Volume of tetrahedron formed by the planes `x+y=0, y+z=0, z+x=0,x+y+z-1=0` is

Options

A. `(1)/(6)`
B. `(1)/(3)`
C. `(2)/(3)`
D. None of these

Explanation

To find the volume of the tetrahedron formed by the planes \( x + y = 0 \), \( y + z = 0 \), \( z + x = 0 \), and \( x + y + z - 1 = 0 \), we will follow these steps:

### Step 1: Find the intersection points of the planes
We need to find the vertices of the tetrahedron by solving the equations of the planes in pairs.

1. **First Vertex (A)**: Solve the equations \( x + y = 0 \), \( y + z = 0 \), and \( z + x = 0 \).
- From \( x + y = 0 \), we have \( y = -x \).
- Substituting \( y = -x \) into \( y + z = 0 \) gives \( -x + z = 0 \) or \( z = x \).
- Substituting \( z = x \) into \( z + x = 0 \) gives \( x + x = 0 \) or \( 2x = 0 \), hence \( x = 0 \).
- Therefore, \( y = 0 \) and \( z = 0 \).
- Thus, the first vertex \( A = (0, 0, 0) \).

2. **Second Vertex (B)**: Solve \( x + y = 0 \), \( y + z = 0 \), and \( x + y + z - 1 = 0 \).
- From \( x + y = 0 \), we have \( y = -x \).
- Substituting \( y = -x \) into \( y + z = 0 \) gives \( -x + z = 0 \) or \( z = x \).
- Substituting \( z = x \) into \( x + y + z - 1 = 0 \) gives \( x - x + x - 1 = 0 \) or \( x - 1 = 0 \), hence \( x = 1 \).
- Therefore, \( y = -1 \) and \( z = 1 \).
- Thus, the second vertex \( B = (1, -1, 1) \).

3. **Third Vertex (C)**: Solve \( x + y = 0 \), \( z + x = 0 \), and \( x + y + z - 1 = 0 \).
- From \( x + y = 0 \), we have \( y = -x \).
- Substituting \( y = -x \) into \( z + x = 0 \) gives \( z = -x \).
- Substituting \( z = -x \) into \( x + y + z - 1 = 0 \) gives \( x - x - x - 1 = 0 \) or \( -x - 1 = 0 \), hence \( x = -1 \).
- Therefore, \( y = 1 \) and \( z = 1 \).
- Thus, the third vertex \( C = (-1, 1, 1) \).

4. **Fourth Vertex (D)**: Solve \( y + z = 0 \), \( z + x = 0 \), and \( x + y + z - 1 = 0 \).
- From \( y + z = 0 \), we have \( z = -y \).
- Substituting \( z = -y \) into \( z + x = 0 \) gives \( -y + x = 0 \) or \( x = y \).
- Substituting \( x = y \) into \( x + y + z - 1 = 0 \) gives \( y + y - y - 1 = 0 \) or \( y - 1 = 0 \), hence \( y = 1 \).
- Therefore, \( x = 1 \) and \( z = -1 \).
- Thus, the fourth vertex \( D = (1, 1, -1) \).

### Step 2: Calculate the volume of the tetrahedron
The volume \( V \) of a tetrahedron given vertices \( A \), \( B \), \( C \), and \( D \) can be calculated using the formula:
\[ V = \frac{1}{6} \left| \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \right| \]
Where:
- \( \vec{AB} = B - A = (1, -1, 1) - (0, 0, 0) = (1, -1, 1) \)
- \( \vec{AC} = C - A = (-1, 1, 1) - (0, 0, 0) = (-1, 1, 1) \)
- \( \vec{AD} = D - A = (1, 1, -1) - (0, 0, 0) = (1, 1, -1) \)

### Step 3: Compute the cross product \( \vec{AC} \times \vec{AD} \)
\[ \vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \hat{i}(1 \cdot -1 - 1 \cdot 1) - \hat{j}(-1 \cdot -1 - 1 \cdot 1) + \hat{k}(-1 \cdot 1 - 1 \cdot 1) \]
\[ = \hat{i}(-1 - 1) - \hat{j}(1 - 1) + \hat{k}(-1 - 1) = -2\hat{i} + 0\hat{j} - 2\hat{k} = (-2, 0, -2) \]

### Step 4: Compute the dot product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \)
\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (1, -1, 1) \cdot (-2, 0, -2) = 1 \cdot -2 + (-1) \cdot 0 + 1 \cdot -2 = -2 + 0 - 2 = -4 \]

### Step 5: Calculate the volume
\[ V = \frac{1}{6} \left| -4 \right| = \frac{4}{6} = \frac{2}{3} \]

Thus, the volume of the tetrahedron is \( \frac{2}{3} \).