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math question (distance)

Started by Luke1 Mar 11, 2003 at 3:17 PM 5 replies 450+ views

Original Post

Mar 11, 2003 at 3:17 PM

I know the distance formula for 2d but im not too sure on the 3d part would some like this work:

    
int GetDist(int x1, int y1, int z1, int x2, int y2, int z2)
{
return sqr(((x1-x2)^2)+((y1-y2)^2)+((z1-z2)^2))
}

is that right or is there another way ? [edited by - Luke1 on March 11, 2003 4:18:26 PM]

Punty50

Mar 11, 2003 at 5:08 PM

Thats not right, because as AP said, ^ is the binary exclusive or operator. Use * because you know that you are squaring, and that way things will be nicely optimized and everything.

int GetDist(int x1, int y1, int z1, int x2, int y2, int z2){return sqr(((x1-x2)*(x1-x2))+((y1-y2)*(y1-y2))+((z1-z2)*(z1-z2)))}

Brendan

Jesper T

116

Jesper T

Mar 11, 2003 at 6:09 PM

And if you use c++ its sqrt(..) or sqrtf(..), not sqr(..).

[edited by - grhodes_at_work on March 12, 2003 12:33:37 PM]

Thunder_Hawk

Mar 11, 2003 at 6:26 PM

To figure out how to apply the distance formula to 3D (which you have correct BTW), imagine an arbitrary rectangular box and assign the letters x, y, and z to the axes of the box. Then use pythagoras to find the length of a diagonal through the middle of the box (hint: it is the hypotenuse of a triangle formed by the diagonal of one side of the box and a connected edge). This will give you the equivalent of the pythagorean theorem in 3D. From there it''s child''s play to derive the 3D distance formula.

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