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 \[ABC).\]You need to login to perform this action.
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