Question

If the edges of a parallelepiped are of unit length and are parallel to non-coplanar unit vectors \[\hat a\], \[\hat b\] and \[\hat c\] such that \[\hat a \cdot \hat b = \hat b \cdot \hat c = \hat c \cdot \hat a = \dfrac{1}{2}\]. Find the volume of the parallelepiped.

Answer 1

Hint: The edges of a parallelepiped are represented by vectors \[\hat a\], \[\hat b\]and \[\hat c\]. We will use here the formula for the volume of a parallelepiped. First, we will find all the unit vectors. Using the value of unit vectors we will find the determinant. Then we will simplify the determinant to find the volume of the parallelepiped.

Formulas used:
We will use the following formulas to solve the question:
1. \[\vec p.\vec q = \left| p \right|\left| q \right|\cos \theta = \vec q.\vec p\] where \[\theta \] is the angle between vectors \[\vec p\] and \[\vec q\] .
2. \[\left[ {\hat a\hat b\hat c} \right] = \left| {\begin{array}{*{20}{c}}{\hat a \cdot \hat a}&{\hat a \cdot \hat b}&{\hat a \cdot \hat c} \\{\hat b \cdot \hat a}&{\hat b \cdot \hat b}&{\hat b \cdot \hat c} \\{\hat c \cdot \hat a}&{\hat c \cdot \hat b}&{\hat c \cdot \hat c}\end{array}} \right|\]
3. \[\left| {\begin{array}{*{20}{l}}d&e&f \\g&h&i \\j&k&l\end{array}} \right| = d\left( {hl - ik} \right) - e\left( {gl - ij} \right) + f\left( {gk - hj} \right)\]

Complete step by step answer:
The volume of a parallelepiped with coterminous edges \[\hat a\], \[\hat b\] and \[\hat c\] is given by \[\left[ {\hat a\hat b\hat c} \right]\] (the scalar triple product or box product of vectors \[\hat a\], \[\hat b\]and \[\hat c\]).
As \[\hat a\], \[\hat b\] and \[\hat c\] are unit vectors, their magnitude is 1. Therefore,
  \[\begin{array}{l}\hat a \cdot \hat a = \hat b \cdot \hat b = \hat c \cdot \hat c = \left| 1 \right|\left| 1 \right|{\mathop{\text cos}\nolimits} 0 \\ \hat a \cdot \hat a = \hat b \cdot \hat b = \hat c \cdot \hat c = 1{\text{ }}\left( 1 \right)\end{array}\]
Let us substitute equation (1) and the values given in the question in the second formula.
\[\left[ {\hat a\hat b\hat c} \right] = \left| {\begin{array}{*{20}{l}}1&{\dfrac{1}{2}}&{\dfrac{1}{2}} \\{\dfrac{1}{2}}&1&{\dfrac{1}{2}} \\{\dfrac{1}{2}}&{\dfrac{1}{2}}&1\end{array}} \right|\]
Here it is given that the edges are of unit length.
Let us substitute \[d = h = l = 1\] and \[e = f = g = i = j = k = \dfrac{1}{2}\] in the third formula.
$ \left[ {\hat a\hat b\hat c} \right] = \dfrac{1}{4}$
\[ \left[ {\hat a\hat b\hat c} \right] = 1\left( {\dfrac{3}{4}} \right) - \dfrac{1}{2}\left( {\dfrac{1}{4}} \right) + \dfrac{1}{2} \left( { - \dfrac{1}{4}} \right) \]
 $ \left[ {\hat a \hat b \hat c} \right] = \dfrac{3}{4} - \dfrac{1}{8} - \dfrac{1}{8}$
Let us simplify the equation:
\[\begin{array}{l}\left[ {\hat a\hat b\hat c} \right] = \dfrac{{6 - 1 - 1}}{8} \\\left[ {\hat a\hat b\hat c} \right] = \dfrac{4}{8} \\ \left[ {\hat a\hat b\hat c} \right] = \dfrac{1}{4}\end{array}\]

Therefore, we found that the volume of the parallelepiped is given by \[\dfrac{1}{4}\].

Note:
1. Cyclic permutation of three vectors does not change the value of the scalar triple product; that is \[\left[ {\hat a\hat b\hat c} \right] = \left[ {\hat b\hat a\hat c} \right] = \left[ {\hat c\hat a\hat b} \right]\]. We must be careful about the fact that \[\left[ {\hat a\hat b\hat c} \right] \ne \left[ {\hat a\hat c\hat b} \right]\] rather, \[\left[ {\hat a\hat b\hat c} \right] = - \left[ {\hat a\hat c\hat b} \right]\].
2. If \[\vec a\], \[\vec b\] and \[\vec c\] are any three vectors, then the scalar product of \[\vec a\] and \[\left( {\vec b \times \vec c} \right)\], that is \[\vec a \cdot \left( {\vec b \times \vec c} \right)\]is called the scalar triple product of \[\vec a\], \[\vec b\] and \[\vec c\] in the same order and is denoted by \[\left[ {\vec a\vec b\vec c} \right]\].