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, $b×c=â\begin{array}{ccc}\vec{i}& \vec{j}& \vec{k}\\ 2& 1& −1\\ 1& 0& 3\end{array}â$
$=\vec{i}(3−0)−\vec{j}(6+1)+\vec{k}(0−1)$
$=3\vec{i}−7\vec{j}−\vec{k}$
Now, $abc=aâ‹\dots b×c$
$=a(b×c)\cos θ$
$=1\sqrt{3^2+7^2+1^2}\cos θ$
$=\sqrt{59}\cos θ$
$⇒ab{c}_{\max }=\sqrt{59}â‹\dots 1$
( $âˆ\mu$ maximum value of $\cos θ$ is 1 )
Hence, maximum value is $\sqrt{59}$.