Question

If the volume of parallelepiped formed by the vectors $\widehat i + \lambda \widehat j + \widehat k$, $\widehat j + \lambda \widehat k$ and $\lambda \widehat i + \widehat k$ is minimum, then $\lambda$ is equal to?
A. $\sqrt 3$
B. $- \dfrac{1}{{\sqrt 3 }}$
C. $\dfrac{1}{{\sqrt 3 }}$
D. $- \sqrt 3$

Answer 1

Hint : In the question, we want to calculate the value of $\lambda$ for which the volume of the parallelepiped is minimum. The volume is calculated by the formula $volume = \left| {\overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right)} \right|$ and the condition for the minimum is by equating the value of $\lambda$ into the volume which will give us the positive value will be the point of minima.

Complete step by step solution:
In the given question, we are asked to calculate the volume of the parallelepiped. A parallelepiped is a solid body having each side to be a parallelogram. And in the question, we are given three vectors. So, the scalar triple product of these vectors gave us the volume of the parallelepiped.
$volume = \left| {\overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c } \right)} \right|$
Let
$\overrightarrow a = \widehat i + \lambda \widehat j + \widehat k$
$\overrightarrow b = \widehat j + \lambda \widehat k$
$\overrightarrow c = \lambda \widehat i + \widehat k$
The volume of the parallelepiped can be calculated by the form

1&\lambda &1 \\\ 0&1&\lambda \\\ \lambda &0&1 \end{array}} \right|$$ Taking the determinant, we get $ \Rightarrow \,\,1\left( {1 - 0} \right) - \lambda \left( {0 - {\lambda ^2}} \right) + 1\left( {0 - \lambda } \right) $ On simplification, we get $ \Rightarrow \,\,V = \left| {1 + {\lambda ^3} - \lambda } \right| $ Now, we are given the question that the volume should be minimum. So, applying the condition for minima which states that $ \dfrac{{dV}}{{d\lambda }} = 0 $ So, differentiating the V $ \Rightarrow \,\,\dfrac{{dV}}{{d\lambda }} = \dfrac{d}{{d\lambda }}\left( {1 + {\lambda ^3} - \lambda } \right) $ $ \Rightarrow \,\,\dfrac{{dV}}{{d\lambda }} = 3{\lambda ^2} - 1 $ Again, differentiate with respect $$\lambda $$, we have $ \Rightarrow \,\,\dfrac{{{d^2}V}}{{d{\lambda ^2}}} = 6\lambda $ Now, Equating the first derivative to zero, $ \Rightarrow \,\,3{\lambda ^2} - 1 = 0 $ Add 1 on both side, then $ \Rightarrow \,\,3{\lambda ^2} = 1 $ Divide both side by 3 $ \Rightarrow \,\,{\lambda ^2} = \dfrac{1}{3} $ Taking square root on both the side $ \Rightarrow \,\,\lambda = \pm \dfrac{1}{{\sqrt 3 }} $ To find which value of $$\lambda $$ is minimum, putting the two values into the second derivative For $ \lambda = \dfrac{1}{{\sqrt 3 }} $, then $ \Rightarrow \,\,\dfrac{{{d^2}V}}{{d{\lambda ^2}}} = 6\left( {\dfrac{1}{{\sqrt 3 }}} \right) $ which is positive And for $ \lambda = - \dfrac{1}{{\sqrt 3 }} $ $ \Rightarrow \,\,\dfrac{{{d^2}V}}{{d{\lambda ^2}}} = 6\left( { - \dfrac{1}{{\sqrt 3 }}} \right) $ which is negative So, $ \lambda = \dfrac{1}{{\sqrt 3 }} $ gives the minimum volume to the parallelepiped Hence, the correct option is C. **So, the correct answer is “Option C”.** **Note** : While taking the cross product, be careful about the signs and mostly students make minor mistakes while doing cross product which lead to wrong questions. The condition of minima should be learned carefully, which states that if the second derivative gives positive value then only the point is minimum.