Suppose, if {111} and {112} are orthogonal to each other, then how to determine the possible set of crystal planes which are perpendicular to both {111} and {112} ?
In cubic systems, the lattice planes perpendicular to both, (h1,k1,l1) and (h2,k2,l2) can be obtained from the vector product of (h1,k1,l1) and (h2,k2,l2). Specifically for the perpendicular planes (1 1 1) and (1 1 -2), the joint perpendicular lattice plane is (-1 1 0), and analogously for all symmetrically equivqlent lattice planes.
For your problem, I would do as follows. Consider a crystal system for which the lattice vectors are a, b and c. The reciprocal lattice is then defined from three vectors a*, b* and c* with:
a*=2 pi b x c / V
b*=2 pi c x a / V
c*=2 pi a x b / V
V= (a x b).c
For a plane of the {hkl} family, the corresponding normal direction ghkl is given by:
ghkl=h a*+k b*+l c*
If you now consider another family (say {uvw}), the normal direction associated with each plane of the {uvw} family is:
kuvw=u a*+v b*+w c*
Therefore, for each set of planes, you can easily determine the corresponding set of normal directions. The set of planes which are perpendicular to both the {hkl} and {uvw} planes is then defined from the set of directions :
ghkl x kuvw
For your specific example (assuming a cubic lattice), you easily observe that the set of planes you are looking for is the {110} set of planes. For instance:
(-110) is perpendicular to both (111) and (11-2) planes.
Hope this helps.
For cubic system indices of a plane and its normal are same. The crystal plane which is perpendicular to the specific planes can be obtained by taking cross product of the vectors corresponding to the plane normal vectors. For the orthogonal planes (111) & (11-2) or (-1-12) & (112), common normal ( for the 1st case) can be found out easily by,
1 1 1 1 1 1
1 1 -2 1 1 -2
equivalent to
Plane also would be (-110)
Thanks to everyone. (-110) is one of the solution. I will try to calculate using above mentioned ways for remaining solutions. Particularly I need solutions of higher order. thanks for the solutions.
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