1

A student just asked me about the possibility of a cross product defined in terms of a nonstandard basis, that is, don't use i, j, and k.

Can anyone point me to an article explaining such a question, if it exists?

David
  • 2,322
  • 1
    The cross product can be defined geometrically (scroll down to Geometric Definition under The Cross Product). Or you do it the algebraic way and just convert to ${i, j, k}$, take the cross, then convert back. –  Nov 13 '15 at 22:33
  • 1
    Or, come to think of it, use approach #3: the matrix approach. You want to compute the cross product $a\times b$, then in ${i,j,k}$ that's the same as the matrix multiplication $[a]\times b$, where [$[a]\times$ is a $3\times 3$ matrix](https://en.wikipedia.org/wiki/Cross_product#Conversion_to_matrix_multiplication). If you'd like to express that linear transformation in a different basis, just do a change of basis transformation on the matrix $[a]_\times$. –  Nov 13 '15 at 22:41
  • For $v=v_1\times v_2$, we have $v$ orthogonal to $v_1$ and $v_2$, and the distance from $v$ to the origin $ O$ is twice the area of the triangle $O v_1 v_2$. This gives 2 possible values for $v$. The usual choice is that $(1,0,0)\times (0,1,0)=(0,0,1)$ and that "$\times$" is continuous in each variable. This uniquely determines all $u\times v$. – DanielWainfleet Nov 13 '15 at 22:44

1 Answers1

1

Yes, there is. See this great discussion and the paper suggested by @Bye_World. Also, if you are just dealing with $\mathbb{R}^n$, then any vector $a,b$ has a representation under a 'non-standard basis', and so you could compute $a,b$ under that non-standard basis and then use those basis vectors instead.

Community
  • 1
9301293
  • 2,304