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

Options

A. `(a/3,a/3,a/3)`
B. `(a/2,a/2,a/2)`
C. `(a,a,a)`
D. `(2a,2a,2a)`

Answer

To find the coordinates of the point that is equidistant from the points \( (a,0,0) \), \( (0,a,0) \), \( (0,0,a) \), and \( (0,0,0) \), we can follow these steps:

### Step 1: Define the Points
Let the points be defined as follows:
- \( A = (a, 0, 0) \)
- \( B = (0, a, 0) \)
- \( C = (0, 0, a) \)
- \( O = (0, 0, 0) \)

Let the point we are looking for be \( P = (x, y, z) \).

### Step 2: Calculate Distances
We need to find the distances from point \( P \) to each of the points \( A \), \( B \), \( C \), and \( O \).

1. **Distance from \( P \) to \( O \)**:
\[ d_{PO} = \sqrt{x^2 + y^2 + z^2} \]

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

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

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

### Step 3: Set Distances Equal
Since point \( P \) is equidistant from all four points, we can set the distances equal to each other. We will start by equating \( d_{PO} \) and \( d_{PA} \):
\[ \sqrt{x^2 + y^2 + z^2} = \sqrt{(x - a)^2 + y^2 + z^2} \]

### Step 4: Square Both Sides
Squaring both sides to eliminate the square roots gives:
\[ x^2 + y^2 + z^2 = (x - a)^2 + y^2 + z^2 \]

### Step 5: Simplify the Equation
This simplifies to:
\[ x^2 = (x - a)^2 \]
Expanding the right side:
\[ x^2 = x^2 - 2ax + a^2 \]
Cancelling \( x^2 \) from both sides gives:
\[ 0 = -2ax + a^2 \]
Rearranging gives:
\[ 2ax = a^2 \quad \Rightarrow \quad x = \frac{a}{2} \]

### Step 6: Repeat for \( y \) and \( z \)
Now, we can repeat the process for \( d_{PO} \) and \( d_{PB} \):
\[ \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2 + (y - a)^2 + z^2} \]
Squaring both sides gives:
\[ x^2 + y^2 + z^2 = x^2 + (y - a)^2 + z^2 \]
This simplifies to:
\[ y^2 = (y - a)^2 \]
Expanding gives:
\[ y^2 = y^2 - 2ay + a^2 \]
Cancelling \( y^2 \) gives:
\[ 0 = -2ay + a^2 \quad \Rightarrow \quad y = \frac{a}{2} \]

Similarly, for \( d_{PO} \) and \( d_{PC} \):
\[ \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2 + y^2 + (z - a)^2} \]
Squaring gives:
\[ x^2 + y^2 + z^2 = x^2 + y^2 + (z - a)^2 \]
This simplifies to:
\[ z^2 = (z - a)^2 \]
Expanding gives:
\[ z^2 = z^2 - 2az + a^2 \]
Cancelling \( z^2 \) gives:
\[ 0 = -2az + a^2 \quad \Rightarrow \quad z = \frac{a}{2} \]

### Step 7: Final Coordinates
Thus, the coordinates of the point \( P \) that is equidistant from all four points are:
\[ P = \left( \frac{a}{2}, \frac{a}{2}, \frac{a}{2} \right) \]