# What's the point of the cross product?

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

• 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? – Blue Apr 14 '19 at 11:04
• @LordSharktheUnknown Sorry! – user168651 Apr 14 '19 at 11:04
• @Blue I forgot to mention that I'm assuming all entries are non-zero. Sorry – user168651 Apr 14 '19 at 11:05
• 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? – Blue Apr 14 '19 at 11:16
• @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 – user168651 Apr 14 '19 at 11:35

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.
• 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 $(bf-ce,cd-af,ae-bd)$. – José Carlos Santos Apr 14 '19 at 11:20