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Question: If P and Q be the middle points of the sides BC and CD of the parallelogram ABCD, then \(\overright...

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

A

AC→\overrightarrow { A C }

B

12AC→\frac { 1 } { 2 } \overrightarrow { A C }

C

23AC→\frac { 2 } { 3 } \overrightarrow { A C }

D

32AC→\frac { 3 } { 2 } \overrightarrow { A C }

Answer

32AC→\frac { 3 } { 2 } \overrightarrow { A C }

Explanation

Solution

AP→=AB→+BP→=AB→+12BC→=AB→+12AD→\overrightarrow { A P } = \overrightarrow { A B } + \overrightarrow { B P } = \overrightarrow { A B } + \frac { 1 } { 2 } \overrightarrow { B C } = \overrightarrow { A B } + \frac { 1 } { 2 } \overrightarrow { A D } …..(i)

AQ→=AD→+DQ→=AD→+12DC→=AD→+12AB→\overrightarrow { A Q } = \overrightarrow { A D } + \overrightarrow { D Q } = \overrightarrow { A D } + \frac { 1 } { 2 } \overrightarrow { D C } = \overrightarrow { A D } + \frac { 1 } { 2 } \overrightarrow { A B } …..(ii)

By (i) and (ii), we get,

AP→+AQ→=32(AB→+AD→)=32(AB→+BC→)=32AC→\overrightarrow { A P } + \overrightarrow { A Q } = \frac { 3 } { 2 } ( \overrightarrow { A B } + \overrightarrow { A D } ) = \frac { 3 } { 2 } ( \overrightarrow { A B } + \overrightarrow { B C } ) = \frac { 3 } { 2 } \overrightarrow { A C }.