# Proving that three vectors in $\mathbb R^3$ are coplanar if one is a linear combination of the other two

I read that three vectors in $\mathbb R^3$ are coplanar if one is a linear combination of the other two. I tried to prove this statement but couldn't do it.

I know that three vectors $\mathbf u,\,\mathbf v,\,\mathbf w \in \mathbb R^3$ are complanar if and only if $$\mathbf u \cdot (\mathbf v \times \mathbf w) = 0.$$

Can I use this fact to prove the above statement? Also is the statement an "if" or an "if and only if"?

Suppose $\;\mathbf u=a\mathbf v+b\mathbf w\;, \;\;a,b\in\Bbb R\;$ , then using the properties of cross product:
$$\mathbf u\cdot(\mathbf v\times\mathbf w)=(a\mathbf v+b\mathbf w)\cdot(\mathbf v\times \mathbf w)=a\mathbf v\cdot(\mathbf v\times\mathbf w)+b\mathbf w\cdot(\mathbf v\times\mathbf w)=0$$
since $\;\mathbf v\times\mathbf w\;$ is a vector perpendicular to both $\;\mathbf v\,,\,\,\mathbf w\;$ .
The iff statement enters only when $\;\mathbf v\,,\,\mathbf w\neq\mathbf 0\;$ , otherwise we get no plane at all.
• Thanks! Your proof is really clear. I see you wrote $\mathbf 0$ instead of $0$ when comparing to vectors. That's what I write as well in my notes. My book instead keeps writing $0$ even in (in)equalities with vectors, but I think it's wrong, isn't it? The book only defined equality between vectors. – rubik Mar 8 '15 at 8:51