If \(|\mathbf{a}|=2\) and \(|\mathbf{b}|=3\) and the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is \(120^{\circ}\), then the length of the vector \(\left|\frac{\mathbf{a}}{2}-\frac{\mathbf{b}}{3}\right|\) is

We know that,
\[
\begin{aligned}
& \mathbf{a} \cdot \mathbf{b}=|\mathbf{a}| \cdot|\mathbf{b}| \cos \theta \\
& \text { So, } \mathbf{a} \cdot \mathbf{b}=(2)(3) \cos 120^{\circ} \quad\left[\because \theta=120^{\circ}\right] \\
& =6 \cos \left(90^{\circ}+30^{\circ}\right)=6\left(-\sin 30^{\circ}\right)=-6 \times \frac{1}{2} \\
& \mathbf{a} \cdot \mathbf{b}=-3 \\
& \text { Now, }\left|\frac{1}{2} \mathbf{a}-\frac{1}{3} b^2\right|^2=\left(\frac{1}{2} \mathbf{a}-\frac{1}{3} \mathbf{b}\right)\left(\frac{1}{2} \mathbf{a}-\frac{1}{3} \mathbf{b}\right) \\
& =\frac{1}{4}|\mathbf{a}|^2-\frac{1}{6} \mathbf{a} \cdot \mathbf{b}-\frac{1}{6} \mathbf{a} \cdot \mathbf{b}+\frac{1}{9}|\mathbf{b}|^2 \\
& =\frac{1}{4} \times 4-\frac{1}{6} \times(-3)-\frac{1}{6}(-3)+\frac{1}{9} \times 9 \\
& =1+\frac{1}{2}+\frac{1}{2}+1=2+1=3 \\
&
\end{aligned}
\]
Hence, option (2) is correct.