Four points A, B , C , D with position vectors $\vec{a},\vec{b}, \vec{c}, \vec{d}$ respectively are coplanar iff $3\vec{a} -2\vec{b}+ \vec{c}-2 \vec{d}=\vec{0}$
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We know that four vectors $\vec{a},\vec{b}, \vec{c}, \vec{d}$ are coplanar iff
$x\vec{a} +y\vec{b}+ z\vec{c}+t \vec{d}=0$ and $x+y+z+t=0$
How can I prove the result. Please help
we have $3\vec{OA}-2\vec{OB}+\vec{OC}-2\vec{OD}=0$
or, Using Chasles relation,
$2\vec{BA}+\vec{DA}+\vec{DC}=0$
thus
$2\vec{BD}+3\vec{DA}+\vec{DC}=0$
and
$\vec{DC}=2\vec{DB}-3\vec{DA}$
which means that the point $C$ is in the plane defined by the points $D , A$ and $B$ or by the vectors
$\vec{DA}$ and $\vec{DB}$.