Find the locus of a point which is equidistant from the points (-1,2,3) and (3,2,1)

Answer

To find the locus of a point that is equidistant from the points A(-1, 2, 3) and B(3, 2, 1), we can follow these steps:

### Step 1: Define the point P
Let the coordinates of the point P be (x, y, z).

### Step 2: Use the distance formula
The distance from point P to point A is given by:
\[ PA = \sqrt{(x + 1)^2 + (y - 2)^2 + (z - 3)^2} \]

The distance from point P to point B is given by:
\[ PB = \sqrt{(x - 3)^2 + (y - 2)^2 + (z - 1)^2} \]

### Step 3: Set the distances equal
Since point P is equidistant from points A and B, we have:
\[ PA = PB \]

This leads to the equation:
\[ \sqrt{(x + 1)^2 + (y - 2)^2 + (z - 3)^2} = \sqrt{(x - 3)^2 + (y - 2)^2 + (z - 1)^2} \]

### Step 4: Square both sides
To eliminate the square roots, we square both sides:
\[ (x + 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z - 1)^2 \]

### Step 5: Simplify the equation
Expanding both sides:
- Left side:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
\[ (z - 3)^2 = z^2 - 6z + 9 \]
Thus, the left side becomes:
\[ x^2 + 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 + y^2 + z^2 + 2x - 4y - 6z + 14 \]

- Right side:
\[ (x - 3)^2 = x^2 - 6x + 9 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
\[ (z - 1)^2 = z^2 - 2z + 1 \]
Thus, the right side becomes:
\[ x^2 - 6x + 9 + y^2 - 4y + 4 + z^2 - 2z + 1 = x^2 + y^2 + z^2 - 6x - 4y - 2z + 14 \]

### Step 6: Set the expanded forms equal
Now we equate the two expanded forms:
\[ x^2 + y^2 + z^2 + 2x - 4y - 6z + 14 = x^2 + y^2 + z^2 - 6x - 4y - 2z + 14 \]

### Step 7: Cancel out common terms
Subtract \(x^2 + y^2 + z^2 + 14\) from both sides:
\[ 2x - 4y - 6z = -6x - 4y - 2z \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ 2x + 6x - 6z + 2z = 0 \]
\[ 8x - 4z = 0 \]

### Step 9: Simplify the equation
Dividing the entire equation by 4:
\[ 2x - z = 0 \]

### Final Answer
Thus, the locus of the point P is given by the equation:
\[ 2x - z = 0 \]

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