What is relationship between scalar and vector product? I am intrested in relationship between scalar and vector product in $\mathbb{R}^3$; I am going to give definitions which I will use in my question.
Scalar product - function $\cdot:\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}$ which satisfay following properties:
a) $(\alpha \vec{u}+\beta\vec{v})\cdot \vec{w} = \alpha(\vec{u}\cdot \vec{w})+\beta(\vec{v}\cdot \vec{w})$
b) $\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}$
c) $\vec{u}\cdot \vec{u} \geq 0$ and $\vec{u}\cdot\vec{u}=0 \iff \vec{u}=\vec{0}$
Vector product - function $\times:\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$ which satisfay following properties:
a) $(\alpha \vec{u}+\beta\vec{v})\times \vec{w} = \alpha(\vec{u}\times\vec{w})+\beta(\vec{v}\times \vec{w})$
b) $\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$
c) $ (\vec{u} \times \vec{v}) \times \vec{w} +(\vec{v} \times \vec{w}) \times \vec{u}+(\vec{w} \times \vec{u}) \times \vec{v}=\vec{0}$
Question: We have endowed $\mathbb{R}^3$ with these two structures (structure of scalar product and structure of vector product). I know that space endowed with dot product is called inner space, is there a name for a vector space endowed with vector product? Are vector product and scalar product (as I have defined them) connected structures? Can we somehow exspress vecotor prodcut via scalar product or slacar product via vector prodcut? Is there some formula which can reduce on product to another? What preciasly is connection between these two structures, and can that connection be generalized to $\mathbb{R}^n$ (I do not know whether this question makes sense)?
Edit: I am hoping that if there is some connection between these two strcutures that answers not just say it, but rather explain why does this connection hold.
Thank you for any help.
 A: You can easily define a scalar product in $\mathbb{R}^n$, for each $n\in\mathbb N$, be the vector product is specific to $\mathbb{R}^3$. There is a generalization to $\mathbb{R}^n$, but that's a map from $(\mathbb{R}^n)^{n-1}$ into $\mathbb{R}^n$.
A connection betwen both operations (assuming that you are dealing with the usual dot product here) is given by the triple product: given $v,w,u\in\mathbb{R}^3$, $\bigl\lvert v.(w\times u)\bigr\rvert$ is the volume of the parallelepiped defined by them.
A: We say $\Bbb R^3$ is an exterior algebra under $\times$; in such terminology it's the wedge product $\land$.
Scalar-vector connections include the scalar triple product $a\cdot b\times c$ being fully antisymmetric, $(a\cdot b)^2+(a\times b)\cdot(a\times b)=(a\cdot a)(b\cdot b)$ and $a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$.
Proofs, with implicit summation over repeated indices:


*

*$a\cdot (b\times c)=\varepsilon_{ijk}a_ib_jc_k$, then use $\varepsilon_{ijk}$ being fully antisymmetric

*$(a\cdot b)^2+(a\times b)\cdot(a\times b)=a_jb_ka_mb_n(\delta_{jk}\delta_{lm}+\varepsilon_{ijk}\varepsilon_{ilm})$, the bracketed coefficient famously being $\delta_{jm}\delta_{kn}$

*The left-hand side's $i$th component is $\varepsilon_{ijk}\varepsilon_{klm}a_jb_mc_n=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})a_jb_mc_n$ as required

A: If you want to relate the two products, note that
$$|a\times b|^2 + (a\cdot b)^2 = |a|^2|b|^2.$$
A: The usual inner (dot) and cross products on $\Bbb R^3$ are defined by
$${\bf x} \cdot  {\bf y} = x_1 y_1 + x_2 y_2 + x_3 y_3, \qquad
  {\bf x} \times {\bf y} = (x_2 y_3 - x_3 y_2, x_3 y_1 - x_1 y_3, x_1 y_2 - x_2 y_1) .$$
They are related by the properties
$$\phantom{(\ast)} \qquad ({\bf x} \times {\bf y}) \cdot {\bf x} = 0, \qquad
  ({\bf x} \times {\bf y}) \cdot ({\bf x} \times {\bf y}) = ({\bf x} \cdot {\bf x})({\bf y} \cdot {\bf y}) - ({\bf x} \cdot {\bf y})^2 , \qquad (\ast)$$
and by the vector triple product identity,
$${\bf x} \times ({\bf y} \times {\bf z}) = ({\bf x} \cdot {\bf z}) {\bf y} - ({\bf x} \cdot {\bf y}) {\bf z}.$$
By taking an appropriate trace one can derive from this formula one that recovers the dot product from the cross product:
$${\bf x} \cdot {\bf y} = -\frac{1}{2} \operatorname{tr}({\bf z} \mapsto {\bf x} \times ({\bf y} \times {\bf z})) .$$
One cannot, however, recover the cross product from the dot product (alone): If one replaces the "right-handed" cross product $\times$ with the "left-handed" cross product ${\bf x} \times' {\bf y} := -{\bf x} \times {\bf y}$, we see that $\times'$ still satisfies all of the above properties. Indeed, the first equation of $(\ast)$ says that ${\bf x} \times {\bf y}$ is a vector mutually orthogonal to ${\bf x}$ and ${\bf y}$, and the second specifies its length, but these two conditions determine ${\bf x} \times {\bf y}$ only up to sign. Thus, we can recover $\times$ from $\cdot$ using $(\ast)$ and a choice of orientation, i.e., a choice of "handedness". (If you know a little group theory, this situation can be viewed another way: The group of linear transformations that preserve the cross product is $SO(3)$, but the group of those that preserve the dot product is $O(3)$, in which $SO(3)$ is an index-$2$ subgroup.)
The structure defined by the three given conditions (a)-(c) satisfied by the cross product is called a Lie algebra, and usually one writes the product of $a, b$ in a Lie algebra as $[a, b]$. These are not in general called cross products however, and most don't have a nice relationship with a particular inner product like $\times$ does. (Indeed, the zero map $\Bbb R^3 \times \Bbb R^3 \to \Bbb R^3$ defines the abelian Lie algebra on $\Bbb R^3$.) Instead, the term cross product is usually reserved for a map $\Bbb R^n \times \cdots \times \Bbb R^n \to \Bbb R^n$ that satisfies $(\ast)$, where we replace the second equation in $(\ast)$ with an appropriate generalization.
To give an interesting example, suppose with identify $\Bbb R^3$ with the vector space of tracefree $2 \times 2$ real matrices, that is, those of the form $$\pmatrix{a&b\\c&-a} .$$ We can define a "cross product" on this set by
$$A \times B = \operatorname{tf}(AB) = A B - \tfrac{1}{2} \operatorname{tr}(AB) I$$---here $\operatorname{tf} C$ just denotes the tracefree part of C. Using the above trace formula gives a "dot product" $A \cdot B = \tfrac{1}{2} \operatorname{tr}(AB)$. Here I put "dot product" in quotation marks because this product is not positive-definite, i.e., doesn't satisfy property (c), but it satisfies the nondegeneracy condition that if ${\bf x} \cdot {\bf y} = 0$ for all ${\bf y}$ then ${\bf x} = 0$.
