Let $\overrightarrow a$ and $\overrightarrow b$ be two vectors
such that $$\left| {2\overrightarrow a + 3\overrightarrow b } \right| = \left| {3\overrightarrow a + \overrightarrow b } \right|$$ and the angle between $\overrightarrow a$ and $\overrightarrow b$ is 60$^\circ$. If ${1 \over 8}\overrightarrow a$ is a unit vector, then $\left| {\overrightarrow b } \right|$ is equal to :

$${\left| {3\overrightarrow a + \overrightarrow b } \right|^2} = {\left| {2\overrightarrow a + 3\overrightarrow b } \right|^2}$$<br><br>$$\left( {3\overrightarrow a + \overrightarrow b } \right).\left( {3\overrightarrow a + \overrightarrow b } \right) = \left( {2\overrightarrow a + 3\overrightarrow b } \right).\left( {2\overrightarrow a + 3\overrightarrow b } \right)$$<br><br>$$9\overrightarrow a .\,\overrightarrow a + 6\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow b = 4\overrightarrow a \,.\,\overrightarrow a + 12\overrightarrow a \,.\,\overrightarrow b + 9\overrightarrow b \,.\,\overrightarrow b $$<br><br>$$5{\left| {\overrightarrow a } \right|^2} - 6\overrightarrow a \,.\,\overrightarrow b = 8{\left| {\overrightarrow b } \right|^2}$$<br><br>$$5{(8)^2} - 6.8\,.\,\left| {\overrightarrow b } \right|\cos 60^\circ = 8{\left| {\overrightarrow b } \right|^2}$$ $\because$ $$\left( \matrix{ {1 \over 8}\left| {\overrightarrow a } \right| = 1 \hfill \cr \Rightarrow \left| {\overrightarrow a } \right| = 8 \hfill \cr} \right)$$<br><br>$$40 - 3\left| {\overrightarrow b } \right| = {\left| {\overrightarrow b } \right|^2}$$<br><br>$$ \Rightarrow {\left| {\overrightarrow b } \right|^2} + 3\left| {\overrightarrow b } \right| - 40 = 0$$<br><br>$\left| {\overrightarrow b } \right| = - 8$, $\left| {\overrightarrow b } \right| = 5$<br><br>(rejected)