Properties of some operator on vectors of $\mathbb{R}^2$ Suppose that $\circ$ is an operation on $\Bbb R^2$ with the following properties:


*

*For any $\vec p,\vec q \in \mathbb{R}^2$, and $t \in \mathbb{R}$, $(t \vec p ) \circ \vec q = t(\vec p \circ \vec q)$ holds. 

*For any $\vec p, \vec q, \vec r \in \mathbb{R}^2$, $\vec p \circ (\vec q + \vec r) = \vec p \circ \vec q + \vec p \circ \vec r$ holds.

*For any $\vec p, \vec q \in \mathbb{R}^2$, $\vec p \circ \vec q = -\vec q \circ \vec p$ holds.

*For any $\vec p, \vec q, \vec r \in \mathbb{R}^2$, $(\vec p \circ \vec q) \circ \vec r = (\vec p \cdot \vec r)\vec q - (\vec q \cdot \vec r)\vec p$ holds.


Why is it then true that $\vec p \circ \vec q = \vec 0$ all the time?
 A: I assume that the operation $\circ$ is inner, i.e.
$$
\circ:\Bbb R^2\times\Bbb R^2\longrightarrow\Bbb R^2
$$
and satisfies rules 1.--4. as stated (else, 4. would make no sense).
By antisimmetricity (rule 3.) $\vec v\circ\vec v=0$ for all $\vec v\in\Bbb R^2$.
Fix an orthonormal basis $\{\vec e_1,\vec e_2\}$. By bilinearity and what we just observed, the operation $\circ$ is left completely determined by the value
$$
\vec w=\vec e_1\circ\vec e_2=a\vec e_1+b\vec e_2.
$$
Now, applying 4.
$$
0=\vec w\circ\vec w=(\vec e_1\circ\vec e_2)\circ\vec w=(\vec e_1\cdot\vec w)\vec e_2-(\vec e_2\cdot\vec w)\vec e_1=a\vec e_2-b\vec e_1.
$$
By linear independence of basis vectors, $a=b=0$, so that $\circ$ needs to be the constant zero function.
BUT: let now $\vec r=c\vec e_1+d\vec e_2\neq\vec0$, so that $(c,d)\neq(0,0)$. Rule 4. would yield
$$
\vec 0=(\vec e_1\circ\vec e_2)\circ\vec r=c\vec e_2-d\vec e_1\neq\vec 0
$$
(again by the linear independence of basis vectors) which is obviously impossible.
THUS no such a function exists, not even the constant zero function.
A: Hint: Take a look at the outer product in $\mathbb R^3$ and restrict one component to be constant. What do you see?
