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Question: If C is the middle point of AB and P is any point outside AB, then...

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

A

PA→+PB→=PC→\overset{\rightarrow}{PA} + \overset{\rightarrow}{PB} = \overset{\rightarrow}{PC}

B

PA→+PB→=2PC→\overset{\rightarrow}{PA} + \overset{\rightarrow}{PB} = 2\overset{\rightarrow}{PC}

C

PA→+PB→+PC→=0\overset{\rightarrow}{PA} + \overset{\rightarrow}{PB} + \overset{\rightarrow}{PC} = 0

D

PA→+PB→+2PC→=0\overset{\rightarrow}{PA} + \overset{\rightarrow}{PB} + 2\overset{\rightarrow}{PC} = 0

Answer

PA→+PB→=2PC→\overset{\rightarrow}{PA} + \overset{\rightarrow}{PB} = 2\overset{\rightarrow}{PC}

Explanation

Solution

PA→+PB→=(PA→+AC→)+(PB→+BC→)−(AC→+BC→)\overset{\rightarrow}{PA} + \overset{\rightarrow}{PB} = (\overset{\rightarrow}{PA} + \overset{\rightarrow}{AC}) + (\overset{\rightarrow}{PB} + \overset{\rightarrow}{BC}) - (\overset{\rightarrow}{AC} + \overset{\rightarrow}{BC})

= PC→+PC→−(AC→−CB→)=2PC→−0,\overset{\rightarrow}{PC} + \overset{\rightarrow}{PC} - (\overset{\rightarrow}{AC} - \overset{\rightarrow}{CB}) = 2\overset{\rightarrow}{PC} - 0, (∵AC→=CB→)(\because\overset{\rightarrow}{AC} = \overset{\rightarrow}{CB})

∴\therefore PA→+PB→=2PC→\overset{\rightarrow}{PA} + \overset{\rightarrow}{PB} = 2\overset{\rightarrow}{PC}.