Motivation behind inner and exterior product of a vector with a k-grade blade in geometric algebra The inner product of a vector with a k-grade blade is given as:
$$ a \cdot A_k = \frac12 \left( aA_k-(-1)^kA_k a\right)$$
And the exterior product as:
$$ a \wedge A_k= \frac12 \left( a A_k+ (-1)^kA_k a \right)$$
What are the motivation for the definitions?
I understand very well the motivation for inner product of two vectors and bivectors. They consider all the information regarding perpendicularity and parallelism. The anti symmetric part shows the area and the symmetric the dot. But here, it seems so that depending on the grade $k$ the wedge becomes symmetric and in other cases anti symmetric.
P.S: Basis r-blades haven't been introduced yet in the book.
 A: As mentioned by Hans, and also detailed in an answer to a related question, these are not fundamental definitions.  Instead, you should view the following as fundamental:
$$   a \cdot A_k = {\left\langle{{ a A_k }}\right\rangle}_{{k-1}},$$
$$   a \wedge A_k = {\left\langle{{ a A_k }}\right\rangle}_{{k+1}},$$
where the angle brackets mean grade selection (i.e. the dot product of a vector and a grade $k$ blade is all the grade $k-1$ components of the product, and the wedge product of the same is all the grade $k+1$ components).  With those definitions in place, one can then go through the exercise of proving that
$${\left\langle{{ a A_k }}\right\rangle}_{{k-1}} = \frac12 \left( aA_k-(-1)^kA_k a\right),$$
and
$${\left\langle{{a  A_k}}\right\rangle}_{{k+1}} = \frac12 \left( a A_k+ (-1)^kA_k a \right).$$
To prove these relations, split the blade $ A_k $ into components that intersect with and are disjoint from $ a $ as follows
$$A_k=\frac{1}{{a}} n_1 n_2 \cdots n_{k-1} + m_1 m_2 \cdots m_k,$$
where the $ m_i, n_i $ all orthogonal to $ a $, and where $ m_i $ are mutually orthogonal, and where $ n_i $ are mutually orthogonal (but the $m_i$'s do not have to be orthogonal to any of the $n_j$'s).  The products of $ A_k $ with $ a $ are
$$\begin{aligned}a A_k&=a \frac{1}{{a}} n_1 n_2 \cdots n_{k-1} + a m_1 m_2 \cdots m_k \\ &=n_1 n_2 \cdots n_{k-1} + a m_1 m_2 \cdots m_k,\end{aligned}$$
and
$$\begin{aligned}A_k a&=\frac{1}{{a}} n_1 n_2 \cdots n_{k-1} a + m_1 m_2 \cdots m_k a \\ &=(-1)^{k-1} n_1 n_2 \cdots n_{k-1} + (-1)^k a m_1 m_2 \cdots m_k \\ &=(-1)^k \left( { - n_1 n_2 \cdots n_{k-1} + a m_1 m_2 \cdots m_k } \right),\end{aligned}$$
or
$$(-1)^k A_k a=- n_1 n_2 \cdots n_{k-1} + a m_1 m_2 \cdots m_k.$$
Respective addition and subtraction gives
$$\begin{aligned}a A_k + (-1)^k A_k a&= 2 a m_1 m_2 \cdots m_k \\ &= 2 {\left\langle{{a A_k}}\right\rangle}_{{k+1}},\end{aligned}$$
and
$$\begin{aligned}a A_k - (-1)^k A_k a&=2n_1 n_2 \cdots n_{k-1} \\ &= 2 {\left\langle{{a A_k}}\right\rangle}_{{k-1}},\end{aligned}$$
completing the proof.
