Let $\vec v_1$, $\vec v_2$, $\vec v_3$ be three vectors in $\mathbb{R^3}$. We say these vectors are coplanar, if there is a plane through the origin in $\mathbb{R^3}$ that contains all of them.
Show that $\vec v_1$, $\vec v_2$, $\vec v_3$ are coplanar if and only if they are linearly independent?
I started by using the definition of linear independence (e.g coefficients equal to zero) but I am somewhat confused how to set up this proof.
Best Answer
I'm going to assume that you meant coplanar if and only if they are linearly dependent because otherwise it is false.
If $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are coplanar, their triple product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ is $0$ since $\mathbf{u}$ must be perpendicular to $\mathbf{v} \times \mathbf{w}$. The triple product can be calculated using the determinant $$\begin{vmatrix} u_0 && u_1 && u_2 \\ v_0 && v_1 && v_2 \\ w_0 && w_1 && w_2 \end{vmatrix} = 0 $$ A determinant of $0$ implies that the rows are linearly dependent. This proof is bi-directional.
If you don't want to use determinant, you can instead use the fact that a plane is a $2$-dimensional subspace of $\mathbb{R}^3$, and since $\{\mathbf{u}, \mathbf{v}, \mathbf{w}\}$ is a list of $3 > 2$ elements, it is linearly dependent.
For the other direction, if $\{\mathbf{u}, \mathbf{v}, \mathbf{w}\}$ is linearly dependent, then one of the vectors(without loss of generality we can assume to be $\mathbf{w}$) is a linear combination of two others. That is, $\lambda\mathbf{u} + \mu\mathbf{v} = \mathbf{w}$. The span of this set is hence equal to the span of $\{\mathbf{u}, \mathbf{v}\}$, a plane.