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