Question

Let $a$ be a unit vector, $b=2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}$ and $c=\overrightarrow{i}+3\overrightarrow{k}$. Then, maximum value of $\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]$ is

• A. $-1$
• B. $\sqrt{10}+\sqrt{6}$
• C. $\sqrt{10}-\sqrt{6}$
• D. $\sqrt{59}$

Answer 1

The correct option is D $\sqrt{59}$

Given, $b=2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}$ and $c=\overrightarrow{i}+3\overrightarrow{k}$
Now, $b\times c=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 2& 1& -1\\ 1& 0& 3\end{array}\right|$
$=\overrightarrow{i}(3-0)-\overrightarrow{j}(6+1)+\overrightarrow{k}(0-1)$
$=3\overrightarrow{i}-7\overrightarrow{j}-\overrightarrow{k}$
Now, $abc=a\cdot b\times c$
$=a(b\times c)\cos \theta$
$=1\sqrt{3^2+7^2+1^2}\cos \theta$
$=\sqrt{59}\cos \theta$
$\Rightarrow {abc}_{max}=\sqrt{59}\cdot 1$
($\because$ maximum value of $\cos \theta$ is $1$)
Hence, maximum value is $\sqrt{59}$.