Question

If the volume of a tetrahedron whose conterminous edges are $\bar{a} + \bar{b}$, $\bar{b} + \bar{c}$, $\bar{c} + \bar{a}$ is 24 cubic units then the volume of parallelopiped whose conterminous edges are $\bar{a}$, $\bar{b}$, $\bar{c}$ is

• A. 48 cubic units
• B. 144 cubic units
• C. 72 cubic units
• D. 10 cubic units

Answer 1

Answer: (C)

72 cubic units

Answer 2

The volume of a tetrahedron with conterminous edges $\bar{a} + \bar{b}$, $\bar{b} + \bar{c}$, and $\bar{c} + \bar{a}$ is given as 24 cubic units.

The volume $V_t$ of a tetrahedron formed by vectors $\bar{u}$, $\bar{v}$, and $\bar{w}$ is:

$$V_t = \frac{1}{6} | \bar{u} \cdot (\bar{v} \times \bar{w}) |$$

In this case, $\bar{u} = \bar{a} + \bar{b}$, $\bar{v} = \bar{b} + \bar{c}$, and $\bar{w} = \bar{c} + \bar{a}$. Thus,

$$V_t = \frac{1}{6} | (\bar{a} + \bar{b}) \cdot ((\bar{b} + \bar{c}) \times (\bar{c} + \bar{a})) |$$

Expanding the cross product and dot product:

$$(\bar{b} + \bar{c}) \times (\bar{c} + \bar{a}) = \bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{c} + \bar{c} \times \bar{a} = \bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{a}$$ $$(\bar{a} + \bar{b}) \cdot (\bar{b} \times \bar{c} + \bar{b} \times \bar{a} + \bar{c} \times \bar{a}) = \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{b} \times \bar{a}) + \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{b} \times \bar{a}) + \bar{b} \cdot (\bar{c} \times \bar{a})$$

Since $\bar{a} \cdot (\bar{b} \times \bar{a}) = 0$, $\bar{a} \cdot (\bar{c} \times \bar{a}) = 0$, $\bar{b} \cdot (\bar{b} \times \bar{c}) = 0$, and $\bar{b} \cdot (\bar{b} \times \bar{a}) = 0$, we have:

$$(\bar{a} + \bar{b}) \cdot ((\bar{b} + \bar{c}) \times (\bar{c} + \bar{a})) = \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{b} \cdot (\bar{c} \times \bar{a}) = \bar{a} \cdot (\bar{b} \times \bar{c}) + \bar{a} \cdot (\bar{b} \times \bar{c}) = 2 [\bar{a} \cdot (\bar{b} \times \bar{c})]$$

So, $V_t = \frac{1}{6} | 2 [\bar{a} \cdot (\bar{b} \times \bar{c})] | = \frac{1}{3} | \bar{a} \cdot (\bar{b} \times \bar{c}) | = 24$.

Therefore, $| \bar{a} \cdot (\bar{b} \times \bar{c}) | = 3 \times 24 = 72$.

The volume of the parallelepiped $V_p$ formed by vectors $\bar{a}$, $\bar{b}$, and $\bar{c}$ is given by:

$$V_p = | \bar{a} \cdot (\bar{b} \times \bar{c}) |$$

Thus, the volume of the parallelepiped is 72 cubic units.