#### Let be two vectors such that and the angle between and is . If is a unit vector, then is equal to :Option: 1Option: 2Option: 3Option: 4

$\left | \frac{\vec{a}}{8} \right |= 1\Rightarrow \left | \vec{a} \right |= 8$
$\left | 2\vec{a}+3\vec{b} \right |^{2}= \left | 3\vec{a}+\vec{b} \right |^{2}$
$\Rightarrow 4\left | \vec{a} \right |^{2}+9\left | \vec{b} \right |^{2}+ 12 \, \vec{a}\cdot \; \vec{b}= 9\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+6\, \vec{a}\,\;\cdot \vec{b}$
$\Rightarrow 5\left | \vec{a} \right |^{2}-8\left | \vec{b} \right |^{2}-6\; \left | \vec{a} \right |\left | \vec{b} \right |\cos 60^{\circ}= 0$
$\Rightarrow 5\times 8^{2}-8\left | \vec{b} \right |^{2}-6\times 8\times \left | \vec{b} \right |\times \frac{1}{2}= 0$
$\Rightarrow \left | \vec{b} \right |^{2}+3 \left | \vec{b} \right |-40= 0$
$\Rightarrow \left | \vec{b} \right |^{2}+8 \left | \vec{b} \right |-5\left | \vec{b} \right |-40= 0$
$\Rightarrow \left | \vec{b} \right |^{2}= -8 \times \left ( Not\: possible \right )$
OR $\left | \vec{b} \right |= 5$
option (4)