Question

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

Hint:

The volume of a tetrahedron can be found using the scalar triple product: \( V = \frac{1}{6} \left| \overrightarrow{a} \cdot (\overrightarrow{b} \times \overrightarrow{c}) \right| \).

Answer 1

Step 1: The formula for the volume of a tetrahedron with coterminous edges \( \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \) is given by: \[ V = \frac{1}{6} \left| \overrightarrow{a} \cdot (\overrightarrow{b} \times \overrightarrow{c}) \right| \] Step 2: First, we compute the cross product \( \overrightarrow{b} \times \overrightarrow{c} \): \[ \overrightarrow{b} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 1 & 2
2 & 1 & 4 \end{vmatrix} \] Expanding this determinant: \[ \overrightarrow{b} \times \overrightarrow{c} = \hat{i} \left( 1 \cdot 4 - 2 \cdot 1 \right) - \hat{j} \left( -1 \cdot 4 - 2 \cdot 2 \right) + \hat{k} \left( -1 \cdot 1 - 1 \cdot 2 \right) \] \[ = \hat{i} (4 - 2) - \hat{j} (-4 - 4) + \hat{k} (-1 - 2) \] \[ = \hat{i} (2) - \hat{j} (-8) + \hat{k} (-3) \] \[ = 2\hat{i} + 8\hat{j} - 3\hat{k} \] Step 3: Next, compute the dot product \( \overrightarrow{a} \cdot (\overrightarrow{b} \times \overrightarrow{c}) \): \[ \overrightarrow{a} \cdot (\overrightarrow{b} \times \overrightarrow{c}) = \left( \hat{i} + 2\hat{j} + 3\hat{k} \right) \cdot \left( 2\hat{i} + 8\hat{j} - 3\hat{k} \right) \] \[ = (1)(2) + (2)(8) + (3)(-3) = 2 + 16 - 9 = 9 \] Step 4: Now calculate the volume: \[ V = \frac{1}{6} \times |9| = \frac{9}{6} = \frac{3}{2} \] Thus, the volume of the tetrahedron is: \[ \boxed{\frac{3}{2}} \]