MHT CET · Maths · Vector Algebra
The value of \(a\), so that 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 3
- C -3
- D \(\sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Volume of parallelepiped \(=\left|\begin{array}{lll}1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1\end{array}\right|=a^3-a+1=V(a)\)
Now \(V^{\prime}(a)=3 a^2-1\)
\(\stackrel{+}{\stackrel{\max .}{1}-\min _1^1+\frac{1}{\sqrt{3}} \frac{1}{\sqrt{3}}}\)
\(\Rightarrow\) volume is minimum at \(x=\frac{1}{\sqrt{3}}\)
Now \(V^{\prime}(a)=3 a^2-1\)
\(\stackrel{+}{\stackrel{\max .}{1}-\min _1^1+\frac{1}{\sqrt{3}} \frac{1}{\sqrt{3}}}\)
\(\Rightarrow\) volume is minimum at \(x=\frac{1}{\sqrt{3}}\)
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