MHT CET · Maths · Vector Algebra
The value of \(\alpha\), so that the volume of parallelopiped formed by \(\hat{i}+\alpha \hat{j}+\hat{k}, \hat{j}+\alpha \hat{k}\) and \(\alpha \hat{\mathrm{i}}+\hat{\mathrm{k}}\) becomes minimum, is
- A \(-3\)
- B \(3\)
- C \(\frac{1}{\sqrt{3}}\)
- D \(-\frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Volume of parallelopiped \(=\left|\begin{array}{ccc}1 & \alpha & 1 \\ 0 & 1 & \alpha \\ \alpha & 0 & 1\end{array}\right|\)
\(\therefore \quad \mathrm{V}=1+\alpha^3-\alpha\)
For maxima or minima,
\(\begin{aligned}
& \frac{\mathrm{dV}}{\mathrm{d} \alpha}=0 \\
& \Rightarrow 3 \alpha^2-1=0 \\
& \Rightarrow \alpha= \pm \frac{1}{\sqrt{3}}
\end{aligned}\)
Now, \(\frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} \alpha^2}=6 \alpha\)
For \(\alpha=\frac{1}{\sqrt{3}}\),
\(\frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} x^2}>0\)
\(\therefore \quad \mathrm{V}\) is minimum at \(\alpha=\frac{1}{\sqrt{3}}\).
\(\therefore \quad \mathrm{V}=1+\alpha^3-\alpha\)
For maxima or minima,
\(\begin{aligned}
& \frac{\mathrm{dV}}{\mathrm{d} \alpha}=0 \\
& \Rightarrow 3 \alpha^2-1=0 \\
& \Rightarrow \alpha= \pm \frac{1}{\sqrt{3}}
\end{aligned}\)
Now, \(\frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} \alpha^2}=6 \alpha\)
For \(\alpha=\frac{1}{\sqrt{3}}\),
\(\frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} x^2}>0\)
\(\therefore \quad \mathrm{V}\) is minimum at \(\alpha=\frac{1}{\sqrt{3}}\).
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