Volume of tetrahedron with vertices at `(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1)` is

Options

A. `(1)/(6)`cu. Units
B. `(1)/(4)` cu. Units
C. `(1)/(3)`cu.units
D. `(1)/(5)`cu. Units

Answer

To find the volume of the tetrahedron with vertices at \( (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1) \), we can follow these steps:

### Step 1: Identify the vertices
Let the vertices be:
- \( A = (0, 0, 0) \)
- \( B = (1, 0, 0) \)
- \( C = (0, 1, 0) \)
- \( D = (0, 0, 1) \)

### Step 2: Find the vectors
We need to find the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \).

- \( \vec{AB} = B - A = (1, 0, 0) - (0, 0, 0) = (1, 0, 0) \)
- \( \vec{AC} = C - A = (0, 1, 0) - (0, 0, 0) = (0, 1, 0) \)
- \( \vec{AD} = D - A = (0, 0, 1) - (0, 0, 0) = (0, 0, 1) \)

### Step 3: Set up the volume formula
The volume \( V \) of a tetrahedron formed by vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \) can be calculated using the formula:
\[ V = \frac{1}{6} | \vec{AB} \cdot (\vec{AC} \times \vec{AD}) | \]
Alternatively, we can use the determinant method:
\[ V = \frac{1}{6} \left| \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \right| \]

### Step 4: Calculate the determinant
Calculating the determinant:
\[ \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1 \cdot (1 \cdot 1 - 0 \cdot 0) - 0 \cdot (0 \cdot 1 - 0 \cdot 0) + 0 \cdot (0 \cdot 0 - 1 \cdot 0) = 1 \]

### Step 5: Calculate the volume
Substituting the determinant back into the volume formula:
\[ V = \frac{1}{6} \times 1 = \frac{1}{6} \]

### Conclusion
The volume of the tetrahedron is \( \frac{1}{6} \) cubic units.