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I don't understand the motivation behind defining cross products the way they're defined.

Given two vectors $\vec{A}$ and $\vec{B}$ in $\mathbb{R^3}$, I can find a third vector $\vec{C}$ such that $\vec{C}$ is normal to $\vec{A}$ and $\vec{B}$ by using the following formula:

$$\vec{C} = r \left\langle \;1,\;\frac{a_xb_z-b_xa_z}{b_ya_z-a_yb_z}, \;\frac{a_x+\frac{a_xb_z-b_xa_z}{b_ya_z-a_yb_z}a_y}{a_z}\;\right\rangle \quad\text{where}\quad r \in \mathbb{R}$$

EDIT: all entries are non-zero

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  • $\begingroup$ For non-zero $r$, your $\vec{C}$ has a non-zero $x$ component. What if $\vec{A} = (1,0,0)$? Also, what happens if any of your denominators are zero? $\endgroup$ – Blue Apr 14 at 11:04
  • $\begingroup$ @LordSharktheUnknown Sorry! $\endgroup$ – user168651 Apr 14 at 11:04
  • $\begingroup$ @Blue I forgot to mention that I'm assuming all entries are non-zero. Sorry $\endgroup$ – user168651 Apr 14 at 11:05
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    $\begingroup$ Have you considered scaling your vector by $b_y a_z-a_y b_z$, simplifying, and comparing the result to the conventional cross product? $\endgroup$ – Blue Apr 14 at 11:16
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    $\begingroup$ @Blue I got the same definiton! that was what I was aiming for in the first place. I should've seen it on my own, thanks $\endgroup$ – user168651 Apr 14 at 11:35
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Yes, $v\times w$ is orthogonal to both $v$ and $w$. But it also has the property that $\lVert v\times w\rVert=\lVert v\rVert.\lVert w\rVert.\sin\theta$, where $\theta$ is the angle between $v$ and $w$. In particular, it provides an easy way to find an unit vector which is orthogonal to two given unit vectors which are already orthogonal to each other.

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  • $\begingroup$ I understand. Assuming I don't know the the definition of the cross product how can I derive it giving the properties you mentioned? $\endgroup$ – user168651 Apr 14 at 11:15
  • $\begingroup$ I don't know. It seems to me that the natural way of getting the cross-preduct $(a,b,c)\times(d,e,f)$ consists in trying to find a vector orthogonal to both $(a,b,c)$ and $(d,e,f)$ and to see that a simple solution is $(bf-ce,cd-af,ae-bd)$. $\endgroup$ – José Carlos Santos Apr 14 at 11:20

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