Question
      
        If $$C$$ is the mid point of $$AB$$  and $$P$$ is any point outside $$AB,$$  then :                                                          
       A.
        $$\overrightarrow {PA}  + \overrightarrow {PB}  = 2\overrightarrow {PC} $$                 
              
       B.
        $$\overrightarrow {PA}  + \overrightarrow {PB}  = \overrightarrow {PC} $$              
       C.
        $$\overrightarrow {PA}  + \overrightarrow {PB}  + 2\overrightarrow {PC}  = \vec 0$$              
       D.
        $$\overrightarrow {PA}  + \overrightarrow {PB}  + \overrightarrow {PC}  = \vec 0$$              
            
                Answer :  
        $$\overrightarrow {PA}  + \overrightarrow {PB}  = 2\overrightarrow {PC} $$      
             Solution :
        $$\eqalign{
  & \overrightarrow {PA}  + \overrightarrow {AP}  = 0{\text{ and }}\overrightarrow {PC}  + \overrightarrow {CP}  = 0  \cr 
  &  \Rightarrow \overrightarrow {PA}  + \overrightarrow {AC}  + \overrightarrow {CP}  = 0{\text{ and }}\overrightarrow {PB}  + \overrightarrow {BC}  + \overrightarrow {CP}  = 0 \cr} $$
Adding, we get $$\overrightarrow {PA}  + \overrightarrow {PB}  + \overrightarrow {AC}  + \overrightarrow {BC}  + 2\overrightarrow {CP}  = 0$$
Since $$\overrightarrow {AC}  =  - \overrightarrow {BC} \,\,\,{\text{&  }}\,\,\overrightarrow {CP}  =  - \overrightarrow {PC} $$
$$ \Rightarrow \overrightarrow {PA}  + \overrightarrow {PB}  - 2\overrightarrow {PC}  = 0$$ 
