Question

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

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

Answer 1

Answer: (D)

Given, $b=2 \vec{i}+\vec{j}-\vec{k} \text { and } c=\vec{i}+3 \vec{k}$
Now, $b \times c=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix}$
$=\vec{i}(3-0)-\vec{j}(6+1)+\vec{k}(0-1)$
$=3 \vec{i}-7 \vec{j}-\vec{k}$
Now, $a b c=a \cdot b \times c$
$=a(b \times c) \cos \theta$
$=1 \sqrt{3^{2}+7^{2}+1^{2}} \cos \theta$
$=\sqrt{59} \cos \theta$
$\Rightarrow a b c_{\max }=\sqrt{59} \cdot 1$
( $\because$ maximum value of $\cos \theta$ is 1 )
Hence, maximum value is $\sqrt{59}$.