Coplanar points

Given any three points one may always find a plane containing all three of them. However, for more than three points, there may not always be a plane containing all of them. Coplanar points are three or more points that lie in the same plane. The problem of determining whether a set of points are coplanar is interesting in the case of more than three points since three or fewer are always coplanar.

One approach to determining whether a given set of vectors is coplanar is to write the equation of the plane for the first three points provided that they are not collinear (i.e. that they do not lie on the same line) and then to check that the remaining points satisfy the derived equation of the plane in the form $\boldsymbol{r} \cdot \boldsymbol{n}=d$.

We may also use the scalar triple product, $\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol{c})$ to determine if $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ lie on the same plane through the origin. Now, $\boldsymbol{b} \times \boldsymbol{c}$ is perpendicular to the plane containing $\boldsymbol{b}$ and $\boldsymbol{c}$ (and the origin). If the scalar triple product, $\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol{c})$ is 0 then the origin, $\boldsymbol{a}, \boldsymbol{b}$, and $\boldsymbol{c}$ are coplanar.

Three vectors $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ are coplanar iff

or

or

for some scalar $\lambda, \mu$. Rearranging (9.34) we obtain

which implies that $\boldsymbol{a}, \boldsymbol{b}$, and $\boldsymbol{c}$ are linearly dependent.

Definition 9.8 Linear dependence and independence

The vectors $\boldsymbol{a}, \boldsymbol{b}$, and $\boldsymbol{c}$ are said to be linearly dependent if there exist real scalars $\alpha$, $\beta, \gamma$, not all zero such that

If $\boldsymbol{a}, \boldsymbol{b}$, and $\boldsymbol{c}$ are not linearly dependent, they are linearly independent. This is expressed by

provided that $\alpha=\beta=\gamma=0$. Example 9.6 Show that $A=(2,3,4), B=(2,3,6), C=(1,1,1)$, and $\mathrm{D}=(1,1,5)$ are coplanar.

Solution We start with the three points, say $A, B$, and $C$ and define two direction vectors (there are more than one way to choose these) as follows

We also denote the position vectors of points $A$ and $D$ as $\boldsymbol{a}$ and $\boldsymbol{d}$, respectively. Using $r_{1}$ and $r_{2}$, the points $A, B, C$ lie on a plane with equation

Next, we need to show that the fourth point, $D$ also lies on the plane. We now define $\overrightarrow{A D}=(-1,-2,1)$ which is equivalent to $(\boldsymbol{d}-\boldsymbol{a})$ and ask if it lies on the same plane as the other 3 points. We proceed by taking the cross product of $\boldsymbol{r}_{\mathbf{1}}$ and $\boldsymbol{r}_{\mathbf{2}}$ which gives a vector that is perpendicular to the plane containing $A, B$, and $C$,

We expect the dot product between $(\boldsymbol{d}-\boldsymbol{a})$ and $\boldsymbol{r}_{\mathbf{1}} \times \boldsymbol{r}_{\mathbf{2}}$ to be 0 if $(\boldsymbol{d}-\boldsymbol{a})$ also lies in the plane. Indeed,

Alternatively, we can show that the four points are coplanar by checking that the dot product of $\boldsymbol{a}$ with $\boldsymbol{r}_{\mathbf{1}} \times \boldsymbol{r}_{\mathbf{2}}$ gives the same result as the dot product of $\boldsymbol{d}$ with $\boldsymbol{r}_{\mathbf{1}} \times \boldsymbol{r}_{\mathbf{2}}$. Since the dot product is a linear operator,

Hence, the two terms on the RHS are equal if and only if the LHS is 0 .