The angle between vector $$\left(\vec{A}\right)$$ and $$\left(\vec{A} - \vec{B}\right)$$ is:

We need to find the angle between the vector $$\vec{A}$$ and the resultant vector $$\vec{A} - \vec{B}$$. From the options, we can infer that the angle between vector $$\vec{A}$$ and vector $$\vec{B}$$ is given as $$\theta = 60^\circ$$.

1. Visualize Vector Subtraction

The vector subtraction $$\vec{R} = \vec{A} - \vec{B}$$ can be treated as the vector addition of $$\vec{A}$$ and $$(-\vec{B})$$:

$$\vec{R} = \vec{A} + (-\vec{B})$$

If the angle between $$\vec{A}$$ and $$\vec{B}$$ is $$\theta$$, then the angle between $$\vec{A}$$ and $$-\vec{B}$$ is $$(180^\circ - \theta)$$.

2. Use the Formula for Direction of Resultant Vector

Let $$\alpha$$ be the angle that the resultant vector $$\vec{R} = \vec{A} - \vec{B}$$ makes with the vector $$\vec{A}$$. Using the standard vector direction formula:

$$\tan \alpha = \frac{B \sin(180^\circ - \theta)}{A + B \cos(180^\circ - \theta)}$$

Using the trigonometric identities $$\sin(180^\circ - \theta) = \sin \theta$$ and $$\cos(180^\circ - \theta) = -\cos \theta$$, the expression simplifies to:

$$\tan \alpha = \frac{B \sin \theta}{A - B \cos \theta}$$

3. Substitute $\theta = 60^\circ$ into the Equation

Let's substitute the values for $$\sin 60^\circ = \frac{\sqrt{3}}{2}$$ and $$\cos 60^\circ = \frac{1}{2}$$ into the expression:

$$\tan \alpha = \frac{B \left(\frac{\sqrt{3}}{2}\right)}{A - B \left(\frac{1}{2}\right)}$$

To eliminate the fractions in the numerator and denominator, multiply both top and bottom by $$2$$:

$$\tan \alpha = \frac{\sqrt{3}B}{2A - B}$$

4. Solve for $\alpha$

Taking the inverse tangent of both sides isolates the angle $$\alpha$$:

$$\alpha = \tan^{-1}\left(\frac{\sqrt{3}B}{2A - B}\right)$$

Final Answer: Option B $$\left[ \tan^{-1}\left(\frac{\sqrt{3}B}{2A - B}\right) \right]$$