If C is the midpoint of AB and P is any point | vedclass

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(A) Given that $C$ is the midpoint of $AB$,we have $\vec{AC} + \vec{CB} = \vec{0}$,which implies $\vec{AC} = -\vec{CB}$.Using the triangle law of vect

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$\vec{PA} + \vec{PB} = 2\vec{PC}$

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(A) Given that $C$ is the midpoint of $AB$,we have $\vec{AC} + \vec{CB} = \vec{0}$,which implies $\vec{AC} = -\vec{CB}$.Using the triangle law of vector addition:$\vec{PA} = \vec{PC} + \vec{CA}$$\vec{PB} = \vec{PC} + \vec{CB}$Adding these two equations:$\vec{PA} + \vec{PB} = (\vec{PC} + \vec{CA}) + (\vec{PC} + \vec{CB})$$\vec{PA} + \vec{PB} = 2\vec{PC} + (\vec{CA} + \vec{CB})$Since $\vec{CA} = -\vec{AC}$ and $\vec{CB} = -\vec{AC}$ (because $C$ is the midpoint),we have $\vec{CA} + \vec{CB} = \vec{0}$.Therefore,$\vec{PA} + \vec{PB} = 2\vec{PC}$.

If $C$ is the midpoint of $AB$ and $P$ is any point outside $AB$,then:

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