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Question

If C is the mid point of AB and P is any point outside AB, then

A
¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB+¯¯¯¯¯¯¯¯PC=0
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B
¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB+2¯¯¯¯¯¯¯¯PC=¯0
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C
¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB=¯¯¯¯¯¯¯¯PC
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D
¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB=2¯¯¯¯¯¯¯¯PC
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Solution

The correct option is D ¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB=2¯¯¯¯¯¯¯¯PC
Let p,a,b,c be the position vectors of points P,A,B,C respectively.
L.H.S=¯¯¯¯¯¯¯¯PA+¯¯¯¯¯¯¯¯PB=pa+pb=2(pa+b2)=2PC

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