Im just wondering for the normal distance forumla its like distance = sqrt((x₁ - x₂)² + (y₁ - y₂)²) with that being true does that mean that distance forumla for 3d is distance = sqrt((x₁ - x₂)² + (y₁ - y₂)² + (z₁ - z₂)²) Killer Eagle Software

yup.

yes it does. and you can even prove that easily. if you know how to calculate the distance in 2d, you can do it for a box in 3d simple: first calculate the 2d diagonal of one side. that builds up a quat in 2d, build up by this diagonal the length you know, and the 3rd dimension of your box. now you calculate its distance..

diagonal = sqrt(a^2 + b^2)
fulldiagonal = sqrt(diagonal^2 + c^2) = sqrt(sqrt(a^2 + b^2)^2 + c^2) = sqrt(a^2 + b^2 + c^2)

that is to get the diagonal of a box.

and that is the distance from one edge to the other. and calculating a,b,c is easy.. the distance vector from one point to the other..

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shouldnt it be n2 - n1, not n1 - n2?

Is there any other formulas for calcuating distance or calclating like intersection of parts ect. If so what ones are there any basic ones u use in your programs?

Killer Eagle Software

quote:

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Original post by MENTAL
shouldnt it be n2 - n1, not n1 - n2?

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It doesn''t matter which way you do it, because you square the answer.

n2 - n1 = -1(n1 - n2)

so think off the 2 equations as

n1 - n2
and
-1(n1 - n2)

when you square both sides it''s

(n1 - n2)^2

and

(-1)^2 * (n1 - n2)^2

which is

1(n1 - n2)^2

and therefore

(n1 - n2)^2

The square of a positive number is the same as the square of its opposite.

it''s called the ''Pythagorean Theorem;'' it''s ancient. Remember, "A^2 + B^2 = C^2"? In 3 dimensions, you incorperate a third leg or dimension to the slope, or ¡î(X©÷+Y©÷+Z©÷)...sweet, i found the character map...

oops, that''s for generating a vector, not distance

I don''t think it matters- isn''t a vector''s magnitude just it''s distance from the origin, or (x^2 + y^2 + z^2)? =)

Yeah, it''s exactly the same (it''s just like you''re measuring the distance between a point originally at the origin translated by the vector, and the origin).