prove the scalar product of AC.DF=0
knowing that AB=a
>so it's possible to expand the dot product the same way as any algebric product ?
Better make this step more detailed.
AD.DH = 0
AD.HG = 0
AD.GF = -2|a|
DC.DH = 0
DC.HG = 2|a|
DC.GF = 0
There are a few ways to solve this problem, some more intuitive than others.
AC = a(-i + j)
DF = a(i + j + k)
AC.DF = -a^2 + a^2 + 0 = 0
An even easier way would be to evaluate it analytically. AC is inverse to DB, and because AC omits the extra dimension that DF posses, then we can determine the dot product between the two vectors to be 0 (with the angle between them being pi/2 radians).
Stop being an ungrateful dick and learn the material for yourself. You aren't going to benefit if people just tell you. Be lucky that people are willing to help you even when you're being a selfish pig.