Question

If the volume of parallelepiped formed by vectors $\hat{i}+\lambda \hat{j}+\hat{k},\hat{j}+\lambda \hat{k}\text{ and }\lambda \hat{i}+\hat{k}$ is minimum, then $\lambda $ is equal to?
\[\begin{align}
  & A.\sqrt{3} \\
 & B.\dfrac{-1}{\sqrt{3}} \\
 & C.\dfrac{1}{\sqrt{3}} \\
 & D.-\sqrt{3} \\
\end{align}\]

Answer 1

Hint: In this question, we are given three vectors and these vectors form a parallelepiped. The vectors are in the form of variable $\lambda $ and it is given that volume is minimum, so we have to find the value of $\lambda $. For this, we will first find the volume of parallelepiped using the given three vectors. Volume will be found in the form of polynomials. Since any figures minimum volume is zero. Therefore, polynomial forms will be kept equal to zero and hence, we will find the value of $\lambda $. Volume of parallelepiped formed by vectors ${{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k},{{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}\text{ and }{{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ is given by
\[\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|\]

Complete step by step answer:
We are given three vectors as $\hat{i}+\lambda \hat{j}+\hat{k},\hat{j}+\lambda \hat{k}\text{ and }\lambda \hat{i}+\hat{k}$. These vectors form a parallelepiped. Volume of parallelepiped formed by vectors ${{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k},{{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}\text{ and }{{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ is given by \[\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|\].
For given vector $\hat{i}+\lambda \hat{j}+\hat{k},\hat{j}+\lambda \hat{k}\text{ and }\lambda \hat{i}+\hat{k}$ volume will be given by \[\left| \begin{matrix}
   1 & \lambda & 1 \\
   0 & 1 & \lambda \\
   \lambda & 0 & 1 \\
\end{matrix} \right|\].
Now, let us calculate value of this determinant using first column, we get:
\[\begin{align}
  & \left| 1\left( 1-0 \right)-0+\lambda \left( {{\lambda }^{2}}-1 \right) \right| \\
 & \Rightarrow \left| 1+{{\lambda }^{3}}-\lambda \right| \\
 & \Rightarrow \left| {{\lambda }^{3}}-\lambda +1 \right| \\
\end{align}\]
Hence, required volume is $\left| {{\lambda }^{3}}-\lambda +1 \right|$ since, parallelopiped's minimum volume should be zero, therefore, ${{\lambda }^{3}}-\lambda +1=0$.
Therefore, the value of $\lambda $ will satisfy the above options. We are given four options as $\sqrt{3},\dfrac{-1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},-\sqrt{3}$.
Putting them one by one in equation we get:
For $\lambda =\sqrt{3},{{\left( \sqrt{3} \right)}^{3}}-\sqrt{3}+1=3\sqrt{3}-\sqrt{3}+1\ne 0$.
Therefore, $\sqrt{3}$ is not the correct answer.
For $\lambda =\dfrac{-1}{\sqrt{3}},{{\left( \dfrac{-1}{\sqrt{3}} \right)}^{3}}-\left( \dfrac{-1}{\sqrt{3}} \right)+1=\dfrac{-1}{3\sqrt{3}}+\dfrac{1}{\sqrt{3}}+1\ne 0$.
Therefore, $\dfrac{-1}{\sqrt{3}}$ is not the correct answer.
For $\lambda =\dfrac{1}{\sqrt{3}},{{\left( \dfrac{1}{\sqrt{3}} \right)}^{3}}-\left( \dfrac{1}{\sqrt{3}} \right)+1=\dfrac{1}{3\sqrt{3}}-\dfrac{1}{\sqrt{3}}+1\ne 0$.
Therefore, $\dfrac{1}{\sqrt{3}}$ is not the correct answer.
For $\lambda =-\sqrt{3},{{\left( -\sqrt{3} \right)}^{3}}-\left( -\sqrt{3} \right)+1=-3\sqrt{3}+\sqrt{3}+1\ne 0$.
Therefore, $-\sqrt{3}$ is not the correct answer.
Hence, none of the options is correct.
So the problem setter to the question is the local minimum value of volume.

So, the correct answer is “Option B”.

Note: Students should take care of signs while calculating determinants. Students can make mistakes while taking 0 as coefficients of $\hat{i}\text{ and }\hat{j}$ in second and third vectors respectively. If options were not given, we had to find the value of $\lambda $. The formed equation is cubic (degree 3) thus, there will be three values of $\lambda $.