Question

Let $\overline{v}=2\overline{i}+\overline{j}-\overline{k}$ and $\overline{w}=\overline{i}+3\overline{k}$. If $\overline{u}$ is any unit vector then the maximum value of the scalar triple product $\left[\overline{u}\overline{v}\overline{w}\right]$ is

• A. 1
• B. $\sqrt{10}+\sqrt{6}$
• C. $\sqrt{59}$
• D. $\sqrt{60}$

Answer 1

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

$\overline{v}=2\overline{i}+\overline{j}-\overline{k}$

$\overline{w}=\overline{i}+3\overline{k}$

$\left[\overline{u}\overline{v}\overline{w}\right]=\overline{u}.(\overline{v}\times \overline{w})$

$=\overline{u}.\left[\right(2\overline{i}+\overline{j}-\overline{k})\times (\overline{i}+3\overline{k}\left)\right]$

$=\overline{u}.[0-6\overline{j}-\overline{k}+3\overline{i}-\overline{j}-0]$

$=\overline{u}.[3\overline{i}-7\overline{j}-\overline{k}]$

Since $\overline{a}.\overline{b}=|\overline{a}||\overline{b}|\cos \theta$,

And here the maximum values of $\cos \theta$ and $|\overline{u}|$ being $1$,

The maximum value of $\left[\overline{u}\overline{v}\overline{w}\right]$ becomes $\sqrt{9+49+1}=\sqrt{59}$