question_answer

If OACB is a parallelogram with \[\overrightarrow{\mathbf{OC}}=\mathbf{\vec{a}}\] and \[\overrightarrow{\mathbf{AB}}=\mathbf{\vec{b}}\], then OA is equal to:

A)  \[\frac{1}{2}\left( \vec{a}+\vec{b} \right)\]                    

B)  \[\frac{1}{2}\left( \vec{a}-\vec{b} \right)\]      

C)  \[\vec{a}-\vec{b}\]                               

D)  None of these

Correct Answer: B

Solution :

[b] Since, we know that diagonal of parallelogram bisect each other. Let diagonal \[\overrightarrow{AB}\] and \[\overrightarrow{OC}\] bisect at P. \[\therefore \overrightarrow{PA}=\overrightarrow{PB}=\frac{\overrightarrow{AB}}{2}=\frac{{\vec{b}}}{2}\] Similarly, \[\overrightarrow{OP}=\overrightarrow{PB}=\frac{\overrightarrow{OC}}{2}=\frac{{\vec{a}}}{2}\] In \[\Delta OAP,\] \[\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OP}\] \[\overrightarrow{OA}=\overrightarrow{OP}-\overrightarrow{AP}\] \[=\frac{{\vec{a}}}{2}-\frac{{\vec{b}}}{2}=\frac{1}{2}(\vec{a}-\vec{b})\] Hence, option [b] is correct.