Find the volume of the tetrahedron whose vertices are A(3,7,4),B(5,-2,3)C(-4,5,6) and D(1,2,3).

Answer

To find the volume of the tetrahedron with vertices A(3,7,4), B(5,-2,3), C(-4,5,6), and D(1,2,3), we can use the formula for the volume of a tetrahedron given by the scalar triple product of vectors originating from one vertex to the other three vertices.

### Step-by-step Solution:

1. **Identify the vertices**:
- A(3, 7, 4)
- B(5, -2, 3)
- C(-4, 5, 6)
- D(1, 2, 3)

2. **Calculate the vectors**:
- **Vector AD**:
\[ \vec{AD} = \vec{A} - \vec{D} = (3 - 1, 7 - 2, 4 - 3) = (2, 5, 1) \]
- **Vector BD**:
\[ \vec{BD} = \vec{B} - \vec{D} = (5 - 1, -2 - 2, 3 - 3) = (4, -4, 0) \]
- **Vector CD**:
\[ \vec{CD} = \vec{C} - \vec{D} = (-4 - 1, 5 - 2, 6 - 3) = (-5, 3, 3) \]

3. **Set up the scalar triple product**:
The volume \( V \) of the tetrahedron can be calculated using the formula:
\[ V = \frac{1}{6} |\vec{AD} \cdot (\vec{BD} \times \vec{CD})| \]

4. **Calculate the cross product \( \vec{BD} \times \vec{CD} \)**:
\[ \vec{BD} \times \vec{CD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -4 & 0 \\ -5 & 3 & 3 \end{vmatrix} \]
Expanding the determinant:
\[ = \hat{i}((-4)(3) - (0)(3)) - \hat{j}((4)(3) - (0)(-5)) + \hat{k}((4)(3) - (-4)(-5)) \]
\[ = \hat{i}(-12) - \hat{j}(12) + \hat{k}(12 - 20) \]
\[ = -12\hat{i} - 12\hat{j} - 8\hat{k} \]

5. **Calculate the dot product \( \vec{AD} \cdot (\vec{BD} \times \vec{CD}) \)**:
\[ \vec{AD} \cdot (-12\hat{i} - 12\hat{j} - 8\hat{k}) = (2)(-12) + (5)(-12) + (1)(-8) \]
\[ = -24 - 60 - 8 = -92 \]

6. **Calculate the volume**:
\[ V = \frac{1}{6} | -92 | = \frac{92}{6} = \frac{46}{3} \]

### Final Answer:
The volume of the tetrahedron is \( \frac{46}{3} \) cubic units.

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