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(B) Let $A$ be the origin and let the position vectors of $B, C,$ and $D$ be $\vec{b}, \vec{c},$ and $\vec{d}$ respectively.Then $\overrightarrow{AB}

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$4$

Vedclass · Answer

(B) Let $A$ be the origin and let the position vectors of $B, C,$ and $D$ be $\vec{b}, \vec{c},$ and $\vec{d}$ respectively.Then $\overrightarrow{AB} = \vec{b}, \overrightarrow{CD} = \vec{d} - \vec{c}, \overrightarrow{BC} = \vec{c} - \vec{b}, \overrightarrow{AD} = \vec{d}, \overrightarrow{CA} = -\vec{c},$ and $\overrightarrow{BD} = \vec{d} - \vec{b}$.Now, consider the expression:$|\overrightarrow{AB} 	imes \overrightarrow{CD} + \overrightarrow{BC} 	imes \overrightarrow{AD} + \overrightarrow{CA} 	imes \overrightarrow{BD}|$$= |\vec{b} 	imes (\vec{d} - \vec{c}) + (\vec{c} - \vec{b}) 	imes \vec{d} - \vec{c} 	imes (\vec{d} - \vec{b})|$$= |\vec{b} 	imes \vec{d} - \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{d} - \vec{b} 	imes \vec{d} - \vec{c} 	imes \vec{d} + \vec{c} 	imes \vec{b}|$$= |-\vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{b}|$$= |-\vec{b} 	imes \vec{c} - \vec{b} 	imes \vec{c}| = |-2(\vec{b} 	imes \vec{c})| = 2|\vec{b} 	imes \vec{c}|$Since the area of $\Delta ABC$ is $\Delta = \frac{1}{2}|\vec{b} 	imes \vec{c}|$, we have $|\vec{b} 	imes \vec{c}| = 2\Delta$.Therefore, the expression equals $2(2\Delta) = 4\Delta$.

If $A, B, C, D$ are any four points in space, then $|\overrightarrow{AB} \times \overrightarrow{CD} + \overrightarrow{BC} \times \overrightarrow{AD} + \overrightarrow{CA} \times \overrightarrow{BD}|$ is equal to (where $\Delta$ denotes the area of $\Delta ABC$) (in $\Delta$)

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