Question

The edges of a parallelepiped are of unit length and are parallel to non-coplanar unit vectors $\hat{a},\hat{b},\hat{c}$ such that $\hat{a}\cdot \hat{b}=\hat{b}\cdot \hat{c}=\hat{c}\cdot \hat{a}=\dfrac{1}{2}$ then the volume of the parallelepiped is,
(a) $\dfrac{1}{\sqrt{2}}$
(b) $\dfrac{1}{2\sqrt{2}}$
(c) $\dfrac{\sqrt{3}}{2}$
(d) $\dfrac{1}{\sqrt{3}}$

Answer 1

Hint: As we know that the volume of parallelepiped is $\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]$ so we will use the formula \[{{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=\left| \begin{matrix}
   \hat{a}\cdot \hat{a} & \hat{a}\cdot \hat{b} & \hat{a}\cdot \hat{c} \\
   \hat{b}\cdot \hat{a} & \hat{b}\cdot \hat{b} & \hat{b}\cdot \hat{c} \\
   \hat{c}\cdot \hat{a} & \hat{c}\cdot \hat{b} & \hat{c}\cdot \hat{c} \\
\end{matrix} \right|\]. This is because it is the only way to get the required value of $\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]$. We will also use $\hat{a}\cdot \hat{a}=\hat{b}\cdot \hat{b}=\hat{c}\cdot \hat{c}=1$, $\hat{b}\cdot \hat{a}=\hat{c}\cdot \hat{b}=\hat{a}\cdot \hat{c}=\dfrac{1}{2}$ values to solve it further. Finally, we will take the help of the formula $\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   a & b & c \\
   d & e & f \\
\end{matrix} \right|=\hat{i}\left( bf-ce \right)-\hat{j}\left( af-dc \right)+\hat{k}\left( ae-bd \right)$.

Complete step-by-step solution:
We will use a clear cut formula to get the right volume of parallelepiped. The formula that we used here is given by $\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]$. We will use \[{{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=\left| \begin{matrix}
   \hat{a}\cdot \hat{a} & \hat{a}\cdot \hat{b} & \hat{a}\cdot \hat{c} \\
   \hat{b}\cdot \hat{a} & \hat{b}\cdot \hat{b} & \hat{b}\cdot \hat{c} \\
   \hat{c}\cdot \hat{a} & \hat{c}\cdot \hat{b} & \hat{c}\cdot \hat{c} \\
\end{matrix} \right|\]…(i). Since, we are given that $\hat{a}\cdot \hat{b}=\hat{b}\cdot \hat{c}=\hat{c}\cdot \hat{a}=\dfrac{1}{2}$ so, by substituting these values to (i) we get
${{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=\left| \begin{matrix}
   \hat{a}\cdot \hat{a} & \dfrac{1}{2} & \hat{a}\cdot \hat{c} \\
   \hat{b}\cdot \hat{a} & \hat{b}\cdot \hat{b} & \dfrac{1}{2} \\
   \dfrac{1}{2} & \hat{c}\cdot \hat{b} & \hat{c}\cdot \hat{c} \\
\end{matrix} \right|$
At this point we can use the fact that $\hat{a}\cdot \hat{a}=\hat{b}\cdot \hat{b}=\hat{c}\cdot \hat{c}=1$ so we can write
${{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=\left| \begin{matrix}
   1 & \dfrac{1}{2} & \hat{a}\cdot \hat{c} \\
   \hat{b}\cdot \hat{a} & 1 & \dfrac{1}{2} \\
   \dfrac{1}{2} & \hat{c}\cdot \hat{b} & 1 \\
\end{matrix} \right|$
As we are already informed that $\hat{a}\cdot \hat{b}=\hat{b}\cdot \hat{c}=\hat{c}\cdot \hat{a}=\dfrac{1}{2}$therefore, we can write $\hat{b}\cdot \hat{a}=\hat{c}\cdot \hat{b}=\hat{a}\cdot \hat{c}=\dfrac{1}{2}$. Thus, the above gets converted into the following,
${{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=\left| \begin{matrix}
   1 & \dfrac{1}{2} & \dfrac{1}{2} \\
   \dfrac{1}{2} & 1 & \dfrac{1}{2} \\
   \dfrac{1}{2} & \dfrac{1}{2} & 1 \\
\end{matrix} \right|$
To solve it further we are going to apply the formula, $\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   a & b & c \\
   d & e & f \\
\end{matrix} \right|=\hat{i}\left( bf-ce \right)-\hat{j}\left( af-dc \right)+\hat{k}\left( ae-bd \right)$ therefore, we get
\[\begin{align}
  & {{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=\left| \begin{matrix}
   1 & \dfrac{1}{2} & \dfrac{1}{2} \\
   \dfrac{1}{2} & 1 & \dfrac{1}{2} \\
   \dfrac{1}{2} & \dfrac{1}{2} & 1 \\
\end{matrix} \right| \\
 & \Rightarrow {{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=1\left( 1\cdot 1-\dfrac{1}{2}\cdot \dfrac{1}{2} \right)-\dfrac{1}{2}\left( 1\cdot \dfrac{1}{2}-\dfrac{1}{2}\cdot \dfrac{1}{2} \right)+\dfrac{1}{2}\left( \dfrac{1}{2}\cdot \dfrac{1}{2}-1\cdot \dfrac{1}{2} \right) \\
 & \Rightarrow {{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=1\left( 1-\dfrac{1}{4} \right)-\dfrac{1}{2}\left( \dfrac{1}{2}-\dfrac{1}{4} \right)+\dfrac{1}{2}\left( \dfrac{1}{4}-\dfrac{1}{2} \right) \\
 & \Rightarrow {{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=\dfrac{3}{4}-\dfrac{1}{2}\cdot \dfrac{1}{4}+\dfrac{1}{2}\left( -\dfrac{1}{4} \right) \\
 & \Rightarrow {{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=\dfrac{3}{4}-\dfrac{1}{8}-\dfrac{1}{8} \\
 & \Rightarrow {{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=\dfrac{3}{4}-\dfrac{2}{8} \\
 & \Rightarrow {{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}=\dfrac{4}{8}=\dfrac{1}{2} \\
\end{align}\]
By taking square root on both the sides we get the following,
\[\sqrt{{{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}}=\sqrt{\dfrac{1}{2}}=\dfrac{1}{\sqrt{2}}\]
Therefore, the required volume of parallelepiped is $\dfrac{1}{\sqrt{2}}$.
Hence, the correct option is (a).

Note: In this question it should be noted that we have used \[{{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}\] for representing the volume of parallelepiped. But we can also use here $\hat{a}\cdot \left( \hat{b}\cdot \hat{c} \right)$ instead of $\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]$. We have used $\hat{a}\cdot \hat{a}=\hat{b}\cdot \hat{b}=\hat{c}\cdot \hat{c}=1$ in this question due to the fact that the edges of the given parallelopiped are of 1 unit in length. One should not forget about taking the square root of ${{\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]}^{2}}$ otherwise, the solution will remain unfinished and we eventually get to the wrong answer. Solving such questions with focus can lead to the correct option.