The volume of the tetrahedron whose vertices are `A(-1, 2, 3), B(3, -2, 1), C(2, 1, 3) and C(-1, -2, 4)`

Options

A. `(2)/(3)` cu. Units
B. `(32)/(3)`cu. Units
C. `(8)/(3)`cu. Units
D. `(16)/((3)`cu. Units

Answer

To find the volume of the tetrahedron with vertices \( A(-1, 2, 3) \), \( B(3, -2, 1) \), \( C(2, 1, 3) \), and \( D(-1, -2, 4) \), we can use the formula for the volume of a tetrahedron given by:

\[ V = \frac{1}{6} | \vec{AB} \cdot (\vec{AC} \times \vec{AD}) | \]

where \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \) are vectors from vertex \( A \) to vertices \( B \), \( C \), and \( D \) respectively.

### Step 1: Find the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \)

1. **Calculate \( \vec{AB} \)**:
\[ \vec{AB} = B - A = (3 - (-1), -2 - 2, 1 - 3) = (3 + 1, -2 - 2, 1 - 3) = (4, -4, -2) \]

2. **Calculate \( \vec{AC} \)**:
\[ \vec{AC} = C - A = (2 - (-1), 1 - 2, 3 - 3) = (2 + 1, 1 - 2, 3 - 3) = (3, -1, 0) \]

3. **Calculate \( \vec{AD} \)**:
\[ \vec{AD} = D - A = (-1 - (-1), -2 - 2, 4 - 3) = (0, -2 - 2, 4 - 3) = (0, -4, 1) \]

### Step 2: Calculate the cross product \( \vec{AC} \times \vec{AD} \)

The cross product of two vectors \( \vec{u} = (u_1, u_2, u_3) \) and \( \vec{v} = (v_1, v_2, v_3) \) is given by the determinant:

\[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} \]

For \( \vec{AC} = (3, -1, 0) \) and \( \vec{AD} = (0, -4, 1) \):

\[ \vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 0 \\ 0 & -4 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} -1 & 0 \\ -4 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 0 \\ 0 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1 \\ 0 & -4 \end{vmatrix} \]

Calculating each of these determinants:

1. \( \begin{vmatrix} -1 & 0 \\ -4 & 1 \end{vmatrix} = (-1)(1) - (0)(-4) = -1 \)

2. \( \begin{vmatrix} 3 & 0 \\ 0 & 1 \end{vmatrix} = (3)(1) - (0)(0) = 3 \)

3. \( \begin{vmatrix} 3 & -1 \\ 0 & -4 \end{vmatrix} = (3)(-4) - (-1)(0) = -12 \)

So we have:

\[ \vec{AC} \times \vec{AD} = (-1)\hat{i} - 3\hat{j} - 12\hat{k} = (-1, -3, -12) \]

### Step 3: Calculate the dot product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \)

Now, we calculate the dot product:

\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (4, -4, -2) \cdot (-1, -3, -12) \]

Calculating this:

\[ = 4 \cdot (-1) + (-4) \cdot (-3) + (-2) \cdot (-12) \]
\[ = -4 + 12 + 24 = 32 \]

### Step 4: Calculate the volume of the tetrahedron

Finally, substituting into the volume formula:

\[ V = \frac{1}{6} |32| = \frac{32}{6} = \frac{16}{3} \]

Thus, the volume of the tetrahedron is:

\[ \boxed{\frac{16}{3}} \text{ cubic units} \]