Question

The volume of tetrahedron with co-terminus edges $\vec{a}$, $\vec{b}$, $\vec{c}$ is $\frac{64}{3}$ cubic units, then volume of parallelopiped considering co-terminus edges given by the vectors $\vec{a} + \vec{b}$, $\vec{b} + \vec{c}$, $\vec{c} + \vec{a}$ is ______ cubic units.

• A. 384
• B. $\frac{128}{3}$
• C. 256
• D. $\frac{32}{3}$

Hint:

Standard Scalar Triple Product shortcut to memorize: $[\vec{a}+\vec{b} \quad \vec{b}+\vec{c} \quad \vec{c}+\vec{a}] = 2[\vec{a} \ \vec{b} \ \vec{c}]$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The volume of a tetrahedron with edges $\vec{a}, \vec{b}, \vec{c}$ is $V_{tet} = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$. The volume of a parallelepiped with edges $\vec{u}, \vec{v}, \vec{w}$ is $V_{par} = |[\vec{u} \vec{v} \vec{w}]|$.

Step 2: Formula Application:
$[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] = 2[\vec{a} \vec{b} \vec{c}]$

Step 3: Explanation:
Given $V_{tet} = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]| = \frac{64}{3}$. Therefore, $|[\vec{a} \vec{b} \vec{c}]| = 6 \times \frac{64}{3} = 128$. Now, the volume of the new parallelepiped is $|[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}]|$. Using the property, $V_{new} = 2 \times |[\vec{a} \vec{b} \vec{c}]| = 2 \times 128 = 256$.

Step 4: Final Answer:
The volume of the parallelepiped is 256 cubic units.