A small particle moves to position $5 \hat{i}-2 \hat{j}+\hat{k}$ from its initial position $2 \hat{i}+3 \hat{j}-4 \hat{k}$ under the action of force $5 \hat{i}+2 \hat{j}+7 \hat{k} \mathrm{~N}$. The value of work done will be __________ J.

The given expression calculates the work done by a force vector $\vec{F} = 5\hat{i} + 2\hat{j} + 7\hat{k}$ when it acts on an object that moves from an initial position vector $\vec{r}_i = 2\hat{i} + 3\hat{j} - 4\hat{k}$ to a final position vector $\vec{r}_f = 5\hat{i} - 2\hat{j} + \hat{k}$. <br/><br/>To find the work done, we use the dot product of the force and displacement vectors : <br/><br/>$$ \begin{aligned} & W=\vec{F} \cdot\left(\vec{r}_f-\vec{r}_{\mathrm{i}}\right) \\\\ & =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot((5 \hat{i}-2 \hat{j}+\hat{k})-(2 \hat{i}+3 \hat{j}-4 \hat{k})) \\\\ & =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot(3 \hat{i}-5 \hat{j}+5 \hat{k}) \\\\ & =15-10+35 \\\\ & =40 \mathrm{~J} \end{aligned} $$