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Work done $W = \vec{F}\; .\vec{S}$ where F is the force vector and S is the displacement vector.

Let us first establish that a vector in 3-dimensions can be broken into 3 components:

The x-axis component $\hat{i}$,

The y-axis component $\hat{j}$, and

The z-axis component $\hat{k}$.

Each component of a vector depicts the magnitude of influence of that vector in a given direction. The $\hat{i}$, $\hat{j}$ and $\hat{k}$ represent unit vectors in the x-, y- and z-direction respectively, and the number that precedes them represent the magnitude of the vector in that direction.

Now, we have a body that can move only along the direction of the z-axis. This means that any distance that we take that this body covers will be in the z($\vec{k}$)-direction. Therefore, the distance that the body travels under the influence of the force can be represented by the displacement vector $\vec{S} = 0\hat{i}+0\hat{j}+4\hat{k}$.

The work done by the force $\vec{F} = -1\hat{i}+2\hat{j}+3\hat{k}$ to move the body by a distance $\vec{S} =4\hat{k}$ is given as the scalar product of the two, i.e.,:

$W = \vec{F}.\vec{S} = \left(-1\hat{i}+2\hat{j}+3\hat{k}\right). \left(4\hat{k}\right)$

$\Rightarrow W = \left(3\hat{k}\right). \left(4\hat{k}\right) = 12\;J$

Therefore, only the z-component of the force contributes to moving the body in the z-direction. Thus, the work done by the force in moving the body through a distance of $4\;m$ is $12\;J$

Remember that the dot product of two vectors results in a scalar quantity and hence is it not directional. Another form of expressing the dot product when instead of the individual components the angle $\theta$ between the two vectors is given is:

$W = \vec{F}.\vec{S} = |F||S|cos\theta$

In the above problem, we consider only $ W = \left(3\hat{k}\right). \left(4\hat{k}\right) $, which means $ W = 4 \times 3 \cos 0^{\circ} = 12\;J$ since $cos0^{\circ} =1$ . This is the same reason why we do not consider $\hat{i}.\hat{j}$ or $\hat{j}.\hat{k}$ or $\hat{i}.\hat{k}$ since for them, $\theta =90^{\circ} \Rightarrow cos 90^{\circ} = 0, \Rightarrow W=0$.

Therefore, the work done is numerically quantified only when the vectors are not perpendicular to each other and the vectors have non-zero components.