A particle which is experiencing a force, given by $$\vec{F} = 3\hat{i} - 12\hat{j}$$, undergoes a displacement of $$\vec{d} = 4\hat{i}$$. If the particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement?

We begin by recalling the work-energy theorem, which states that the work $$W$$ done by the net force on a particle equals the change in its kinetic energy $$\Delta K$$. In symbols, $$W = K_{\text{final}} - K_{\text{initial}}.$$

Next, we find the work done by the given force during the specified displacement. The work by a constant force is the dot product of the force vector $$\vec F$$ and the displacement vector $$\vec d$$:

$$W = \vec F \cdot \vec d.$$

Substituting the given vectors, we have

$$\vec F = 3\hat i - 12\hat j,$$

$$\vec d = 4\hat i.$$

Carrying out the dot product component-wise,

$$W = (3\hat i - 12\hat j) \cdot (4\hat i)$$

$$\;\; = (3)(4)\, (\hat i \cdot \hat i) \;+\; (-12)(4)\, (\hat j \cdot \hat i).$$

Because the unit vectors are orthogonal, $$\hat i \cdot \hat i = 1$$ and $$\hat j \cdot \hat i = 0$$. Therefore,

$$W = (3)(4)(1) + (-12)(4)(0)$$

$$\;\; = 12 + 0$$

$$\;\; = 12 \text{ J}.$$

So the force does 12 J of work on the particle.

Now, the particle’s initial kinetic energy is given as $$K_{\text{initial}} = 3 \text{ J}.$$

Applying the work-energy theorem:

$$K_{\text{final}} - K_{\text{initial}} = W$$

$$\Rightarrow K_{\text{final}} = K_{\text{initial}} + W$$

$$\Rightarrow K_{\text{final}} = 3 \text{ J} + 12 \text{ J}$$

$$\Rightarrow K_{\text{final}} = 15 \text{ J}.$$

Hence, the correct answer is Option B.