Question

The volume of a tetrahedron whose vertices are \( A = (-1, 2, 3) \), \( B = (3, -2, 1) \), \( C = (2, 1, 3) \), and \( D = (-1, -2, 4) \) is

• A. \( \frac{14}{3} \) cu. units
• B. \( \frac{16}{3} \) cu. units
• C. \( \frac{17}{3} \) cu. units
• D. \( \frac{15}{3} \) cu. units

Hint:

For calculating the volume of a tetrahedron, use the determinant method with the coordinates of the vertices.

Answer 1

The Correct Option is B

Step 1: Use the formula for the volume of a tetrahedron.
The volume \( V \) of a tetrahedron with vertices at \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), \( C(x_3, y_3, z_3) \), and \( D(x_4, y_4, z_4) \) is given by: \[ V = \frac{1}{6} \left| \text{det} \begin{pmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \\ \end{pmatrix} \right| \] Substitute the coordinates of points \( A, B, C, D \) into this determinant formula.
Step 2: Compute the determinant.
The determinant calculation gives the volume as \( \frac{16}{3} \) cubic units.
Step 3: Conclusion.
Thus, the volume of the tetrahedron is \( \frac{16}{3} \) cu. units.