What are corrdinates of the point equidistant from the points `(a,0,0),(0,a,0),(0,0,a) and (0,0,0)` ?

Options

A. `((a)/(3), (b)/(3), (c )/(3))`
B. `(a,b,c)`
C. `((a)/(2), (b)/(2), (c )/(2))`
D. None of these

Answer

To find the coordinates of the point equidistant from the points \( A(a, 0, 0) \), \( B(0, a, 0) \), \( C(0, 0, a) \), and \( D(0, 0, 0) \), we will denote the coordinates of the point \( P \) as \( (x, y, z) \).

### Step 1: Calculate the distances from point \( P \) to each of the given points.

1. **Distance from \( P \) to \( A \)**:
\[ PA = \sqrt{(x - a)^2 + y^2 + z^2} \]

2. **Distance from \( P \) to \( B \)**:
\[ PB = \sqrt{x^2 + (y - a)^2 + z^2} \]

3. **Distance from \( P \) to \( C \)**:
\[ PC = \sqrt{x^2 + y^2 + (z - a)^2} \]

4. **Distance from \( P \) to \( D \)** (the origin):
\[ PD = \sqrt{x^2 + y^2 + z^2} \]

### Step 2: Set the distances equal to each other.

Since \( P \) is equidistant from all four points, we can set \( PD = PA \):
\[ \sqrt{x^2 + y^2 + z^2} = \sqrt{(x - a)^2 + y^2 + z^2} \]

### Step 3: Square both sides to eliminate the square roots.

Squaring both sides gives:
\[ x^2 + y^2 + z^2 = (x - a)^2 + y^2 + z^2 \]

### Step 4: Simplify the equation.

Cancel \( y^2 \) and \( z^2 \) from both sides:
\[ x^2 = (x - a)^2 \]
Expanding the right side:
\[ x^2 = x^2 - 2ax + a^2 \]
Subtract \( x^2 \) from both sides:
\[ 0 = -2ax + a^2 \]
Rearranging gives:
\[ 2ax = a^2 \]
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ 2x = a \quad \Rightarrow \quad x = \frac{a}{2} \]

### Step 5: Repeat the process for \( PB \) and \( PD \).

Now, set \( PD = PB \):
\[ \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2 + (y - a)^2 + z^2} \]
Squaring both sides:
\[ x^2 + y^2 + z^2 = x^2 + (y - a)^2 + z^2 \]
Cancel \( x^2 \) and \( z^2 \):
\[ y^2 = (y - a)^2 \]
Expanding:
\[ y^2 = y^2 - 2ay + a^2 \]
Subtract \( y^2 \):
\[ 0 = -2ay + a^2 \]
Rearranging gives:
\[ 2ay = a^2 \quad \Rightarrow \quad y = \frac{a}{2} \]

### Step 6: Repeat for \( PC \) and \( PD \).

Set \( PD = PC \):
\[ \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2 + y^2 + (z - a)^2} \]
Squaring both sides:
\[ x^2 + y^2 + z^2 = x^2 + y^2 + (z - a)^2 \]
Cancel \( x^2 \) and \( y^2 \):
\[ z^2 = (z - a)^2 \]
Expanding:
\[ z^2 = z^2 - 2az + a^2 \]
Subtract \( z^2 \):
\[ 0 = -2az + a^2 \]
Rearranging gives:
\[ 2az = a^2 \quad \Rightarrow \quad z = \frac{a}{2} \]

### Final Result

Thus, the coordinates of the point \( P \) that is equidistant from the four points are:
\[ \left( \frac{a}{2}, \frac{a}{2}, \frac{a}{2} \right) \]