{$\vec{a},\vec{b}$}, ($\vec{a}\times\vec{b}$)$\cdot\vec{c}=0$ prove that $\vec{c} \in \operatorname{span}\{\vec a,\vec b\}$

Question

let $\vec{c} \in \Bbb{R^3}$ such that for a linearly independent set {$\vec{a},\vec{b}$}, ($\vec{a}\times\vec{b}$)$\cdot\vec{c}$\vec{a},\vec{b}}$

I have failed to find a counterexample which leads me to believe that this is true, but unfortunately I'm stuck on how to show that $\vec{c}$ can be written as a linear combination of $\vec{a},\:\vec{b}$.

Answer 1

Note that by definition of cross product $\vec v= \vec{a}\times\vec{b}$ is orthogonal both to $\vec{a}$ and $\vec{b}$, thus $\vec v$ is orthogonal to the plane spanned by $\vec a$ and $\vec b$.

Since by the definition given $(\vec{a}\times\vec{b})\cdot \vec c=0,$ $\vec c$ is orthogonal to $\vec{a}\times\vec{b}$ then $\vec c$ belongs to the span$(\vec a,\vec b)$.

Answer 2

The visual intuition is clear from a picture, draw $a$ and $b$ linearly independent. Draw their cross product which is orthogonal to both, and note that $c$ is orthogonal to the cross product. In $\mathbb{R}^3$, this doesn't leave anywhere for $c$ to go other than parallel to the plane spanned by $a$ and $b$.

Proving it, note that $(a\times b)\cdot c$ can be written as the determinant of following matrix $$ \begin{bmatrix} c_1&c_2&c_3\\ a_1&a_2&a_3\\ b_1&b_2&b_3 \end{bmatrix} $$ now if this determinant is $0$, what can you say about the vectors inside?

Answer 3

Dropping the vector-arrow clutter.

The one dimensional subspace $\langle a\times b\rangle$ is exactly the orthogonal complement of $\langle a,b\rangle$. That is, $\langle a\times b\rangle=\langle a,b\rangle ^\perp$.

To say that $(a\times b)\cdot c=0$ means that $c\in \langle a\times b\rangle ^\perp=\langle a, b\rangle ^{\perp\perp}=\langle a, b \rangle$.

Or for a different intuition, $\{a,b,a\times b\}$ is a basis for $\mathbb R^3$. Then $c$ is in the span... but it has no component in the direction of $a\times b$, so the only nonzero components must be in $\langle a,b \rangle$.