Proving linear dependence when the triple product is zero How would I prove the following statement:
$\vec{A}\cdot(\vec{B}\times \vec{C})=0 \Leftrightarrow \vec{A}, \vec{B}, \vec{C} \text{ is linearly dependent.}$
The inverse is quite straightforward. However, going from left to right is kind of difficult. I just can't find out how to start the proof.
I am trying to show for a general coordinate system, so calculation using elements is not what I need.
 A: If $B$ and $C$ are linearly dependent, then we are done.
If $B$ and $C$ are linearly independent, then they span a two-dimensional plane in $\Bbb{R}^3$. Note that $B \times C$ is normal to this plane. If $A \cdot (B \times C) = 0$, then $A$ is perpendicular to this normal vector, meaning that it must be contained in this plane. In other words, $A \in \operatorname{span}(B, C)$.
In either case, if $A \cdot (B \times C) = 0$, we have linear dependence of $A, B, C$.
A: $َA \cdot B \times  C$ is the volume of a parallelepiped whose edges are formed by three vectors.  The only way the volume can be zero is if all three vectors are co-planar and every three vectors on a plane are linearly independent.
A: This triple product of the three vectors is basically the determinant formed by the three given vectors taken as the columns of the matrix associated with this determinant. Now, think about a standard property of the determinant function: If any column is a linear combination of the other columns, then the determinant is zero. Conversely, the determinant zero means the given matrix is not invertible that means the rank of the associated matrix is not 3. That also means that the given three vectors don't span the whole space R^3. Hence, these three vectors span a two or one-dimensional subspace of R^3. Both cases imply that the given three vectors are not linearly independent. Please think about it.
