question_answer

If P and Q be the middle points of the sides BC and CD of the parallelogram ABCD, then \[\overrightarrow{AP}+\overrightarrow{AQ}=\]

A) \[\overrightarrow{AC}\]       

B) \[\frac{1}{2}\overrightarrow{AC}\]

C) \[\frac{2}{3}\overrightarrow{AC}\]   

D) \[\frac{3}{2}\overrightarrow{AC}\]

Correct Answer: D

Solution :

\[\overrightarrow{AP}\,=\overrightarrow{AB}+\overrightarrow{BP}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}\]                  ?..(i) \[\overrightarrow{AQ}=\overrightarrow{AD}+\overrightarrow{DQ}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{DC}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}\]   ?..(ii) By (i) and (ii), we get, \[\overrightarrow{AP}+\overrightarrow{AQ}=\frac{3}{2}(\overrightarrow{AB}+\overrightarrow{AD})=\frac{3}{2}(\overrightarrow{AB}+\overrightarrow{BC})=\frac{3}{2}\overrightarrow{AC}\].