# Diagonals and coordinates in 3-space

• Jan 21st 2011, 06:53 PM
Lprdgecko
Diagonals and coordinates in 3-space
It's easier if I just post a picture of what I've worked out:

http://i161.photobucket.com/albums/t...cko/img039.jpg

I'm a little confused on how to find the diagonal. Rather, which vertices do I use to find it?

Also, am I drawing this / plotting the points correctly? I know it's not to scale, but it seems off...
• Jan 21st 2011, 06:58 PM
Ackbeet
Everything looks good to me.
• Jan 22nd 2011, 05:39 AM
HallsofIvy
If you are using the standard "two-dimensional" Pythagorean theorem, you can think of (0, 0, 0) to (2, 4, 5) as being the hypotenuse of the right triangle with right angle at (2, 4, 0). One leg of that is the line from (2, 4, 0) to (2, 4, 5) which, obviously, has length 5. The other leg is the line from (0, 0, 0) to (2, 4, 0). To find its length think of it as being the hypotenuse of the right triangle with right angle at (0, 4, 0) (or (2, 0, 0)- either way works). The legs of the right triangle with vertices at (0, 0, 0), (0, 4, 0), and (2, 4, 0) are the lines from (0, 0, 0) to (0, 4, 0), which has length 4, and from (0, 4, 0) to (2, 4, 0), which has length 2. By the Pythagorean theorem, the hypotenuse, from (0, 0, 0) to (2, 4, 0) has length $\displaystyle \sqrt{2^2+ 4^2}$ so that the hypotenuse of the first triangle, from (0, 0, 0) to (2, 4, 5) has length $\displaystyle \sqrt{\left(\sqrt{2^2+ 4^2}\right)^2+ 5^2}= \sqrt{2^2+ 4^2+ 5^2}$.