The locus of a point equidistant from two points with position vectors `veca` and `vecb` is

Options

A. `[r-(1)/(2)(a+b)]*(a+b)=0`
B. `[r-(1)/(2)(a+b)]*(a-b)=0`
C. `[r-(1)/(2)(a+b)]*a=0`
D. `[r-(a+b)]*b=0`

Answer

To find the locus of a point \( P \) that is equidistant from two points with position vectors \( \vec{A} \) and \( \vec{B} \), we can follow these steps:

### Step 1: Set Up the Equation
Since point \( P \) is equidistant from points \( A \) and \( B \), we can express this condition mathematically as:
\[ PA = PB \]
Where \( PA \) is the distance from \( P \) to \( A \) and \( PB \) is the distance from \( P \) to \( B \).

### Step 2: Express Distances in Vector Form
The distance \( PA \) can be expressed as:
\[ PA = |\vec{R} - \vec{A}| \]
And the distance \( PB \) can be expressed as:
\[ PB = |\vec{R} - \vec{B}| \]
Where \( \vec{R} \) is the position vector of point \( P \).

### Step 3: Square Both Sides
Since both distances are equal, we can square both sides to eliminate the square root:
\[ |\vec{R} - \vec{A}|^2 = |\vec{R} - \vec{B}|^2 \]

### Step 4: Expand Both Sides
Expanding both sides gives us:
\[ (\vec{R} - \vec{A}) \cdot (\vec{R} - \vec{A}) = (\vec{R} - \vec{B}) \cdot (\vec{R} - \vec{B}) \]
This simplifies to:
\[ \vec{R} \cdot \vec{R} - 2\vec{R} \cdot \vec{A} + \vec{A} \cdot \vec{A} = \vec{R} \cdot \vec{R} - 2\vec{R} \cdot \vec{B} + \vec{B} \cdot \vec{B} \]

### Step 5: Cancel Common Terms
Notice that \( \vec{R} \cdot \vec{R} \) appears on both sides, so we can cancel it:
\[ -2\vec{R} \cdot \vec{A} + \vec{A} \cdot \vec{A} = -2\vec{R} \cdot \vec{B} + \vec{B} \cdot \vec{B} \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ 2\vec{R} \cdot \vec{B} - 2\vec{R} \cdot \vec{A} = \vec{B} \cdot \vec{B} - \vec{A} \cdot \vec{A} \]
Factoring out \( 2\vec{R} \):
\[ 2\vec{R} \cdot (\vec{B} - \vec{A}) = \vec{B} \cdot \vec{B} - \vec{A} \cdot \vec{A} \]

### Step 7: Solve for \( \vec{R} \)
Now, we can solve for \( \vec{R} \):
\[ \vec{R} = \frac{1}{2} (\vec{B} + \vec{A}) + k(\vec{B} - \vec{A}) \]
Where \( k \) is a scalar parameter.

### Step 8: Identify the Locus
The equation represents a line that is the perpendicular bisector of the segment joining points \( A \) and \( B \). Therefore, the locus of point \( P \) is the line that is equidistant from points \( A \) and \( B \).

### Conclusion
The locus of a point equidistant from two points with position vectors \( \vec{A} \) and \( \vec{B} \) is the perpendicular bisector of the line segment joining \( A \) and \( B \).

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