Question

Prove that the volume of a tetrahedron with coterminous edges \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) is \( \frac{1}{6} |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| \). Hence, find the volume of tetrahedron whose coterminous edges are \( \mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k} \), \( \mathbf{b} = -\hat{i} + \hat{j} + 2\hat{k} \), and \( \mathbf{c} = 2\hat{i} + \hat{j} + 4\hat{k} \).

Hint:

Use scalar triple product for tetrahedron volume; ensure correct vector components.

Answer 1

For a tetrahedron with coterminous edges \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), volume is: \[ V = \frac{1}{6} |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|. \] Proof: The scalar triple product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \) gives the volume of the parallelepiped formed by \( \mathbf{a}, \mathbf{b}, \mathbf{c} \). A tetrahedron is \( \frac{1}{6} \) of this volume.
For given vectors: \[ \mathbf{a} = (1, 2, 3), \quad \mathbf{b} = (-1, 1, 2), \quad \mathbf{c} = (2, 1, 4). \] \[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 1 & 2
2 & 1 & 4 \end{vmatrix} = \hat{i}(4 - 2) - \hat{j}(-4 - 4) + \hat{k}(-1 - 2) = (2, 8, -3). \] \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (1, 2, 3) \cdot (2, 8, -3) = 2 + 16 - 9 = 9. \] \[ V = \frac{1}{6} |9| = \frac{9}{6} = \frac{3}{2}. \] Answer: Volume = \( \frac{3}{2} \) cubic units.