3d vectors

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3d vectors

Postby joyrider » Wed Sep 08, 2004 11:06 pm

I have read a little about 3d vectors & i was wondering whether anyone can verify that this is correct:

I have 3 points on a graph - (0,0,0), (0,0,4), (4,0,0). Would it be true to state that the straight line going from (0,0,4) to (4,0,0) equals 32^.5?

This is how I did it: first I found the distances between (0,0,0) & (4,0,0) on the x-axis = 4 units, & between (0,0,0) & (0,0,4) on the z axis = 4 units. After that, I used Pythagoras's theorem to find the straight line from (0,0,4) to (4,0,0), which involves the z & x axis.

Here is the calculation for that:
Code: Select all
a^2 + b^2 = c^2 /* a = dist (0,0,0)to(4,0,0); b = dist (0,0,0)to(0,0,4); c = dist (0,0,4)to(4,0,0)*/

4^2 + 4^2 = c^2
c = 32^.5
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Postby Blueteeth » Wed Sep 08, 2004 11:14 pm

I don't see anything wrong , it's totally right .
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Postby Kybo Ren » Wed Sep 08, 2004 11:22 pm

I believe the distance formula for 3 dimensions is as follows: Image

That leaves us with
" sqrt( (4)^2 + (0)^2 + (-4)^2 )"

Which simplifies to

" sqrt( 32 ) "

So I belive that you are correct.
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Postby leas5040 » Thu Sep 09, 2004 2:17 am

Kybo_Ren wrote:I believe the distance formula for 3 dimensions is as follows: Image


You would be correct. The distance formula carries into the 3rd dimension. Way to think this one through though joyrider :D
"Given enough time, man can do anything with a bit of string and some Tinker toys." Bruce Bolden, Senior Instructor at the University of Idaho.
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