The inner product of a Clifford Algebra Any  Clifford algebra $\operatorname{Cl}(k, p)$ carries an induced inner product, which is the "trace" on its 0-blade: $\langle AB\rangle_0$ for given elements $A, B$ of the algebra. 
This inner product, when restricted to the generating vector space $V$,  gives back the inner product on $V$.
Now, my question (maybe entirely trivial, but I could not find it in the standard literature, for instance in Lounesto's great book Clifford algebras and spinors):

What is the interplay of the Clifford product and its induced inner product? Are there any formal laws? 

PS. I have done some little googling, and I came up with some references on associative algebras with inner product, but unfortunately there seems to be a certain latitude as far as their definition.
PPS In light of the comments below, I realize that the phrasing of my post is not as clear as it should. Please refer to the comments below and my additional replies for the proper context. 
 A: $
\newcommand\lcontr{\,\lrcorner\,}
\newcommand\rcontr{\,\llcorner\,}
\newcommand\lcontrr{{\rfloor}}
\newcommand\rcontrr{{\lfloor}}
\newcommand\form[1]{\langle#1\rangle}
\newcommand\Ext\bigwedge
\newcommand\rev\widetilde
$
I hope that this is in the vein of what you were looking for.
In a review (Marcel Riesz’s Work on Clifford Algebras, 1993)
of Marcel Riesz's lecture notes (Clifford Numbers and Spinors, 1958)
on Clifford algebras, Pertti Lounesto demonstrates a natural linear isomorphism
between $\Ext(V)$ and $Cl(V)$.
Given a symmetric bilinear form $\form{\cdot,\cdot}$ on a vector space $V$
over a field $K$ with characteristic $\not=2$,
this extends to a bilinear form on $\Ext V$.
On $k$-blades (i.e. simple $k$-vectors), we define
$$
  \form{x_1\wedge\cdots\wedge x_k, y_1\wedge\cdots\wedge y_k}
  = \det\form{x_i, y_j},
$$
which is to say that we take the determinant over the matrix with entries
$a_{ij} = \form{x_i, y_j}$.
For blades $A, B \in \Ext V$ with different grades we define $\form{A, B} = 0$,
and then extend by linearity to the entirety of $\Ext V$.
Though maybe not obvious when described like this,
this extension of the form on $V$ is natural,
see here and here.
With this form on $\Ext V$,
the adjoints of the exterior product
$$
  \form{X \wedge Y, Z} = \form{Y, X \lcontr Z},\quad
  \form{X \wedge Y, Z} = \form{X, Z \rcontr Y}
$$
(where $X, Y, Z \in \Ext V$ are arbitrary multivectors)
are found to be the left ($\lcontr$) and right ($\rcontr$) contractions
of the Clifford algebra defined by $\form{\cdot,\cdot}$ on $V$.
They are indispensable when studying Clifford algebras for applications.
There are various different conventions that can be used when defining them;
see the appendix of this preprint
(Compendium on Multivector Contractions, 2022)
by André Mandolesi.
It's worth noting that the alternative contractions
$$
  X\lcontrr Y = \rev X\lcontr Y,\quad X\rcontrr Y = X\rcontr\rev Y,
$$
where $\rev X$ is the reverse of $X$, are particularly popular;
see Lounesto or this article
(The Inner Products of Geometric Algebra, 2002)
by Leo Dorst for an exposition on their basic properties.
We may note that $a \lcontr b = a \rcontr b = \form{a, b}$ for $a, b \in V$.
Now define the products
$$
  aX = a\lcontr X + a\wedge X,\quad
  Xa = X\rcontr a + X\wedge a,
$$
which can be seen to be exactly the Clifford algebra products of $a$ and $X$.
By this definition, this product is self-adjoint:
$\form{aX, Y} = \form{X, aY}$ with an analogous formula for $a$ on the right.
But any blade $A$ can be written as $A = a_1a_2\cdots a_k$ for some $k$
and $a_1,\dotsc, a_k \in V$ using Clifford products (which are associative);
hence
$$
  \form{AX, Y}
  = \form{a_1a_2\cdots a_kX, Y}
  = \form{a_2a_3\cdots a_kX, a_1Y}
  = \form{a_3a_4\cdots a_kX, a_2a_1Y}
  = \form{X, \rev AY}.
$$
By linearity, the same holds for when $A$ is an arbitrary multivector.
Thus,
$$
  \form{X, Y} = \form{X\cdot 1, Y} = \form{1, \rev XY} = \form{\rev XY}_0.
$$

It's worth noting that the contractions
have a very direct connection with the Clifford product.
For any $X, Y \in \Ext V$,
$$
  X\lcontr Y = \sum_{j=0}^n\sum_{k=0}^n\form{\form{\rev X}_j\form{Y}_k}_{k-j},\quad
  X\rcontr Y = \sum_{j=0}^n\sum_{k=0}^n\form{\form{X}_j\form{\rev Y}_k}_{j-k},
$$
where $\form{\cdot}_l = 0$ if $l < 0$.
For $a \in V$ and $X \in \Ext V$ in particular,
$$
  a\lcontr X = \frac12(aX - \hat Xa),\quad
  X\rcontr a = \frac12(Xa - a\hat X),
$$
where $\hat X$ is the grade involution (i.e. main involution) applied to $X$.
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