Compatibility of cross and inner product on $\mathbb{R}^3$

Question

Consider the following operations on a three-dimensional vector space $\mathbb{R}^3$.

The cross product $\times: \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$ can be defined by the following properties:

1. Linearity in the first argument: $$(\alpha a + \beta b) \times c = \alpha (a \times c) + \beta (b \times c),$$
2. Anticommutativity: $$a \times b = – b \times a,$$
3. Jacobi identity: $$(a \times b) \times c + (b \times c) \times a + (c \times a) \times b = 0.$$

From these, one can deduce bilinearity, alternativity etc.

The inner product $\cdot: \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}$ can be defined by the following properties:

1. Linearity in the first argument: $$(\alpha a + \beta b) \cdot c = \alpha (a \cdot c) + \beta (b \cdot c),$$

2. Commutativity: $$a \cdot b = b \cdot a,$$

3. Positive-definiteness: $$a \cdot a \geq 0, \quad a \cdot a = 0 \iff a = 0.$$

From these, one can deduce bilinearity etc.

The question: It seems to me that the above properties are not sufficient to deduce the identity $$a \cdot (a \times b) = 0,$$ which is equivalent to the cyclicity of the triple scalar product. If I am correct, this would imply

1. There exists a realization of cross and inner product, different from the standard Euclidean ones in which $a \cdot (a \times b) \neq 0$. What would be a geometrical example of this?
2. The identity $a \cdot (a \times b) = 0$ can be thought of as a some sort of compatibility condition between the cross and inner product. Is there some deeper insight to this structure?

Answer 1

You are right: those properties are not enough to deduce the identity$$a.(a\times b)=0.\tag1$$For instance, if $\times$ stands for the usual cross-product and if$$(a_1,a_2,a_3).(b_1,b_2,b_3)=a_1b_1+2a_2b_2+a_3b_2+a_2b_3+a_3b_3,$$then this is an inner product, but $(1)$ doesn't hold; for instance, $(0,0,1).\bigl((0,0,1)\times(1,0,0)\bigr)=1$.