Wednesday

July 30, 2014

July 30, 2014

Posted by **Greg** on Friday, September 14, 2012 at 2:35pm.

a - pick 2 arbitrary points in 3 dimensions, (x1, y1, and Z1) and (x2,y2,z2) and plot these points. Not that there are 0- degrees between the xy, xz, and yz axes.

b- using your sketch from part a, find the distance between these 2 points, generalize the above equation to 3 dimensions.

- Precalculus -
**Steve**, Friday, September 14, 2012 at 2:49pmfirst, the axes are perpendicular, the there are 90°, not 0° between axes.

Second, can't do plots here.

Third, once you have picked your points, just plug them into the 3D version of the formula:

d^2 = (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2

To see that this is true, consider just the distances in the plane z=z1.

w^2 = (x2-x1)^2+(y2-y1)^2

That's just a diagonal line of length w. To get from (x1,y1,z1) to (x2,y2,z2), consider the right triangle from (x1,y1,z1) to (x2,y2,z1) with length w, and go from there to (x2,y2,z2) of length (z2-z1).

If that's confusing, just look online for references to 3D distances.

http://tonto.stanford.edu/~brian/euclidean_distance_in.html

is a good place to start

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