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 :
Solution
$$
\begin{aligned}
& {[\vec{b}-\vec{a} \,\,\,\,\,\vec{c}-\vec{a} \,\,\,\,\,\vec{d}-\vec{a}]=0} \\\\
& (\vec{b}-\vec{a}) \cdot[(\vec{c}-\vec{a}) \times(\vec{d}-\vec{a})]=0 \\\\
& (\vec{b}-\vec{a}) \cdot(\vec{c} \times \vec{d}-\vec{c} \times \vec{a}-\vec{a} \times \vec{d})=0 \\\\
& {[\vec{b}\,\,\,\,\, \vec{c} \,\,\,\,\,\vec{d}]-[\vec{b} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{a}]-[\vec{b} \,\,\,\,\,\vec{a} \,\,\,\,\,\vec{d}]-[\vec{a} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{d}]=0} \\\\
& \therefore [\vec{a} \,\,\,\,\,\vec{b} \,\,\,\,\,\vec{c}]=[\vec{b} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{d}]+[\vec{a} \,\,\,\,\,\vec{b} \,\,\,\,\,\vec{d}]+[\vec{a} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{c}] \\\\
& \quad=[\vec{d} \,\,\,\,\,\vec{c} \,\,\,\,\,\vec{a}]+[\vec{b} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{a}]+[\vec{c} \,\,\,\,\,\vec{d} \,\,\,\,\,\vec{b}]
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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