We need to first find appropriate superlattice vectors starting from the face-centered cubic lattice vectors:

    /  0  a/2 a/2 \
R = | a/2  0  a/2 |
    \ a/2 a/2  0  /

as discussed in detail in the Pt(100) setup page.

For a (110) surface, we want the surface normal along a direction which combines two cubic axes with equal coefficients. Note that the third lattice vector i.e. combination [ 0 0 1 ] already satisfies this property, so we only need to make the other two lattice vectors perpendicular to this direction. There are many ways to do this, but the combinations [ 1 -1 0 ] and [ 1 1 -1 ] satisfy the conditions and produce the smallest possible supercell. This yields the lattice transformation matrix:

    /  1  1  0 \
M = | -1  1  0 |
    \  0 -1  1 /

and the superlattice vectors:

        / -a/2  0  a/2 \
Rsup =  |  a/2  0  a/2 |
        \   0   a   0  /

If you observe carefully, these matrices have the same columns as the Pt(100) setup case, but just in a different order. This works because the in-plane lattice vectors in our 100 surface setup happen to both be 110 directions.

Once again, we have two atoms and hence two layers per supercell. For the second atom, we can pick [ 1 0 0 ] as the unit cell fractional coordinate. Taking that as a column vector and multiplying by inv(M) yields the supercell fractional coordinates [ 0.5 0.5 0.5 ]. Also note that the three superlattice vectors are orthogonal with lengths a/sqrt(2), a and a/sqrt(2). That is technically tetragonal, but in order to keep the lattice vectors in this order (lattice Tetragonal will make teh first two lattice vectors of equal length), we can specify it as orthorhombic. Therefore, we arrive at the supercell input:

lattice Orthorhombic 5.23966 7.41 5.23966  #a/sqrt(2), a and a/sqrt(2) in bohrs
ion Pt  0.0  0.0  0.0   1
ion Pt  0.5  0.5  0.5   1

This time the layer spacing is (a/sqrt(2))/2 = 2.62 bohrs, so five layers occupy approximately 13 bohrs. Therefor we need the third lattice vector to be at least 28 bohrs long to get a 15 bohr vacuum spacing.

We can stretch the third direction by a factor of 28/5.23966, scaling the third fractional coordinates by 5.23966/28 = 0.18713. We can then repeat the supercell in both directions with offsets of 0.18713 in the third fractional coordinate, till we get at least five atomic layers centered symmetrically across zero, and then drop the extra layers. The resulting geometry for the 110 surface is then:

#Save the following to 110.lattice:
lattice Orthorhombic 5.23966 7.41 28


#Save the following to 110.ionpos:
ion Pt  0.0  0.0 -0.187130   1
ion Pt  0.5  0.5 -0.093565   1
ion Pt  0.0  0.0  0.000000   1
ion Pt  0.5  0.5  0.093565   1
ion Pt  0.0  0.0  0.187130   1

Note that since we are breaking periodicity in the third direction anyway, we do not need to keep the final supercell equal to an integer number of atomic layers. We did that in the 110 setup only for simplicity.

Visualize this geometry using testGeometry.in, JDFTx dry run, createXSF and VESTA as discussed at the end of the Pt(100) setup page to get an image similar to the one above.