## plotting normal vector in 3d

on 30 Jul 2013

### Matt Kindig (view profile)

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

on 30 Jul 2013

doc quiver3

Jack_111

### Jack_111 (view profile)

on 31 Jul 2013
The answer is larger than -8.278e-25
Matt Kindig

### Matt Kindig (view profile)

on 31 Jul 2013
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

on 1 Aug 2013