow case in your question let's look again:

Now examine my old slide:
For your plane wave with a projected period on the surface equal to the [1, 1] direction, we can see that it passes TWO rows of equally spaced atoms. The reason that the selection rule says this diffraction is forbidden is that diffraction from the alternating rows of atoms cancel each other; they are 180 degrees out of phase.


Trying to get the correct [hkl] notation for directions of the hexagonal lattice on the fcc(111) surface. These seem to :
 
Sources left: Diffusive Atomistic Dynamics of Edge Dislocations in Two Dimensions, right: Reconstruction of steps on the Cu(111) surface induced by sulfur
