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/
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