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
    9,898
    1,722
    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.
     
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