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(D) Since the four points with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar,the vectors $(\vec{b}-\vec{a}), (\vec{c}-\vec{a}),$

Vedclass · Accepted Answer

$[\vec{b} \vec{c} \vec{d}]+[\vec{d} \vec{a} \vec{c}]+[\vec{d} \vec{b} \vec{a}]$

Vedclass · Answer

(D) Since the four points with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar,the vectors $(\vec{b}-\vec{a}), (\vec{c}-\vec{a}),$ and $(\vec{d}-\vec{a})$ are coplanar.Therefore,their scalar triple product is zero: $[\vec{b}-\vec{a}, \vec{c}-\vec{a}, \vec{d}-\vec{a}] = 0$.Expanding this,we get $(\vec{b}-\vec{a}) \cdot ((\vec{c}-\vec{a}) 	imes (\vec{d}-\vec{a})) = 0$.Using the properties of the scalar triple product,this expands to $[\vec{b} \vec{c} \vec{d}] - [\vec{b} \vec{c} \vec{a}] - [\vec{b} \vec{a} \vec{d}] - [\vec{a} \vec{c} \vec{d}] = 0$.Rearranging the terms,we get $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{d}] + [\vec{d} \vec{a} \vec{c}] + [\vec{d} \vec{b} \vec{a}]$.Comparing this with the given options,option $D$ is correct.

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

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