question_answer

If\[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\]are the position vectors of a\[\Delta ABC,\] then\[\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}\]is equal to

A)  \[\overrightarrow{0}\]                

B)  \[2\text{ }\overrightarrow{AB}\]

C)  \[2\text{ }\overrightarrow{BA}\]            

D)  None of these

Correct Answer: A

Solution :

 \[\because \]\[\overrightarrow{a},\text{ }\overrightarrow{b},\text{ }\overrightarrow{c}\] are the position vectors of\[\Delta ABC\] \[\therefore \]\[\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}\] \[=(\overrightarrow{b}-\overrightarrow{a})+(\overrightarrow{c}-\overrightarrow{b})+(\overrightarrow{a}-\overrightarrow{c})\] \[=\overrightarrow{0}\]