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.

• Since the statement to prove is coordinate-independent, it actually suffices to prove the statement in any one coordinate system you choose. – Travis Willse Sep 23 at 5:31
• Also, it would be useful for you to list which facts about the cross and dot products you have available. The left-hand side is $\det\pmatrix{{\bf A} & {\bf B} & {\bf C}}$, from which both directions follow from the fact that a square matrix has determinant $0$ iff its columns are linearly dependent. – Travis Willse Sep 23 at 5:34
• That was a very helpful insight. Then, I suppose I can just prove in Cartesian coordinate system. So the left-hand side is equal to det(A B C). How would we prove det(A B C)=0 implies linear dependence? – biology12323 Sep 23 at 5:52

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 $$A, B, C$$.
$$َA \cdot B \times C$$ 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.