Question

If \( A(-1,2,3) \), \( B(3,-2,1) \), \( C(2,1,3) \) and \( D(-1,-2,4) \) are the vertices of a tetrahedron, then its volume is

• A. \( \dfrac{16}{3} \) cu. units
• B. \( \dfrac{13}{6} \) cu. units
• C. \( \dfrac{16}{31} \) cu. units
• D. \( \dfrac{31}{6} \) cu. units

Hint:

The volume of a tetrahedron can always be found using the scalar triple product of three edges meeting at a vertex.

Answer 1

The Correct Option is A

Step 1: Use the volume formula of a tetrahedron.
The volume of a tetrahedron with vertices \( A, B, C, D \) is given by \[ V = \frac{1}{6} \left| \det(\vec{AB}, \vec{AC}, \vec{AD}) \right| \] Step 2: Find the vectors.
\[ \vec{AB} = (4,-4,-2), \quad \vec{AC} = (3,-1,0), \quad \vec{AD} = (0,-4,1) \] Step 3: Form the determinant.
\[ \det \begin{vmatrix} 4 & -4 & -2 \\ 3 & -1 & 0 \\ 0 & -4 & 1 \end{vmatrix} \] Step 4: Evaluate the determinant.
\[ = 4(-1\cdot1 - 0\cdot(-4)) + 4(3\cdot1 - 0\cdot0) - 2(3\cdot(-4) - (-1)\cdot0) \] \[ = 4(-1) + 12 + 24 = 32 \] Step 5: Find the volume.
\[ V = \frac{1}{6} \times 32 = \frac{16}{3} \] Step 6: Conclusion.
The volume of the tetrahedron is \( \dfrac{16}{3} \) cubic units.