2D dist = pythagoras, 3D dist =/= pythagoras?

[MadPumpkin]

When I use Pythagorean theorem I always get same answer as when I use this:

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sqrt( (x2-x1)^2 + (y2-y1)^2 )

Which makes perfect sense, one just checks based on sides, and one checks using another method. However, when I do this:

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sqrt( (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)
OR
cuberoot( (x1-x1)^2 + (y2-y1)^2 + (z2-z1)^2)

Either way I get a different answer than what Pythagoras yields. I'm not sure which ones right, any help? Alright well I said that wrong, I AM sure which ones right, Pythagoras is god. I just want to know, what I can do to make the other way of doing it correct too, or another similar way anyways.

[Ginto8]

for 3D it should be the first, just subtract the 2 z's instead of adding them.

[MadPumpkin]

for 3D it should be the first, just subtract the 2 z's instead of adding them.

Yea I had just barely fixed that >.< I meant to put subtraction. I might have just been doing my math on paper wrong. So I'm about to do it.

Well my god, thanks Ginto8. In my math, I squared x and y but forgot to square z before adding them.