I am trying to solve this cross product; 3 vertices p1=(0,0,3), p2=(4,0,0) & p3=(0,4,3) given, I need to find the normal of the surface bounded by these 3 vertices. I was told answer is (3,0,4) but my attempts kept leading me to (12,0,16) here's what I did. normal = (p1-p2) x (p3-p2) = [(0,0,3)-(4,0,0)]x[(4,0,0)-(0,4,3)] = (-4,0,3)X(4,-4,-3) normal = (12,0,16) where did i go wrong?
Isn't (12,0,16) = 4 * (3,0,4) so those two vectors are colinear. I think the direction is the important thing and not the magnitude. I could be wrong though.
i agree in vectors the above two values can be treated as the correct answer since a normal has direction not magnitude if i am not wrong u can have a answer (-3,0,-4)
NO guys! Cross product gives a magnitude that is sometimes very significant! Although, one may not care about it. You can cross product 2 vectors, not 3. p1*p2=[0,12,0] and that cross p3 = [36,0,0] where "*" represents a cross product Note: 3 arbitrary 3d vectors may NOT even be on a plane! I still am not sure what you need to know so good luck.
kourosh - I think p1,p2,p3 are co-ordinates on an x,y,z plane, as they are just three vectors from the origin. Hence, they define the plane that all three are on. i agree too that the two answers (3,0,4) and (12,0,16) are equally valid.