If $ (a \times b) \cdot c = 0$, then prove algebraically that $c \in span(a,b) $ Let $a,b,c$ be vectors in $\mathbb{R}^{3}$. Let the set $\{a,b \}$ be known to be linearly independent. 
If $ (a \times b) \cdot c = 0$, then prove algebraically that $c \in span(a,b) $
This is actually pretty  easy to show using determinants, since the triple scalar product is the determinant of the matrix formed by $a,b,c$. I am looking for an algebraic proof, that only uses the definitions of the dot and cross products, and preferably nothing more. Is this possible?
 A: I'm not sure precisely what criteria you're looking for, but here's one approach.
If ${a,b,d}$ are linearly independent, you have a formula
$$ x = \frac{(a \times b) \cdot x}{(a \times b) \cdot d} d
+ \frac{(b \times d) \cdot x}{(b \times d) \cdot a} a
+ \frac{(d \times a) \cdot x}{(d \times a) \cdot b} b
$$
whose truth can be verified by observing the right hand side is linear in $x$ and has the correct values at $x=a$, $x=b$, and $x=d$. The identity $(u \times v)\cdot v = 0$ is useful in this proof.
With this formula, it's easy to see that $x \in \operatorname{Span}\{a,b\}$ if and only if $(a \times b) \cdot x = 0$.
A: Since $a \times b$ is by definition orthogonal to both $a$ and $b$, this forms a basis of $\mathbb{R}^3$, hence we can write 
$$c = \lambda_1 a + \lambda_2 b + \lambda_3 (a \times b).$$
Computing the inner product of $c$ with $a \times b$ and using linearity of the dot product, we find:
$$(a \times b) \cdot c = \lambda_1 (a \times b) \cdot a + \lambda_2 (a \times b) \cdot b + \lambda_3 (a \times b) \cdot (a \times b)$$
and because $(a\times b) \cdot a = 0$ and $(a\times b) \cdot b = 0$, we find that 
$$0 = (a \times b) \cdot c = \lambda_3 (a \times b) \cdot (a \times b).$$
Since $(a \times b) \cdot (a \times b) = \|a \times b\| \neq 0$ (because $a,b$ are linearly independent, showing that $a \times b \neq 0$), we must have that $\lambda_3 = 0$ and therefore, we find that $c \in \text{span}(a,b)$.
