Problem 264. Find the "ordinary" or Euclidean distance between A and Z
A, B and Z define three points in the 3D Euclidean space of the form: A = [x1;y1;0]; B = [x2;y2;0]; Z = [x2;y2;z];
Find the Euclidean distance between A and Z where
A = [1,0,0]; B = [5,3,0]; Z=[5,3,3];
>> euclidean(A,B,Z)
ans = 5.830951894845301
Your function should be able to handle 1 x 3 vectors or 3 x 1 vectors for all input parameters: A,B and Z. Z need not be 1 x 3 if A and B are. So 1x3,1x3,3x1 inputs, corresponding A, B and Z, are possible function input vectors.
HINT: use the Pythagorean formula.
Solution Stats
Problem Comments

6 Comments
what's the point of having 3 coordinates given? you're only asking for the distance between Z and A, right?
You need x,y and z to to get the distance metric in 3D space. But you can reduce the problem to just 3 values by taking the differences between x values and y values and then just taking the 3rd element of z. So there is unnecessary padding in the vectors and you could reduce them all to 3 scalars. Essentially the problem boils down to: sqrt(x^2 + y^2 + z^2).
looks like after you changed the test suite, it worked out anyway. thanks!
Your solution is the best and yes I did alter/correct the test suite. Thanks.
To repeat bmtran's question: Does B have anything to do with the problem?
B is a red herring. I originally conceived the problem in 2D and then extended it to 3D. So you really only need A and Z vectors. However, B can still be used but this creates a more complex solution which you don't want. Sorry if you felt misled, this was my first attempt at trying to upload an interesting problem that was hopefully not too difficult to solve. :)
Solution Comments
Show commentsProblem Recent Solvers159
Suggested Problems

7178 Solvers

161 Solvers

383 Solvers

Find the index of the largest value in any vector X=[4,3,4,5,9,12,0,4.....5]
368 Solvers

Find the dimensions of a matrix
505 Solvers
More from this Author16
Problem Tags
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!