A, B, C, D are any four points,then overright | vedclass

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(D) Let the position vectors of points $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Then $\overrightarrow{AB} = \vec{b} - \vec{a}

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(D) Let the position vectors of points $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Then $\overrightarrow{AB} = \vec{b} - \vec{a}$,$\overrightarrow{BC} = \vec{c} - \vec{b}$,$\overrightarrow{CD} = \vec{d} - \vec{c}$,$\overrightarrow{AD} = \vec{d} - \vec{a}$,$\overrightarrow{CA} = \vec{a} - \vec{c}$,and $\overrightarrow{BD} = \vec{d} - \vec{b}$.Substituting these into the expression:$\overrightarrow{AB} \cdot \overrightarrow{CD} + \overrightarrow{BC} \cdot \overrightarrow{AD} + \overrightarrow{CA} \cdot \overrightarrow{BD}$$= (\vec{b} - \vec{a}) \cdot (\vec{d} - \vec{c}) + (\vec{c} - \vec{b}) \cdot (\vec{d} - \vec{a}) + (\vec{a} - \vec{c}) \cdot (\vec{d} - \vec{b})$$= (\vec{b} \cdot \vec{d} - \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{d} + \vec{a} \cdot \vec{c}) + (\vec{c} \cdot \vec{d} - \vec{c} \cdot \vec{a} - \vec{b} \cdot \vec{d} + \vec{b} \cdot \vec{a}) + (\vec{a} \cdot \vec{d} - \vec{a} \cdot \vec{b} - \vec{c} \cdot \vec{d} + \vec{c} \cdot \vec{b})$Grouping the terms:$= (\vec{b} \cdot \vec{d} - \vec{b} \cdot \vec{d}) + (-\vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{b}) + (-\vec{a} \cdot \vec{d} + \vec{a} \cdot \vec{d}) + (\vec{a} \cdot \vec{c} - \vec{c} \cdot \vec{a}) + (\vec{c} \cdot \vec{d} - \vec{c} \cdot \vec{d}) + (\vec{b} \cdot \vec{a} - \vec{a} \cdot \vec{b})$$= 0 + 0 + 0 + 0 + 0 + 0 = 0$.

$A, B, C, D$ are any four points,then $\overrightarrow{AB} \cdot \overrightarrow{CD} + \overrightarrow{BC} \cdot \overrightarrow{AD} + \overrightarrow{CA} \cdot \overrightarrow{BD} = $

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