I am teaching Pre-Cal at a community college. In all Pre-Cal textbooks, we have this definition of cross product of two vectors:

If $\textbf{u} = \langle a_1, a_2, a_3 \rangle$ and $\textbf{v} = \langle b_1, b_2, b_3 \rangle$ are two 3-dimensional vectors, then the cross product of the two is the vector: $$\textbf{u} \times \textbf{v} = \langle a_2b_3-a_3b_2, a_3b_1-a_1b_3, a_1b_2-a_2b_1 \rangle.$$

But so far I haven't seen any textbooks that suggest using this definition to solve a cross product problems, instead they invariably suggest using the determinant of matrix of order two.

And here is my question: Is this simply because the determinant method is easier to remember, or is this because there is some other deeper reason? I am posting this question because one of my students use this definition when solving a problem on test.

Thank you for your time and effort.

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    $\begingroup$ It is easier to remember the determinant method. I just realised that you were the teacher. Why do you want to use the formula method? was this the way you were taught? $\endgroup$ – Chinny84 Nov 19 '15 at 15:16
  • $\begingroup$ The determinant $\begin{vmatrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{vmatrix}$ isn't even a genuine determinant (how can you define the determinant of a matrix with some vector entries and some scalar entries?). It is only ever meant as a way to remember the actual formula which you've given in your question. (Even though I have the cross product components memorized at this point) I personally would rather just remember the algebraic rules (like anticommutativity) of the cross product and derive that formula whenever needed. $\endgroup$ – user137731 Nov 19 '15 at 15:21
  • $\begingroup$ There are actually a few ways to attach meaning to the determinant formula as discussed here $\endgroup$ – rVitale Nov 19 '15 at 15:25
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    $\begingroup$ Ha! I love mathematicians. Every time someone comes up with a simple notation that's just supposed to reduce clutter you guys have to come along and go "but could we attach a rigorous meaning to it?". ;) $\endgroup$ – user137731 Nov 19 '15 at 15:30
  • $\begingroup$ @Bye_World Whilst I agree with your statement, being an ex-physicist myself and all, I think the OP was looking for a deeper meaning! My "simple" mind can not compute this and only approximate at best :). $\endgroup$ – Chinny84 Nov 19 '15 at 15:38

Just a comment on this:

In school it took me so long to get that formula into my brain. Then I was shown that you can combine this definition of the cross product with a very illustrative scheme. From this scheme also the name cross product makes sense, I'm sure everyone will remember this easily. Just write down your vectors $a=[a_1,a_2,a_3]$ and $b=[b_1,b_2,b_3]$ twice and do as follows:Scheme for the cross product


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