# Vector cross product

Discussion in 'Math' started by 0001T, Mar 20, 2007.

1. ### 0001T Thread Starter Member

Feb 28, 2007
15
0
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?

2. ### Papabravo Expert

Feb 24, 2006
10,137
1,786
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.

3. ### recca02 Senior Member

Apr 2, 2007
1,211
0
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)

4. ### kourosh New Member

Aug 24, 2007
5
0
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.

5. ### star Member

Dec 18, 2006
19
0
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.