A) \[\frac{1}{\sqrt{3}}\]
B) \[3\]
C) \[-3\]
D) \[1\]
Correct Answer: A
Solution :
Let V be the volume of the parallelepiped formed by the given vectors. Let \[\alpha =i+aj+k,\,\,\beta =j+ak\] and \[\gamma =ai+k\] \[\therefore \] \[V=[\alpha \beta \gamma ]=\left| \begin{matrix} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \\ \end{matrix} \right|\] \[=1(1-0)-a\,(0-{{a}^{2}})+1(0-a)\] \[\therefore \] \[V=1+{{a}^{3}}-a\] On differentiating w. r. t. a, we get \[\frac{dV}{da}=3{{a}^{2}}-1\] Put \[\frac{dV}{da}=0,\] \[\therefore \] \[3{{a}^{2}}-1=0\,\,\,\,\Rightarrow \,\,\,a=\pm \frac{1}{\sqrt{3}}\] Now, \[\frac{{{d}^{2}}V}{d{{a}^{2}}}=6a\] At \[a=\frac{1}{\sqrt{3}},\] \[\frac{{{d}^{2}}V}{d{{a}^{2}}}=\frac{6}{\sqrt{3}}>0,\] (mnima) Hence, volume of parallelepiped is minimum when \[a=\frac{1}{\sqrt{3}}\].
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