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I have a basic wedge product question, I'm very new to this and have been looking at examples but still have trouble computing the following. If $u=(a_1,a_2,a_3,a_4)$ and $v=(b_1,b_2,b_3,b_4)$ and $w=(c_1,c_2,c_3,c_4)$, compute $dx_1 \wedge dx_2 \wedge dx_3 (u,v,w)$.

I have no clue where to begin. I can find $dx_1 \wedge dx_2(u,v)=(a_1b_2-a_2b_1)$ but how to continu from there?

Help would be much appreciated.

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  • $\begingroup$ A hint: It's going to be a $3 \times 3$ determinant. $\endgroup$ – Hans Lundmark May 1 '18 at 11:36
  • $\begingroup$ but how do you wedge product a $\mathbb{R}$ with a $\mathbb{R}^3$? $\endgroup$ – Mathbeginner May 1 '18 at 13:20
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I found the correct answer, which is: $a_1b_2c_3+a_2b_3c_1+a_3b_1c_2-a_1b_3c_2-a_2b_1c_3-a_3b_2c_1$, the alternating plus and minus signs are due to the amounts of 'swapping' the order.

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