- #1

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[tex] =<a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1> [/tex].

A thing I find Annoying is that most books just say, we define the dot product in component form as "" or we define the cross product in component form as " ". What nonsense, whos to say what its defined as. The equal sign demands a REASON why, not just a definition. Stewart provided a good geometric proof of why this is true in component form for the dot product. Here is what he has for the cross product:

[tex] \begin{multline*} a \times b = (a_1i+a_2j+a_3k) \times (b_1i+b_2j+b_3k) \\

= a_1b_1i \times i + a_1b_2i \times j + a_1b_3i \times k \\ + a_2b_1 j \times i + a_2b_2 j \times j + a_2b_3 j \times k \\ + a_3b_1k \times i + a_3b_2k \times j + a_3b_3k \times k \\

= a_1b_2k+a_1b_1(-j)+a_2b_1(-k)+a_2b_3i+a_3b_1j+a_3b_2(-i) \\

=(a_2b_3-a_3b_2)i+(a_3b_1-a_1b_3)j+(a_1b_2-a_2b_1)k \end{multline*}[/tex]

O.K., this is fine by me, but one problem, we ASSUMED that the distributive law holds for [tex] (a_1i+a_2j+a_3k) \times (b_1i+b_2j+b_3k) [/tex]

So to make me happy, I would like to see a proof of why the distributive law holds for this.

Ive seen a few ppl prove it by proving A x (B+D) = (A x B ) + (A x D). This is fine, I dont see any problem with that. But when they do go about proving it, they use the determinant method for components, but then that is using the very method you are setting out to prove! That does not seem right...