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For Miller Indices of planes if we encounter a case where the plane passes through the origin then I have heard that the only remedy can be arrived by shifting the origin to a different point.However I do not understand whether or not I am supposed to change the index of the other planes whose origin was not passing through the previous origin.Please give me a bit of justification for the case.

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    $\begingroup$ Miller indices are not a property of planes at all. Rather, it is a property of systems of parallel planes. $\endgroup$ – Ivan Neretin Oct 9 '18 at 19:47
  • $\begingroup$ @IvanNeretin Does that mean that I have to change the indexing of the planes that I have already named with a separate origin? $\endgroup$ – gateprep Oct 9 '18 at 19:48
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    $\begingroup$ Or, put slightly differently, all lattice points are the same, so moving the origin to another point changes nothing. $\endgroup$ – Jon Custer Oct 9 '18 at 19:49
  • $\begingroup$ What is meant by symmetry of planes/directions in Miller Indices?For cubic it says that the principal directions are symmetric while for tetragonal it excludes the one perpendicular to the basal plane...Something similar is for the planes as well.Please describe this as vividly as possible. $\endgroup$ – gateprep Oct 9 '18 at 20:04
  • $\begingroup$ Symmetry is another thing altogether. You'd better ask a new question to that effect. $\endgroup$ – Ivan Neretin Oct 9 '18 at 20:11
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Since a comment does not allow the inclusion of a picture, perhaps this way is helpful:

Have a look at the following picture, derived from an illustration about the Miller Indices on wikipedia. It depicts a simplified situation of a unit cell which may occur if your crystal is of the cubic lattice system, i.e., the three vectors defining the unit cell are orthogonal to each other, and all are of same length. In the drawing, the author choosed $\vec{a_1}$, $\vec{a_2}$, and $\vec{a_3}$ for the unit cell's dimensions.

enter image description here

Not only the plane closest to the coordinate system's origin drawn is a valid Miller plane, equally so the second, farther one drawn. For each set of Miller indices $(hkl)$, the corresponding Miller planes are parallel to each other. The distance between these planes depends both on the indices, as well as on the symmetry of the crystal.

Since a crystal consists of many unit cells adjacent to each other, it is perfectly reasonable to draw Miller planes either in "positive direction" along vectors $\vec{a_1}$, $\vec{a_2}$, and $\vec{a_3}$ -- as shown here -- or in opposite direction along $-\vec{a_1}$, $-\vec{a_2}$, and $-\vec{a_3}$ with the same set of $(hkl)$.

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