# How do you determine if the vectors are coplanar: a = [-2,-1,4], b = [5,-2,5], and c = [3,0,-1]?

Jul 28, 2016

They are coplanar.

#### Explanation:

Evaluate the determinant of the matrix with these vectors as rows:

$\left\mid \begin{matrix}- 2 & - 1 & 4 \\ 5 & - 2 & 5 \\ 3 & 0 & - 1\end{matrix} \right\mid$

$= - 2 \left\mid \begin{matrix}- 2 & 5 \\ 0 & - 1\end{matrix} \right\mid - 1 \left\mid \begin{matrix}5 & 5 \\ - 1 & 3\end{matrix} \right\mid + 4 \left\mid \begin{matrix}5 & - 2 \\ 3 & 0\end{matrix} \right\mid$

$= - 2 \left(2\right) - 1 \left(20\right) + 4 \left(6\right)$

$= - 4 - 20 + 24$

$= 0$

Since the determinant is $0$, these vectors only span at most a $2$ dimensional space. That is, they are coplanar.

Jul 28, 2016

See below

#### Explanation:

Two non linearly dependent vectors conveniently parametrized, plus a point, generate a plane

$\Pi \to {p}_{0} + {\lambda}_{1} \vec{a} + {\lambda}_{2} \vec{b}$.

with the plane tangent space given by.

${t}_{\Pi} \to {\lambda}_{1} \vec{a} + {\lambda}_{2} \vec{b}$ with $\left\{{\lambda}_{1} , {\lambda}_{2}\right\} \in {\mathbb{R}}^{2}$

Now if $\vec{c} \in {t}_{\Pi}$ then $\exists \left\{{\lambda}_{1} , {\lambda}_{2}\right\} \in {\mathbb{R}}^{2} | {\lambda}_{1} \vec{a} + {\lambda}_{2} \vec{b} = \vec{c}$

Solving for ${\lambda}_{1} , {\lambda}_{2}$ we have

{ (-2 lambda_1 + 5 lambda_2 = 3), (-lambda_1 - 2 lambda_2 = 0), (4 lambda_1 + 5 lambda_2 = -1) :}

${\lambda}_{1} = - \frac{2}{3} , {\lambda}_{2} = \frac{1}{3}$

so $\vec{c} \in {t}_{\Pi}$