If V ¯ = 2 i ¯ + j ¯ - k ¯ ,   W ¯ = i ¯ + 3 k ¯ and U ¯ is a…

It is given that,

$\overrightarrow{V}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$

$\overrightarrow{W}=\overrightarrow{\mathrm{i}}+3\hat{\mathrm{k}}$

The cross product is,

$\overrightarrow{V}\mathit{\times }\overrightarrow{W}\mathit{=}\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 1& -1\\ 1& 0& 3\end{array}\right|$

$=\hat{\mathrm{i}}\left(3-0\right)-\hat{\mathrm{j}}\left(6+1\right)+\hat{\mathrm{k}}\left(0-1\right)$

$=3\hat{\mathrm{i}}-7\hat{\mathrm{j}}-\hat{\mathrm{k}}$

Therefore,

$\left[\overrightarrow{U}\overrightarrow{V}\overrightarrow{W}\right]=U·(\overrightarrow{V}\times \overrightarrow{W})$

$=\left|\overrightarrow{U}\right||\overrightarrow{V}\times \overrightarrow{W}|\cos \theta$

$=\left(1\right)\left(\sqrt{9+49+1}\right)\mathrm{cos\theta }$

$=1\left(\sqrt{59}\right)\cos \theta$

The maximum value of $\left[\overrightarrow{U}\overrightarrow{V}\overrightarrow{W}\right]$ is,

$\left[\overrightarrow{U}\overrightarrow{V}\overrightarrow{W}\right]=\left(\sqrt{59}\right)\cos 0°$

$=\sqrt{59}$