# Show that four points are coplanar

I read all posts online regarding how to show four points are coplanar. However, none of them discuss the idea behind the method. Can someone explain how the triple scalar product works?

You know that three points $A,B,C$ (two vectors $\vec{AB}$, $\vec{AC}$) form a plane. If you want to show the fourth one $D$ is on the same plane, you have to show that it forms, with any of the other point already belonging to the plane, a vector belonging to the plane (for instance $\vec{AD}$).
Since the cross product of two vectors is normal to the plane formed by the two vectors ($\vec{AB} \times \vec{AC}$ is normal to the plane $ABC$), you just have to prove your last vector $\vec{AD}$ is normal to this cross product, hence the triple product that should be equal to $0$:
$\vec{AD} \cdot(\vec{AB} \times \vec{AC})=0$
• This is also equivalent to saying that the volume of the parallelepiped built on the vectors $AB, AC, AD$ is zero (because it is flat). Jun 18, 2015 at 16:04