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As far as I understand the cross product between two vectors $\mathbf{a},\mathbf{b}\in\mathbb{R}^{3}$ is defined as a vector $\mathbf{c}=\mathbf{a}\times\mathbf{b}$ that is orthogonal to the plane spanned by $\mathbf{a}$ and $\mathbf{b}$. What I was wondering though is, if this is the case, what is the motivation for such a definition? (or is the property of orthogonality to $\mathbf{a}$ and $\mathbf{b}$ something that can be proven?)

Is it related at all to the fact that one can define a plane if one has a displacement vector $\mathbf{r}-\mathbf{r}_{0}$ in that plane (where $\mathbf{r}$ and $\mathbf{r}_{0}$ defines the positions of two points in the plane) and a vector $\mathbf{n}$ that is normal to that plane? My reasoning for this is that if we do have $\mathbf{r}-\mathbf{r}_{0}$ in the plane and $\mathbf{n}$ normal to that plane, then it follows that $$ (\mathbf{r}-\mathbf{r}_{0})\cdot\mathbf{n} =0$$ and from this we can determine the explicit forms of vectors in this plane. Does the notion of the cross product follow from taking the opposite approach, i.e. assuming that we know the form of two linearly independent vectors $\mathbf{a},\mathbf{b}\in\mathbb{R}^{3}$ and therefore we know that they span a plane in $\mathbb{R}^{3}$. Now suppose that we wish to determine a vector orthogonal to this plane, how do we do this? We define a new vector $\mathbf{c}=\mathbf{a}\times\mathbf{b}$ and demand that it satisfy $$\mathbf{a}\cdot\mathbf{c}=0,\qquad \mathbf{b}\cdot\mathbf{c}=0$$ Thus the operation $\mathbf{a}\times\mathbf{b}$ results in a new vector $\mathbf{c}=\mathbf{a}\times\mathbf{b}\in\mathbf{R}^{3}$ that is orthogonal to both $\mathbf{a}$ and $\mathbf{b}$.

Would something like this be a correct motivation?

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Yes, it is.

Another motivation is the following: in three dimensions (and only in three dimensions) a plane has only one normal direction. The angle between the normal direction and any vector $v$ has a geometric interpretation as "exposure" of the plane to $v$.

For example, a photo film is maximally exposed to the sun if its normal vector is parallel to the sun rays, which means that the amount of light impressed on the film will be proportional to $(a\times b)\cdot s$, where $a,b$ are the sides of the film, and $s$ is the light ray.

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  • $\begingroup$ Ah ok, that's good to know. Glad my understanding is at least somewhat correct! How does this relate to the more general concept of the wedge product? I know that this generality defines the notion of volume spanned by one-forms in $n$-dimensions (hence why $\alpha\wedge\alpha =0$ as a one-form wedged with itself spans zero area - it is collinear with itself), but how does the cross-product relate to it. I've heard that it's related by the Hodge dual $\star (\alpha\wedge\beta)=\alpha\times\beta$, but I struggle to understand this intuitively?! $\endgroup$ – Will Jul 3 '15 at 12:29
  • $\begingroup$ The wedge product is a little more natural than the cross product, but less intuitive. The wedge product of two vectors gives you the directed area spanned by the two. The cross product gives you a vector normal to that area, whose length is equal to the area. Clearly the latter is possible only in 3 dimensions. This "being normal to" is the geometric way of visualizing Hodge duality. $\endgroup$ – geodude Jul 3 '15 at 12:38

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