If four distinct points with position vectors a,b,c and d are coplanar, then [a, , b, , c]…

Question

If four distinct points with position vectors $\vec{a},\vec{b},\vec{c}$ and $\vec{d}$ are coplanar, then $[\vec{a}\, \, \vec{b}\, \, \vec{c}]$ is equal to :

Accepted Answer

$\begin{gathered} [\vec{b}-\vec{a}\, \, \, \, \, \vec{c}-\vec{a}\, \, \, \, \, \vec{d}-\vec{a}]=0 \ (\vec{b}-\vec{a})\cdot [(\vec{c}-\vec{a})	imes (\vec{d}-\vec{a})]=0 \ (\vec{b}-\vec{a})\cdot (\vec{c}	imes \vec{d}-\vec{c}	imes \vec{a}-\vec{a}	imes \vec{d})=0 \ [\vec{b}\, \, \, \, \, \vec{c}\, \, \, \, \, \vec{d}]-[\vec{b}\, \, \, \, \, \vec{c}\, \, \, \, \, \vec{a}]-[\vec{b}\, \, \, \, \, \vec{a}\, \, \, \, \, \vec{d}]-[\vec{a}\, \, \, \, \, \vec{c}\, \, \, \, \, \vec{d}]=0 \ 	herefore [\vec{a}\, \, \, \, \, \vec{b}\, \, \, \, \, \vec{c}]=[\vec{b}\, \, \, \, \, \vec{c}\, \, \, \, \, \vec{d}]+[\vec{a}\, \, \, \, \, \vec{b}\, \, \, \, \, \vec{d}]+[\vec{a}\, \, \, \, \, \vec{d}\, \, \, \, \, \vec{c}] \ \, =[\vec{d}\, \, \, \, \, \vec{c}\, \, \, \, \, \vec{a}]+[\vec{b}\, \, \, \, \, \vec{d}\, \, \, \, \, \vec{a}]+[\vec{c}\, \, \, \, \, \vec{d}\, \, \, \, \, \vec{b}] \end{gathered}$

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