What is the radius of the sphere `x^2+y^2+z^2-x-y-z=0`?

Options

A. `sqrt(3/4)`
B. `sqrt(1/2)`
C. `sqrt(3/2)`
D. `1/3`

Explanation

To find the radius of the sphere given by the equation \( x^2 + y^2 + z^2 - x - y - z = 0 \), we can follow these steps:

### Step 1: Rewrite the equation in standard form
The standard form of a sphere's equation is given by:
\[ x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 \]
We start with the given equation:
\[ x^2 + y^2 + z^2 - x - y - z = 0 \]
We can rearrange it as:
\[ x^2 - x + y^2 - y + z^2 - z = 0 \]

### Step 2: Identify coefficients
Now, we can compare the given equation with the standard form. We can rewrite it as:
\[ x^2 + y^2 + z^2 + (-1)x + (-1)y + (-1)z + 0 = 0 \]
From this, we can identify:
- \( 2u = -1 \) ⇒ \( u = -\frac{1}{2} \)
- \( 2v = -1 \) ⇒ \( v = -\frac{1}{2} \)
- \( 2w = -1 \) ⇒ \( w = -\frac{1}{2} \)
- \( d = 0 \)

### Step 3: Use the radius formula
The formula for the radius \( r \) of the sphere is:
\[ r = \sqrt{u^2 + v^2 + w^2 - d} \]
Substituting the values we found:
\[ r = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 - 0} \]

### Step 4: Calculate the radius
Calculating the squares:
\[ r = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Conclusion
Thus, the radius of the sphere is:
\[ \frac{\sqrt{3}}{2} \]

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