plotting normal vector in 3d

I have three points P0=[x0,y0,z0], P1=[x1,y1,z1] P2=[x2,y2,z2] and I want to calculate the normal out of them what I did is

    normal = cross(P0-P1, P0-P2);

and then I wanted to plot the normal so what I did is,

    c = normal + P0 %end position of normal vector

quiver3(P0(1), P0(2), P0(3), c(1), c(2), c(3)); but it didn't work any suggestions please

%......Edited

Answer 1

[Matt Kindig]

doc quiver3

[Jack_111]

it didn't work

[Matt Kindig]

What didn't work about it? Did you read the documentation for quiver3?

[Jack_111]

yes I did but the line that I get is not the normal of the plane

[Jack_111]

I tried to insure if it is the normal or not by calculating the dot product but it wasn't zero dot(P0 - P1, normal)) and it was small number but not zero

[Matt Kindig]

How close to zero was the dot product? Within numerical precision?

[Jack_111]

-8.271806125530277e-025 that was the answer but it must be zero

[Jack_111]

is it more helpful to put the entire code?

[Matt Kindig]

-8.27180e-025 is probably as close to zero as you're going to get with Matlab (or for that matter, any computer program). In particular, it is probably within the eps() for your vectors.

Try this:

eps(P0)

what is this number? If it is larger than -8.278e-25, then you are within the precision of the calculation, and the answer cannot, in effect, be closer to zero.

[Jack_111]

The answer is larger than -8.278e-25

[Matt Kindig]

Then what I said above is correct. Your dot product is as close to zero as you'll be able to calculate. In other words, your calculated vector (from the cross product) is as close to normal to the plane as you'll be able to get.

I'm not really sure what else to say. Your code is correct, your answer is correct, and that's that. Read http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html for why exact floating-point calculations are not possible with a computer.

[Jack_111]

Thank you for your support