geometric algebra: dot products of vectors with vectors vs bivectors The book "Matrix Gateway to Geometric Algebra" by Garret Sobczyk gives the following identity:
$a\cdot(b \wedge c)=(a\cdot b)c - (a\cdot c)b$
Using the identities
$a\cdot b= (ab+ba)/2$
$a\wedge b=(ab-ba)/2$
to expand the proposed identity, including the step, e.g.,
$$a\cdot (b\wedge c)=(a(b\wedge c) + (b\wedge c) a)/2$$
I get $$abc + bca - acb - cba = abc + cba - acb - bca$$
which does not appear to check.
Is there a step I am missing?
On the same page 31 the author writes (equation 2.14)
$$a\cdot(b\wedge c) = (a(b\wedge c) - (b\wedge c)a)/2$$
which is the opposite sign to the basic "identity" I relied upon. Why is there sign minus in this instance? Is not $x=(b\wedge c)$, where x is an element, subject to the identity $a\cdot x = (ax+xa)/2$? There is a similar reversal of sign in equation 2.15.
Thanks,
Gary
 A: The identities
$$a\cdot b= (ab+ba)/2$$
$$a\wedge b=(ab-ba)/2$$
are correct, but they are only for vectors $ a, b $.  Eq 2.14 from the text:
$$a\cdot(b\wedge c) = (a(b\wedge c) - (b\wedge c)a)/2$$
(with the negative sign), is also correct.  In general, if $ M $ is a k-vector, and $ a $ is a vector, then one has
$$a \cdot M=\frac{1}{{2}} \left( { a M + (-1)^{k-1} M a } \right).$$
I don't know how the Matrix Gateway book derives eq 2.14.  Here's one way to show it, rewriting the dot product as a grade 1 selection (i.e. take just the vector parts of any multivector products)
$$\begin{aligned}a \cdot \left( { b \wedge c } \right)&={\left\langle{{ a \left( { b \wedge c } \right) }}\right\rangle}_{1} \\    &={\left\langle{{ a \left( { b c - b \cdot c } \right) }}\right\rangle}_{1} \\ &={\left\langle{{ a b c }}\right\rangle}_{1} - a \left( { b \cdot c } \right),\end{aligned}$$
however, using $ b a = 2 a \cdot b - b a $, we have
$$\begin{aligned}{\left\langle{{ a b c }}\right\rangle}_{1}&={\left\langle{{ \left( { 2 a \cdot b - b a } \right) c }}\right\rangle}_{1} \\ &=2 \left( { a \cdot b } \right) c - {\left\langle{{ b a c }}\right\rangle}_{1} \\ &=2 \left( { a \cdot b } \right) c - {\left\langle{{ b \left( { 2 a \cdot c - c a } \right) }}\right\rangle}_{1} \\ &=2 \left( { a \cdot b } \right) c - 2 \left( { a \cdot c } \right) b + {\left\langle{{ b c a }}\right\rangle}_{1} \\ &=2 \left( { a \cdot b } \right) c - 2 \left( { a \cdot c } \right) b + \left( { b \wedge c } \right) \cdot a + \left( { b \cdot c } \right) a.\end{aligned}$$
Putting the pieces together, we have
$$a \cdot \left( { b \wedge c } \right)=2 \left( { a \cdot b } \right) c - 2 \left( { a \cdot c } \right) b + \left( { b \wedge c } \right) \cdot a.$$
Finally, note that the reverse of vector is a vector ($\tilde{a} = a$), so if we apply the reversion operator to the last term:
$$\begin{aligned}\left( { b \wedge c } \right) \cdot a&=a \cdot \left( { c \wedge b } \right) \\ &=-a \cdot \left( { b \wedge c } \right),\end{aligned}$$
and then substitute this back in, we are most of the way towards a derivation of eq 2.14 from the text:
$$a \cdot \left( { b \wedge c } \right)=2 \left( { a \cdot b } \right) c - 2 \left( { a \cdot c } \right) b - a \left( { b \wedge c } \right),$$
so after some trivial rearrangement, we have:
$$a \cdot \left( { b \wedge c } \right)=\left( { a \cdot b } \right) c - \left( { a \cdot c } \right) b.$$
You could also derive this from 2.14, using the reversion argument above, since the trivector terms in that antisymmetric sum cancel.
A: Assuming that we are given a multiplication that is associative
and bilinear, but not commutative, then define
$$a\cdot b := \frac12(ab+ba),\qquad
a\wedge b := \frac12(ab-ba),$$
which apply, at least for $1$-vectors, with perhaps some sign
changes for general $k$-vectors. No equation such as
$$a\cdot(b \wedge c)=(a\cdot b)c -
  (a\cdot c)b$$
can be an identity because, as you computed,
the left side is equal to
$\,\frac14(abc + cba - acb - bca)\,$ while
$\, (a\cdot b)c = \frac12(abc+bac)\,$ and
$\, a(b\cdot c) = \frac12(abc+acb).\,$
There is no way to get $\,\frac12(abc+cba)\,$
no matter what scalar factors are introduced.
The proof is by linear algebra using as the basis $$(abc,acb,bac,bca,cab,cba).$$
