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Let $\{a, b\}$ be a linearly independent set in $\mathbb{R}^3$ If $c \in \mathbb{R}^3$ such that $(a \times b) \cdot c = 0$ then $c \in \text{span } \{a, b \}$

I know this is true.

Technically we have a plane $P = \text{span } \{a, b\}$ and $a \times b = n$ is the normal vector.

What the statement is saying is that $c$ is orthogonal to $n$.

Any hints?

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    Hint: given the premise, $a,b,a \times b$ are linearly independent in $\mathbb{R}^3$ so $c = \lambda \,a + \mu \,b + \nu \,a\times b$. Now use the condition that $(a\times b) \cdot c = 0\,$. – 2017-02-06
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    @dxiv, how can you say $a, b, a\times b$ are linearly independent? Why is the cross product LI to the others? – 2017-02-06
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    Because it's orthogonal to the plane of $a,b$ as you said. Or else suppose $a \times b = \lambda a + \mu b$ then derive a contradiction using the triple product property that $(u \times v) \cdot v = 0$. – 2017-02-06

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Take the subspace generated by $n$ (it has dimension 1), call it $$. Because we are in $\mathbb{R}^3$, its orthogonal subspace has dimension 3-1=2. So the subspace orthogonal to $$ has dimension 2. At the same time, if $a$ and $b$ are linearly independent, and because both are orthogonal to $n$, they form a basis of the subspace orthogonal to $n$. So any vector that is orthogonal to $n$, such as $c$, can be written as a linear combination of the elements of the basis, $a$ and $b$.