Find the least positive integral value of x for which the angle between vectors $\overrightarrow{\mathrm{a}}=\mathrm{x}\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}-\widehat{\mathrm{k}}$ and  $\overrightarrow{b}=2x\widehat{i}+x\widehat{j}-\widehat{k}$ is acute

Let $\overrightarrow{\mathrm{a}}=\mathrm{x}\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}-\widehat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{b}}=2\mathrm{x}\widehat{\mathrm{i}}+\mathrm{x}\widehat{\mathrm{j}}-\widehat{\mathrm{k}}$ be the adjacent sides of the parallelogram 

Now angle between $\overrightarrow{\mathrm{a}}\text{ and }\overrightarrow{\mathrm{b}}$ is acute, i.e., 

$\begin{array}{l}|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|>|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|\\ \Rightarrow |3\mathrm{x}\widehat{\mathrm{i}}+(\mathrm{x}-3)\widehat{\mathrm{j}}-2\widehat{\mathrm{k}}{|}^2>|-\mathrm{x}\widehat{\mathrm{i}}-(\mathrm{x}+3)\widehat{\mathrm{j}}{|}^2\\ \mathrm{or} 9{\mathrm{x}}^2+(\mathrm{x}-3{)}^2+4>{\mathrm{x}}^2+(\mathrm{x}+3{)}^2\end{array}$

$\begin{array}{l}\mathrm{or} 8{\mathrm{x}}^2-12\mathrm{x}+4>0\\ \mathrm{or} 2{\mathrm{x}}^2-3\mathrm{x}+1>0\\ \mathrm{or} (2\mathrm{x}-1)(\mathrm{x}-1)>0\\ \Rightarrow \mathrm{x}<1/2\text{ or }\mathrm{x}>1\end{array}$

Hence, the least positive integral value is 2.