Vectors 3 Question 6
6. The value of $a$, so that the volume of parallelopiped formed by $\hat{\mathbf{i}}+a \hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{j}}+a \hat{\mathbf{k}}$ and $a \hat{\mathbf{i}}+\hat{\mathbf{k}}$ become minimum, is
(2003, 1M)
(a) -3
(b) 3
(c) $1 / \sqrt{3}$
(d) $\sqrt{3}$
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Answer:
Correct Answer: 6. (c)
Solution:
- We know that, volume of parallelopiped whose edges are $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}=[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$.
$\therefore \quad[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=\left|\begin{array}{ccc}1 & a & 1 \ 0 & 1 & a \ a & 0 & 1\end{array}\right|=1+a^{3}-a$
Let $\quad f(a)=a^{3}-a+1$
$\Rightarrow \quad f^{\prime}(a)=3 a^{2}-1$
$\Rightarrow \quad f^{\prime \prime}(a)=6 a$
For maximum or minimum, put $f^{\prime}(a)=0$
$\Rightarrow a= \pm \frac{1}{\sqrt{3}}$, which shows $f(a)$ is minimum at $a=\frac{1}{\sqrt{3}}$ and maximum at $a=-\frac{1}{\sqrt{3}}$.
