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Question: The edges of a parallelepiped are of unit length and are parallel to non-coplanar unit vectors \(\ha...

The edges of a parallelepiped are of unit length and are parallel to non-coplanar unit vectors a^,b^,c^\hat{a},\hat{b},\hat{c} such that a^⋅b^=b^⋅c^=c^⋅a^=12\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) 12\dfrac{1}{\sqrt{2}}
(b) 122\dfrac{1}{2\sqrt{2}}
(c) 32\dfrac{\sqrt{3}}{2}
(d) 13\dfrac{1}{\sqrt{3}}

Explanation

Solution

As we know that the volume of parallelepiped is [a^  b^  c^]\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right] so we will use the formula [a^  b^  c^]2=∣a^⋅a^a^⋅b^a^⋅c^ b^⋅a^b^⋅b^b^⋅c^ c^⋅a^c^⋅b^c^⋅c^ ∣{{\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 [a^  b^  c^]\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]. We will also use a^⋅a^=b^⋅b^=c^⋅c^=1\hat{a}\cdot \hat{a}=\hat{b}\cdot \hat{b}=\hat{c}\cdot \hat{c}=1, b^⋅a^=c^⋅b^=a^⋅c^=12\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 ∣i^j^k^ abc def ∣=i^(bf−ce)−j^(af−dc)+k^(ae−bd)\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 [a^  b^  c^]\left[ \hat{a}\,\,\hat{b}\,\,\hat{c} \right]. We will use [a^  b^  c^]2=∣a^⋅a^a^⋅b^a^⋅c^ b^⋅a^b^⋅b^b^⋅c^ c^⋅a^c^⋅b^c^⋅c^ ∣{{\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 a^⋅b^=b^⋅c^=c^⋅a^=12\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
[a^  b^  c^]2=∣a^⋅a^12a^⋅c^ b^⋅a^b^⋅b^12 12c^⋅b^c^⋅c^ ∣{{\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 a^⋅a^=b^⋅b^=c^⋅c^=1\hat{a}\cdot \hat{a}=\hat{b}\cdot \hat{b}=\hat{c}\cdot \hat{c}=1 so we can write
[a^  b^  c^]2=∣112a^⋅c^ b^⋅a^112 12c^⋅b^1 ∣{{\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 a^⋅b^=b^⋅c^=c^⋅a^=12\hat{a}\cdot \hat{b}=\hat{b}\cdot \hat{c}=\hat{c}\cdot \hat{a}=\dfrac{1}{2}therefore, we can write b^⋅a^=c^⋅b^=a^⋅c^=12\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,
[a^  b^  c^]2=∣11212 12112 12121 ∣{{\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, ∣i^j^k^ abc def ∣=i^(bf−ce)−j^(af−dc)+k^(ae−bd)\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

& {{\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{aligned}$$ By taking square root on both the sides we get the following, $$\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.