Webpage title: Understanding Crystal Planes in Materials Science

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Is plane (1 1 2) identical to (-1 -1 -2), or are they just parallel?
 
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I think you need more information...

The two vectors you supplied lie along the same line, right? But that's not enough to know if the planes are the same or not.

-Kerry
 
They are not vectors, but two planes in a crystal.
 
(1, 1, 2) and (-1, -1, -2) aren't planes. They can represent either vectors, or simply two separate points in three-space.

I agree with Kerry, you need more information to know how the planes in which these vectors/points lie relate to each other.
 
superwolf said:
They are not vectors, but two planes in a crystal.

Ahh... you should have mentioned that you are referring to materials science. We thought you were talking about mathematical planes...

Anyway, those are vectors, they're just used to identify crystalline planes. It's been a few years since I've had materials science, but I would say you probably still don't have enough information. They might be the same plane, they might be parallel, but why does it matter? What are you trying to solve? If you're trying to determine how a particular material will fail, then I would say that it doesn't matter. The vector refers to a direction, or a family of planes, not one specific plane.

-Kerry
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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