Question

If \(C\) is the midpoint of the line segment \(AB\) and \(P\) is any point outside the line \(AB\), then:

• A. \(\overrightarrow{PA}+\overrightarrow{PB}+2\overrightarrow{PC}=\vec0\)
• B. \(\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}=\vec0\)
• C. \(\overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC}\)
• D. \(\overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{PC}\)

Hint:

If \(M\) is the midpoint of \(AB\), then \[ \overrightarrow{PA}+\overrightarrow{PB} = 2\overrightarrow{PM} \] for any point \(P\). This is a standard midpoint vector identity.

Answer 1

The Correct Option is C

Step 1: Use the midpoint formula in vector form.
Let the position vectors of \(A,B,C\) and \(P\) be \[ \vec a,\quad \vec b,\quad \vec c,\quad \vec p \] Since \(C\) is the midpoint of \(AB\), \[ \vec c=\frac{\vec a+\vec b}{2} \]

Step 2: Express the vectors from point \(P\).
\[ \overrightarrow{PA}=\vec a-\vec p \] \[ \overrightarrow{PB}=\vec b-\vec p \] \[ \overrightarrow{PC}=\vec c-\vec p \]

Step 3: Add \(\overrightarrow{PA}\) and \(\overrightarrow{PB}\).
\[ \overrightarrow{PA} + \overrightarrow{PB} = (\vec a-\vec p) + (\vec b-\vec p) \] \[ = \vec a+\vec b-2\vec p \] Using \[ \vec a+\vec b=2\vec c \] we obtain \[ \overrightarrow{PA} + \overrightarrow{PB} = 2\vec c-2\vec p \] \[ = 2(\vec c-\vec p) \] \[ = 2\overrightarrow{PC} \]

Step 4: Final conclusion.
Hence, \[ \boxed{\overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC}} \]