question_answer

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

A)                 \[2\Delta \]

B)                 \[4\Delta \]

C)                 \[3\Delta \]

D)                 \[5\Delta \] (where D denotes the area of \[\Delta ABC\])

Correct Answer: B

Solution :

           Let \[A\] be the origin and let the position vectors of \[B,\,C\] and \[D\] be \[\mathbf{b},\,\mathbf{c}\] and \[\mathbf{d}\] respectively.                    Then \[\overrightarrow{AB}=\mathbf{b},\] \[\overrightarrow{CD}=\mathbf{d}-\mathbf{c},\] \[\overrightarrow{BC}=\mathbf{c}-\mathbf{b},\] \[\overrightarrow{AD}=\mathbf{d},\] \[\overrightarrow{CA}=-\mathbf{c}\] and \[\overrightarrow{BD}=\mathbf{d}-\mathbf{b}.\]            \[\therefore \,\,|\overrightarrow{AB}\times \overrightarrow{CD}+\overrightarrow{BC}\times \overrightarrow{AD}+\overrightarrow{CA}\times \overrightarrow{BD}|\]            \[=\,|\mathbf{b}\times (\mathbf{d}-\mathbf{c})+(\mathbf{c}-\mathbf{b})\times \mathbf{d}-\mathbf{c}\times (\mathbf{d}-\mathbf{b})|\]            \[=\,|\mathbf{b}\times \mathbf{d}-\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{d}-\mathbf{b}\times \mathbf{d}-\mathbf{c}\times \mathbf{d}+\mathbf{c}\times \mathbf{b}|\]            \[=\,|-\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{b}|\,=\,|-2\mathbf{b}\times \mathbf{c}|\,=2|\mathbf{b}\times \mathbf{c}|\]                 \[=4\](area of triangle \[ABC).\]