When four points are coplanar (let be A B C D) then why $AB.BC\times AD=0$?
Or just [AB AC AD]=0?
I know dot product means $AB.AC=(AB)(AC)\cos\theta$ when $\theta$ is angle between vectors AB and AC.
And $AB\times AC= (AB)(AC)\sin\alpha$ where $\alpha$ is angle between AB and AC.
Thanks!
Because of the algebric form of a plane. If we have a perpendicular vector $\vec{n}$ to a plane and two points $(a,b,c)$ and $(x,y,z)$ on the same plane then:
$$\vec {n} \cdot (x-a,y-b,z-c)=0 \quad (1)$$
On your case, by definition, $\vec {BC} \times \vec {AD} $ is perpendicular vector to the plane that contain both vectors. Then if
$$\vec {AB} \cdot (\vec {BC} \times \vec {AD}) =0$$
It means that the vector $\vec {AB}$ respect the equation like $(1)$ and then it belongs to the same plane defined by $\vec {BC}$ and $\vec {AD}$.