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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?

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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 $0$:

$\vec{AD} \cdot(\vec{AB} \times \vec{AC})=0$

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    $\begingroup$ 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). $\endgroup$ Jun 18, 2015 at 16:04
  • $\begingroup$ @MarkBennet This is true, but I think this is more advanced thinking and knowledge than just finding out normal vectors. $\endgroup$
    – Martigan
    Jun 18, 2015 at 16:15
  • $\begingroup$ Is it possible to just solve the equation of the plane ABC, then see if the other point D is included in that plane? I think it'll be simpler. $\endgroup$ Jan 17, 2016 at 16:29
  • $\begingroup$ @user3932000 In fact this is the same. When you compute the triple product, you are doing as many calculus as finding out the equation and verifying that D belongs to the plane. $\endgroup$
    – Martigan
    Jan 19, 2016 at 11:37

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