Key Concepts and Formulas

• Triangle Law of Vector Addition: For any three points $X, Y, Z$, the vector $\overrightarrow{XZ}$ can be expressed as the sum of vectors $\overrightarrow{XY}$ and $\overrightarrow{YZ}$: $\overrightarrow{XZ} = \overrightarrow{XY} + \overrightarrow{YZ}$.
• Midpoint Property in Vectors: If $C$ is the midpoint of a line segment $AB$, then the vector from $C$ to $A$ and the vector from $C$ to $B$ are equal in magnitude and opposite in direction, i.e., $\overrightarrow{CA} = -\overrightarrow{CB}$. This implies $\overrightarrow{CA} + \overrightarrow{CB} = \vec{0}$. Alternatively, for any point $P$, the position vector of the midpoint $C$ is given by $\overrightarrow{PC} = \frac{\overrightarrow{PA} + \overrightarrow{PB}}{2}$.

Step-by-Step Solution

We are given that $C$ is the midpoint of the line segment $AB$, and $P$ is any point outside $AB$. We need to find the relationship between $\overrightarrow{PA}$, $\overrightarrow{PB}$, and $\overrightarrow{PC}$.

Step 1: Express $\overrightarrow{PA}$ and $\overrightarrow{PB}$ using the Triangle Law of Vector Addition with $C$ as an intermediate point. We can use the triangle law to express the vectors originating from $P$ and terminating at $A$ and $B$, by passing through $C$.

• For the vector $\overrightarrow{PA}$, we can write: $\overrightarrow{PA} = \overrightarrow{PC} + \overrightarrow{CA}$ Explanation: This step applies the triangle law to $\triangle PCA$, where the vector from $P$ to $A$ is the sum of the vector from $P$ to $C$ and the vector from $C$ to $A$.

• Similarly, for the vector $\overrightarrow{PB}$, we can write: $\overrightarrow{PB} = \overrightarrow{PC} + \overrightarrow{CB}$ Explanation: This step applies the triangle law to $\triangle PCB$, where the vector from $P$ to $B$ is the sum of the vector from $P$ to $C$ and the vector from $C$ to $B$.

Step 2: Add the expressions for $\overrightarrow{PA}$ and $\overrightarrow{PB}$. Now, we sum the two equations obtained in Step 1: $(\overrightarrow{PA}) + (\overrightarrow{PB}) = (\overrightarrow{PC} + \overrightarrow{CA}) + (\overrightarrow{PC} + \overrightarrow{CB})$ $\overrightarrow{PA} + \overrightarrow{PB} = \overrightarrow{PC} + \overrightarrow{PC} + \overrightarrow{CA} + \overrightarrow{CB}$ $\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC} + (\overrightarrow{CA} + \overrightarrow{CB})$ Explanation: By adding the left-hand sides and right-hand sides of the vector equations, we group the common terms.

Step 3: Utilize the Midpoint Property. We are given that $C$ is the midpoint of $AB$. By the definition of a midpoint in vector terms, the vector from $C$ to $A$ is the negative of the vector from $C$ to $B$. $\overrightarrow{CA} = -\overrightarrow{CB}$ Therefore, their sum is the zero vector: $\overrightarrow{CA} + \overrightarrow{CB} = \vec{0}$ Explanation: This step directly uses the property of a midpoint, which states that the vectors from the midpoint to the endpoints are equal and opposite.

Step 4: Substitute and Simplify. Substitute the result from Step 3 into the equation from Step 2: $\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC} + \vec{0}$ $\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC}$ Explanation: By substituting the zero vector for $(\overrightarrow{CA} + \overrightarrow{CB})$, the equation simplifies to the final relationship between $\overrightarrow{PA}$, $\overrightarrow{PB}$, and $\overrightarrow{PC}$. This shows that the sum of the vectors from an external point $P$ to the endpoints $A$ and $B$ is twice the vector from $P$ to the midpoint $C$ of $AB$.

This derived relationship matches option (A).

Common Mistakes & Tips

• Vector Direction: Always be mindful of the direction of vectors. $\overrightarrow{XY}$ is not the same as $\overrightarrow{YX}$; they are opposite vectors ($\overrightarrow{XY} = -\overrightarrow{YX}$). Incorrectly handling vector directions is a common source of errors.
• Midpoint Definition: Remember that for $C$ being the midpoint of $AB$, $\overrightarrow{AC} = \overrightarrow{CB}$ and $\overrightarrow{CA} = -\overrightarrow{CB}$. The latter form is particularly useful when summing vectors that originate from the midpoint.
• Visual Representation: Sketching the points $A, B, C,$ and $P$ can greatly aid in visualizing the vector relationships and confirming the directions. For instance, drawing $P$ away from the line segment $AB$ helps to see that $\overrightarrow{PA}$ and $\overrightarrow{PB}$ are distinct vectors whose sum is related to $\overrightarrow{PC}$.

Summary

The problem involves finding a vector relationship between an external point $P$ and the endpoints $A, B$ of a line segment, given that $C$ is the midpoint of $AB$. By applying the triangle law of vector addition to express $\overrightarrow{PA}$ and $\overrightarrow{PB}$ in terms of $\overrightarrow{PC}$ and vectors along the segment $AB$, and then using the property that $\overrightarrow{CA} + \overrightarrow{CB} = \vec{0}$ because $C$ is the midpoint, we arrive at the identity $\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC}$. This identity is a fundamental result in vector algebra, often referred to as the midpoint theorem for vectors.

The final answer is $\boxed{\overrightarrow {PA} + \overrightarrow {PB} = 2\overrightarrow {PC} }$, which corresponds to option (A).