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 Question:
We are given the volume of a tetrahedron formed by three vectors. We must use scalar triple product properties to find the volume of a parallelepiped formed by the sum of these vectors.

Step 2: Detailed Explanation:
The volume of a tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by:
$V_{\text{tetrahedron}} = \frac{1}{6} |[\vec{a} \ \vec{b} \ \vec{c}]|$
We are given $V_{\text{tetrahedron}} = \frac{64}{3}$.
$\frac{1}{6} |[\vec{a} \ \vec{b} \ \vec{c}]| = \frac{64}{3}$
Multiply by 6:
$|[\vec{a} \ \vec{b} \ \vec{c}]| = \frac{64}{3} \times 6 = 128$
The volume of a parallelepiped with coterminous edges $\vec{A}, \vec{B}, \vec{C}$ is given by $|[\vec{A} \ \vec{B} \ \vec{C}]|$.
Here, the edges are $\vec{a}+\vec{b}$, $\vec{b}+\vec{c}$, and $\vec{c}+\vec{a}$.
The scalar triple product is:
$[\vec{a}+\vec{b} \quad \vec{b}+\vec{c} \quad \vec{c}+\vec{a}]$
By standard vector expansion properties, this triple sum expands exactly to:
$2 [\vec{a} \ \vec{b} \ \vec{c}]$
Therefore, the volume of the new parallelepiped is:
$V_{\text{parallelepiped}} = 2 \times |[\vec{a} \ \vec{b} \ \vec{c}]|$
$V_{\text{parallelepiped}} = 2 \times 128 = 256$

Step 3: Final Answer:
The volume is 256 cubic units, matching option (c).