# Coplanarity

Coplanarity is concept that applies to vectors in $\displaystyle \mathbb{R}^{3}$. A set of vectors is coplanar if they all lie on the same plane. Just as a single vector is always collinear with itself, a pair of vectors are always coplanar. With three vectors, sometimes they are and sometimes they aren’t.

Three vectors $\displaystyle \vec{u}$, $\displaystyle \vec{v}$, and $\displaystyle \vec{w}$ are coplanar if and only if there exists real numbers $\displaystyle a$ and $\displaystyle b$ such that

$\displaystyle \vec{u} = a \vec{v} + b \vec{w}$.

In other words, it must be possible to express each as a linear combination of the other two. To determine this, you should use the following method:

- Write the linear combination expression as shown above. It doesn’t matter which of the three vectors is alone on the left-hand side.
- Extract three linear equations from the vector equation.
- Choose two of the equations and use them to solve for $\displaystyle a$ or $\displaystyle b$.
- Back-substitute the variable you just found into one of the two chosen equations,
*not*into the unused one, and solve for the remaining variable. - Perform an LS/RS verification of the as yet unused equation with the values of $\displaystyle a$ and $\displaystyle b$ that you just found.
- If $\displaystyle \text{LS} = \text{RS}$, the system is consistent and the vectors are coplanar. If not, the system is inconsistent and the vectors are non-coplanar.

## Example

Are the vectors $\displaystyle \left [ - 1 , 2 , 3 \right ]$, $\displaystyle \left [ 4 , 1 , - 2 \right ]$, and $\displaystyle \left [ - 14 , - 1 , 16 \right ]$ coplanar?

First, we write the linear combination equation:

$\displaystyle \left [ - 14 , - 1 , 16 \right ] = a \left [ - 1 , 2 , 3 \right ] + b \left [ 4 , 1 , - 2 \right ]$.

That gives us the three equations

$\displaystyle - 14 = - a + 4 b , \quad - 1 = 2 a + b , \quad 16 = 3 a - 2 b$.

Solving the system defined by the first two equations tells us that

$\displaystyle a = \frac{10}{9} \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad b = - \frac{29}{9}$.

If we substitute those values into the third equation, we get

$\displaystyle \text{LS} = 16 \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad \text{RS} = \frac{88}{9}$.

Since $\displaystyle \text{LS} \ne \text{RS}$, the system is inconsistent and therefore the vectors $\displaystyle \left [ - 1 , 2 , 3 \right ]$, $\displaystyle \left [ 4 , 1 , - 2 \right ]$, and $\displaystyle \left [ - 14 , - 1 , 16 \right ]$ are non-coplanar.