What is the equivalence of the Bragg condition in vectorial form?

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thefireman
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Hello, I have a quick question regarding the bragg condition.

I know that it is most often stated as [tex]2dSin \theta=n\lambda[/tex]

But I have come across a case (Kittel chp9 pg 255, where it is written as ([tex]\vec{k}[/tex]+[tex]\vec{G}[/tex])[tex]^{2} = k[/tex]

I cannot really see how the vectorial case is the same as the simpler former one.
Could someone elucidate?

Thanks
 
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Or look in the same book (Kittel -Chapter 2) where the equivalence of the two formulas is discussed explicitly.
 
nasu said:
Or look in the same book (Kittel -Chapter 2) where the equivalence of the two formulas is discussed explicitly.

good call!