2. Find the volume of the tetrahedron whose vertices are the points \( A(2,-1,-3), B(4,1,3) \) \( C(3,2,-1) \) and \( D(1,4,2) \). Ans. \( 7 \frac{1}{3} \)

To find the volume of the tetrahedron with vertices \( A(2,-1,-3) \), \( B(4,1,3) \), \( C(3,2,-1) \), and \( D(1,4,2) \), we can use the scalar triple product formula. This formula calculates the volume \( V \) of a tetrahedron formed by three vectors emanating from one vertex. **Step 1: Determine the Vectors** First, choose one vertex as the reference point. Let's choose vertex \( A \). Then, compute the vectors from \( A \) to the other three vertices: \[ \begin{align*} \vec{AB} &= B - A = (4-2, \, 1-(-1), \, 3-(-3)) = (2, \, 2, \, 6) \\ \vec{AC} &= C - A = (3-2, \, 2-(-1), \, -1-(-3)) = (1, \, 3, \, 2) \\ \vec{AD} &= D - A = (1-2, \, 4-(-1), \, 2-(-3)) = (-1, \, 5, \, 5) \\ \end{align*} \] **Step 2: Compute the Cross Product \( \vec{AB} \times \vec{AC} \)** The cross product of \( \vec{AB} \) and \( \vec{AC} \) is calculated as follows: \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 6 \\ 1 & 3 & 2 \\ \end{vmatrix} \] \[ = \mathbf{i}(2 \cdot 2 - 6 \cdot 3) - \mathbf{j}(2 \cdot 2 - 6 \cdot 1) + \mathbf{k}(2 \cdot 3 - 2 \cdot 1) \] \[ = \mathbf{i}(4 - 18) - \mathbf{j}(4 - 6) + \mathbf{k}(6 - 2) \] \[ = -14\mathbf{i} + 2\mathbf{j} + 4\mathbf{k} \] So, \[ \vec{AB} \times \vec{AC} = (-14, \, 2, \, 4) \] **Step 3: Compute the Scalar Triple Product** The scalar triple product \( (\vec{AB} \times \vec{AC}) \cdot \vec{AD} \) gives six times the volume of the tetrahedron: \[ (\vec{AB} \times \vec{AC}) \cdot \vec{AD} = (-14, \, 2, \, 4) \cdot (-1, \, 5, \, 5) \] \[ = (-14)(-1) + (2)(5) + (4)(5) = 14 + 10 + 20 = 44 \] **Step 4: Calculate the Volume** Finally, divide the absolute value of the scalar triple product by 6 to get the volume: \[ V = \frac{|44|}{6} = \frac{44}{6} = \frac{22}{3} = 7 \frac{1}{3} \] **Answer:** The volume of the tetrahedron is \( 7 \frac{1}{3} \) cubic units.