Question

If the volume of parallelepiped formed by the vectors \(\hat i + \lambda \hat j + \hat k,\;\hat j + \lambda \hat k,{\rm{\;and\;}}\lambda \hat i + {\rm{\hat k}}\) is minimum, then λ is equal to:
1. \(- \frac{1}{{\sqrt 3 }}\)
2. \(\frac{1}{{\sqrt 3 }}\)
3. √3
4. \(- \sqrt 3\)

Answer 1

Correct Answer - Option 2 : \(\frac{1}{{\sqrt 3 }}\)

Let’s consider the given vectors as,

\({\rm{\vec a}} = \hat i + \lambda \hat j + \hat k\) 

\({\rm{\vec b}} = \hat j + \lambda \hat k\) 

\(\overrightarrow {\rm{c}} = \lambda \hat i + {\rm{\hat k}}\) 

The volume of the parallelepiped is given by the formula:

\(V = \left| {\vec a\vec b\vec c} \right|\) 

Now,

\(\Rightarrow V = \left| {\begin{array}{*{20}{c}} 1&\lambda &1\\ 0&1&\lambda \\ \lambda &0&1 \end{array}} \right|\) 

⇒ V = 1(1 – 0) – λ(0 – λ²) + 1(0 – λ)

⇒ V = 1 + λ³ – λ     ----(1)           

Now, we need to find the minimum value of the volume of the parallelepiped.

So, we need to differentiate the volume.

\(\Rightarrow \frac{{dV}}{{d\lambda }} = 3{\lambda ^2} - 1 = 0\) 

⇒ 3λ² = 1

\(\Rightarrow {\lambda ^2} = \frac{1}{3}\) 

\(\therefore \lambda = \pm \frac{1}{{\sqrt 3 }}\) 

On substituting the value of λ in equation (1), we can get that the value of volume of the parallelepiped is minimum when the value of λ is \(+ \frac{1}{{\sqrt 3 }}\).