Coplanar points

If there is a plane that contains every point of a given set, those points are called coplanar.

We'll be using vectors and specifically the cross product and dot product.

We want to check if the points $A,$ $B,$ $C$ and $D$ are coplanar.

First consider the plane $ABC.$ The cross product of the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is normal to the plane they are on.

Therefore, if the vector $\overrightarrow{AD}$ is normal to $\overrightarrow{AB} \times \overrightarrow{AC},$ all four points are on the same plane. That is, we want to prove:

$$\left(\overrightarrow{AB} \times \overrightarrow{AC}\right) \cdot \overrightarrow{AD} = 0 .$$

Are $A =(1, 2, 2),$ $B=(3, -2, 4),$ $C= (5, -1, 5),$ and $D = (3, 1, 3)$ on the same plane?

First we want to find the cross product $\overrightarrow{AB} \times \overrightarrow{AC} .$ The vectors are $\overrightarrow{AB} = (3-1)i + (-2-2)j + (4-2)k = 2i -4j + 2k$ and $\overrightarrow{AC} = (5-1)i + (-1-2)j + (5-2)k = 4i - 3j + 3k .$

The cross product is then $\left((-4)(3)-(2)(-3)\right)i+\left((2)(4)-(2)(3)\right)j +\left((2)(-3)-(-4)(4)\right)k = -6i + 2j + 10k .$

The vector $\overrightarrow{AD}$ is $(3-1)i + (1-2)j + (3-2)k = 2i -1j +1k .$ We then just need the dot product of this with $-6i + 2j + 10k.$

$$(2)(-6) + (-1)(2) + (1)(10) = -12 - 2 + 10 = -4$$

Since $-4 \neq 0 ,$ the four points are not coplanar.

The plane containing $A, B,$ and $C,$ with $D$ marked in blue.