If A,B are the points ( -2,2,3),(13,-3,13) respectively , find the locus of P such that ` 3|PA| = 2|PB| `

Answer

To find the locus of point P such that \( 3|PA| = 2|PB| \), where \( A(-2, 2, 3) \) and \( B(13, -3, 13) \), we will follow these steps:

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

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

The distance \( PB \) from point P to point B is given by:
\[ PB = \sqrt{(x - 13)^2 + (y - (-3))^2 + (z - 13)^2} = \sqrt{(x - 13)^2 + (y + 3)^2 + (z - 13)^2} \]

### Step 3: Set up the equation based on the given condition
According to the problem, we have:
\[ 3|PA| = 2|PB| \]
Squaring both sides to eliminate the square roots gives:
\[ 9|PA|^2 = 4|PB|^2 \]

### Step 4: Substitute the expressions for PA and PB
Substituting the expressions we derived for \( PA \) and \( PB \):
\[ 9 \left( (x + 2)^2 + (y - 2)^2 + (z - 3)^2 \right) = 4 \left( (x - 13)^2 + (y + 3)^2 + (z - 13)^2 \right) \]

### Step 5: Expand both sides
Expanding both sides:
- Left side:
\[ 9 \left( (x + 2)^2 + (y - 2)^2 + (z - 3)^2 \right) = 9 \left( x^2 + 4x + 4 + y^2 - 4y + 4 + z^2 - 6z + 9 \right) = 9x^2 + 36x + 81 + 9y^2 - 36y + 9z^2 - 54z \]

- Right side:
\[ 4 \left( (x - 13)^2 + (y + 3)^2 + (z - 13)^2 \right) = 4 \left( x^2 - 26x + 169 + y^2 + 6y + 9 + z^2 - 26z + 169 \right) = 4x^2 - 104x + 676 + 4y^2 + 24y + 36 + 4z^2 - 104z + 676 \]

### Step 6: Combine and simplify
Combining both sides gives:
\[ 9x^2 + 9y^2 + 9z^2 + 36x - 36y - 54z + 81 = 4x^2 + 4y^2 + 4z^2 - 104x + 24y - 104z + 1392 \]

Rearranging this equation:
\[ (9x^2 - 4x^2) + (9y^2 - 4y^2) + (9z^2 - 4z^2) + (36x + 104x) + (-36y - 24y) + (-54z + 104z) + (81 - 1392) = 0 \]

This simplifies to:
\[ 5x^2 + 5y^2 + 5z^2 + 140x - 60y + 50z - 1311 = 0 \]

### Step 7: Finalize the locus equation
Dividing the entire equation by 5 gives:
\[ x^2 + y^2 + z^2 + 28x - 12y + 10z - 262.2 = 0 \]

This is the locus of point P.