Vector Algebra Question 47
Question: If P and Q be the middle points of the sides BC and CD of the parallelogram ABCD, then $ \overrightarrow{AP}+\overrightarrow{AQ}= $
Options:
A) $ \overrightarrow{AC} $
B) $ \frac{1}{2}\overrightarrow{AC} $
C) $ \frac{2}{3}\overrightarrow{AC} $
D) $ \frac{3}{2}\overrightarrow{AC} $
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Answer:
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} $ .
BETA
