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 $$\overrightarrow {AC} = - \overrightarrow {BC} \,\,\,{\text{& }}\,\,\overrightarrow {CP} = - \overrightarrow {PC} $$
$$ \Rightarrow \overrightarrow {PA} + \overrightarrow {PB} - 2\overrightarrow {PC} = 0$$