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Three-Dimensional Cross Product Derivation


Date: 07/26/99 at 08:45:22
From: Rohin Wood
Subject: Derivation of 3-dimensional cross product.

G'day,

I am a 17 year old, year 12, maths 2 student who completely dislikes 
the idea of accepting a formula without knowing and understanding its 
derivation and working principles. So naturally when my textbook gave 
the equation for 3D cross product in terms of determinants without 
explanation and my teacher was unable to assist, I was rather annoyed.

I would be grateful for any help you could give me,
Rohin Wood


Date: 07/26/99 at 14:32:28
From: Doctor Anthony
Subject: Re: Derivation of 3-dimensional cross product.

The easiest proof uses components   a = a1.i + a2.j + a3.k
                                    b = b1.i + b2.j + b3.k
                                    c = c1.1 + c2.j + c3.k

To show  a x (b x c) = (a.c)b - (a.b)c  we have

           b x c = |i    j    k|
                   |b1  b2   b3|
                   |c1  c2   c3|

                 = i(b2.c3-b3.c2) - j(b1.c3-b3.c1) + k(b1.c2-b2.c1)

     a x (b x c) = |  i                j                 k |
                   |  a1              a2                 a3|
                   |b2.c3-b3.c2   b3.c1-b1.c3   b1.c2-b2.c1|

                 = i(a2.b1.c2-a2.b2.c1-a3.b3.c1+a3.b1.c3)   
........(1)

and similar results for the j and k components.

Now compare (1) with the i component of:

     (a.c)b1 - (a.b)c1  = (a1.c1+a2.c2+a3.c3)b1 - 
(a1.b1+a2.b2+a3.b3)c1

          = a1.b1.c1+a2.b1.c2+a3.b1.c3 - (a1.b1.c1+a2.b2.c1+a3.b3.c1)

          = a2.b1.c2-a2.b2.c1-a3.b3.c1+a3.b1.c3   
..................(2)

and we can see that expressions (1) and (2) are identical. We could do 
the same for the j and k components to show that

     a x (b x c) = (a.c)b - (a.b)c

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Linear Algebra

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