A parallelogram is construct $3\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\text{ and }\overrightarrow{\mathrm{a}}-4\overrightarrow{\mathrm{b}},\text{ where }\left|\overrightarrow{\mathrm{a}}\right|=6\text{ and }\left|\overrightarrow{\mathrm{b}}\right|=8\text{, and }\overrightarrow{\mathrm{a}}\text{ and }\overrightarrow{\mathrm{b}}$ are anti parallel. Then the length of the longer diagonal is  

$\begin{array}{l}(3\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\cdot (\overrightarrow{\mathrm{a}}-4\overrightarrow{\mathrm{b}})\\ =3|\overrightarrow{\mathrm{a}}{|}^2-11\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}-4|\overrightarrow{\mathrm{b}}{|}^2\\ =3\times 36-11\times 6\times 8\cos \mathrm{\pi }-4\times 64>0\end{array}$

Therefore, the angle between $\overrightarrow{\mathrm{a}}\text{ and }\overrightarrow{\mathrm{b}}$ is acute. The longer diagonal is given by 

$\begin{array}{l}\overrightarrow{\mathrm{\alpha }}=(3\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{a}}-4\overrightarrow{\mathrm{b}})=4\overrightarrow{\mathrm{a}}-3\overrightarrow{\mathrm{b}}\\ \mathrm{now}, |\overrightarrow{\mathrm{\alpha }}{|}^2=|4\overrightarrow{\mathrm{a}}-3\overrightarrow{\mathrm{b}}{|}^2\\ =16|\overrightarrow{\mathrm{a}}{|}^2+9|\overrightarrow{\mathrm{b}}{|}^2-24\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}\\ =16\times 36+9\times 64-24\times 6\times 8\cos \mathrm{\pi }\\ =16\times 144\\ \mathrm{or} |4\overrightarrow{\mathrm{a}}-3\overrightarrow{\mathrm{b}}|=48\end{array}$