Chapter 9.5, Problem 36E

Rubik’s Tetrahedron Rubik’s Cube, a puzzle craze of the 1980s that remains popular to this day, inspired many similar puzzles. The one illustrated in the figure is called Rubik’s Tetrahedron; it is in the shape of a regular tetrahedron, with each edge $\sqrt{2}$ inches long. The volume of a regular tetrahedron is one-sixth the volume of the parallelepiped determined by any three edges that meet at a corner.

1. (a) Use the triple product to find the volume of Rubik’s Tetrahedron. [Hint: See Exercise 50 in Section 9.4, which gives the corners of a tetrahedron that has the same shape and size as Rubik’s Tetrahedron.]
2. (b) Construct six identical regular tetrahedra using modeling clay. Experiment to see how they can be put together to create a parallelepiped that is determined by three edges of one of the tetrahedra (thus confirming the above statement about the volume of a regular tetrahedron).

To determine

The volume of Rubik’s Tetrahedron with the use of triple product.

Answer to Problem 36E

The volume of Rubik’s Tetrahedron is $\frac{1}{2\sqrt{6}}cubicinches$.

Explanation of Solution

Given:

A Rubik’s Tetrahedron with edge $\sqrt{2}$ inches long is given.

Formula used:

The volume of regular tetrahedron is $V=\frac{1}{6}\left(\text{Volume}\text{of}\text{parallelepiped}\right)$.

Volume of Parallelepiped is $V=\left|a.\left(b\times c\right)\right|$.

Calculation:

Each edge of the tetrahedron is $\sqrt{2}$ inches long

Find the magnitude of the vector $b\times c$.

$\left|b\times c\right|=\left|b\right|\left|c\right|\sin \theta$ (1)

The angle between the two edges of a regular tetrahedron is $60°$

Substitute $60°$ for $\theta$ in (1),

$\begin{array}{c}\left|b\times c\right|=\sqrt{2}.\sqrt{2}\sin 60°\\ =2\frac{\sqrt{3}}{2}\\ =\sqrt{3}\end{array}$

The direction of $b\times c$ is perpendicular to b and c, it is aligned at $30°$ with a

Evaluate the magnitude $\left|a.\left(b\times c\right)\right|$.

$\begin{array}{c}\left|a\left(b\times c\right)\right|=\left|a\right|\left|b\times c\right|\cos \theta \\ =\sqrt{2}.\sqrt{3}\cos 30°\\ =\frac{\sqrt{6}}{2}\\ =1.224\end{array}$

The volume of regular tetrahedron is

$\begin{array}{c}V=\frac{1}{6}\left|a.\left(b\times c\right)\right|\\ =\frac{1}{6}.\frac{\sqrt{6}}{2}\\ =\frac{1}{2\sqrt{6}}\end{array}$

Thus, the volume of Rubik’s tetrahedron is $\frac{1}{2\sqrt{6}}\text{cubic}\text{inches}$.