My question is with regards to the normalization of the projection of a vector connecting two parallel planes onto a normal vector to the (hkl) planes. we know that the unit normal vector is n=(h, k, l)/sqrt(h^2+k^2+l^2). now we choose a point on a next plane one lattice parameter a away, such that r=a * n. to obtain the right answer, we have to divide this vector r again by sqrt(h^2+k^2+l^2), normalizing it another time, because only r dot n yields a. how can i physically interpret the reason we have to divide it again by the square root factor? is this square root factor the numeber of spacings that you obtain from the plane equation?
This isn’t quite right: \mathbf{r} = a\hat{n} doesn’t belong to the next plane. Rather if a Miller plane passes through \hat{x}/h, \hat{y}/k, \hat{z}/l, then the next one passes through 2\hat{x}/h, 2\hat{y}/k, 2\hat{z}/l.
Thank you for your response. computing the distance between the plane that contains the site (0, 0, 0) of the conventional unit cell and a plane defined by the (hkl) indices would then give a vector connecting these two planes r=ax/h +ay/k +az/l. the projection of this vector onto the normal vector n=hx+ky+lz divided by the length of this normal vector times the direction of this normal vector, would then give 3a/(h^2+k^2+l^2) in the direction [hkl] right? or how would you suggest approaching this problem differently?
Best regards
The logic you’re describing seems correct, but the result does not pass a correctness check. For example, imagine the [1,1,1] Miller plane. The formula you have would give a, but the distance between the origin and this Miller plane is something like a/\sqrt{3}.