Is it true that three vectors $\vec{u}, \vec{v} , \vec{w} $ lie on the same plane if and only if there exists constants $A,B,C,D$ for which $A\vec{u} + B\vec{v} + C\vec{w} +D =\vec{0} $ ?
If so, how can I prove it ? I know that three vectors lie on the same plane if and only if $\vec{u}\cdot (\vec{v}\times \vec{w} ) =0 $ . Does it help ?
Thanks in advance
On
Your equation is inconsistant: you are adding vectors and numbers.
The good condition is: $\vec u, \vec v$ and $\vec w$ are copolanar iff there exists three real numbers $A,B,C$, not all zero, such as : $A\vec u + B\vec v + C \vec w =0$.
In linear algebra, it is the definition of "three vectors are coplanar".
Two non collinear lines form a linearly independent basis that will span a plane, however introducing a third coplanar vector means the three vectors will be linearly dependent since the third can be written as a linear combination of the previous two. Therefore the definition of linearly dependence and coplanar are equivalent for $\mathbb{R}^3$.
Therefore you need an $A,B,C$ not all zero such that $$A\mathbf{u}+B\mathbf{v}+C\mathbf{w}=0$$ as this is the common test for linear dependence.