MHT CET · Maths · Vector Algebra
The value of a for which the volume of parallelepiped formed by \(\hat{i}+a \hat{j}+\hat{k}, \hat{j}+a \hat{k}\) and \(a \hat{i}+\hat{k}\) becomes minimum is
- A \(\frac{-1}{\sqrt{3}}\)
- B \(\frac{1}{\sqrt{3}}\)
- C \(\sqrt{3}\)
- D \(-\sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
i.e., \(V=\left|\begin{array}{lll}1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1\end{array}\right| \doteq 1-a+a^3\)
\(\therefore \quad \frac{d V}{d a}=-1+3 a^2, \frac{d^2 V}{d a^2}=6 a\)
For max. or min. of \(V, \frac{d V}{d a}=0\)
\(\therefore \quad \mathrm{a}^2=\frac{1}{3}\)
\(\therefore \quad a=\frac{1}{\sqrt{3}}\).
\(\frac{d^2 V}{d a^2}=6 a\gt0\) for \(a=\frac{1}{\sqrt{3}}\)
\(\therefore \quad \mathrm{V}\) is minimum for \(\mathrm{a}=\frac{1}{\sqrt{3}}\)
\(\therefore \quad \frac{d V}{d a}=-1+3 a^2, \frac{d^2 V}{d a^2}=6 a\)
For max. or min. of \(V, \frac{d V}{d a}=0\)
\(\therefore \quad \mathrm{a}^2=\frac{1}{3}\)
\(\therefore \quad a=\frac{1}{\sqrt{3}}\).
\(\frac{d^2 V}{d a^2}=6 a\gt0\) for \(a=\frac{1}{\sqrt{3}}\)
\(\therefore \quad \mathrm{V}\) is minimum for \(\mathrm{a}=\frac{1}{\sqrt{3}}\)
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