Step 1: Volume of a parallelepiped is given by the scalar triple product:
$\text{Volume} = |\vec{a} \cdot (\vec{b} \times \vec{c})|$
Step 2: Suppose the vectors defining the parallelepiped are:
$\vec{a} = \langle a_1, a_2, a_3 \rangle$
$\vec{b} = \langle b_1, b_2, b_3 \rangle$
$\vec{c} = \langle c_1, c_2, c_3 \rangle$
Step 3: Compute the cross product:
$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{vmatrix}$
Step 4: Compute the scalar triple product:
$\vec{a} \cdot (\vec{b} \times \vec{c})$
Step 5: The result of this scalar product is a scalar value, which gives the volume of the parallelepiped.
Step 6: In the given problem, after performing the required calculations, we find:
$\text{Volume} = 81$
Final Answer: (C): 81 cu. units
Let \( V_1 \) be the volume of the parallelopiped with coterminous edges \( \mathbf{a} \times \mathbf{b}, \mathbf{b} \times \mathbf{c}, \) and \( \mathbf{c} \times \mathbf{a} \). We are given that \( V_1 = [ \mathbf{a} \times \mathbf{b} \ \ \mathbf{b} \times \mathbf{c} \ \ \mathbf{c} \times \mathbf{a} ] = 9 \) cu. units.
We know that \( [ \mathbf{a} \times \mathbf{b} \ \ \mathbf{b} \times \mathbf{c} \ \ \mathbf{c} \times \mathbf{a} ] = (\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}))^2 = [\mathbf{a} \ \mathbf{b} \ \mathbf{c}]^2 \). Therefore, \( [\mathbf{a} \ \mathbf{b} \ \mathbf{c}]^2 = 9 \), which implies \( [\mathbf{a} \ \mathbf{b} \ \mathbf{c}] = \pm 3 \).
Now, let \( V_2 \) be the volume of the parallelopiped with coterminous edges \( (\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c}), (\mathbf{b} \times \mathbf{c}) \times (\mathbf{c} \times \mathbf{a}), \) and \( (\mathbf{c} \times \mathbf{a}) \times (\mathbf{a} \times \mathbf{b}) \). We need to find \( V_2 = [ (\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c}) \ \ (\mathbf{b} \times \mathbf{c}) \times (\mathbf{c} \times \mathbf{a}) \ \ (\mathbf{c} \times \mathbf{a}) \times (\mathbf{a} \times \mathbf{b}) ] \).
Using the vector triple product formula \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \), we have:
\( (\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c}) = [(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}] \mathbf{b} - [(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b}] \mathbf{c} = [\mathbf{a} \ \mathbf{b} \ \mathbf{c}] \mathbf{b} - 0 \mathbf{c} = [\mathbf{a} \ \mathbf{b} \ \mathbf{c}] \mathbf{b} \)
Similarly,
\( (\mathbf{b} \times \mathbf{c}) \times (\mathbf{c} \times \mathbf{a}) = [\mathbf{b} \ \mathbf{c} \ \mathbf{a}] \mathbf{c} = [\mathbf{a} \ \mathbf{b} \ \mathbf{c}] \mathbf{c} \)
\( (\mathbf{c} \times \mathbf{a}) \times (\mathbf{a} \times \mathbf{b}) = [\mathbf{c} \ \mathbf{a} \ \mathbf{b}] \mathbf{a} = [\mathbf{a} \ \mathbf{b} \ \mathbf{c}] \mathbf{a} \)
So,
\( V_2 = [ [\mathbf{a} \ \mathbf{b} \ \mathbf{c}] \mathbf{b} \ \ [\mathbf{a} \ \mathbf{b} \ \mathbf{c}] \mathbf{c} \ \ [\mathbf{a} \ \mathbf{b} \ \mathbf{c}] \mathbf{a} ] = [\mathbf{a} \ \mathbf{b} \ \mathbf{c}]^3 [\mathbf{b} \ \mathbf{c} \ \mathbf{a}] = [\mathbf{a} \ \mathbf{b} \ \mathbf{c}]^3 [\mathbf{a} \ \mathbf{b} \ \mathbf{c}] = [\mathbf{a} \ \mathbf{b} \ \mathbf{c}]^4 \)
Since \( [\mathbf{a} \ \mathbf{b} \ \mathbf{c}]^2 = 9 \), we have \( [\mathbf{a} \ \mathbf{b} \ \mathbf{c}]^4 = 9^2 = 81 \)
Therefore, the volume of the parallelopiped with \( (\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c}), (\mathbf{b} \times \mathbf{c}) \times (\mathbf{c} \times \mathbf{a}), \) and \( (\mathbf{c} \times \mathbf{a}) \times (\mathbf{a} \times \mathbf{b}) \) as coterminous edges is 81 cubic units.
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