QJEEM23Pha806: JEE Main 2023 Physics - Class 11 JEEM23Pha806 Solution: A small particle moves to position from its initial position…

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 $\vec{r}_f = 5\hat{i} - 2\hat{j} + \hat{k}$ .

To find the work done, we use the dot product of the force and displacement vectors :

$\begin{aligned}& W=\vec{F} \cdot\left(\vec{r}_f-\vec{r}_{\mathrm{i}}\right) \\\\& =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot((5 \hat{i}-2 \hat{j}+\hat{k})-(2 \hat{i}+3 \hat{j}-4 \hat{k})) \\\\& =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot(3 \hat{i}-5 \hat{j}+5 \hat{k}) \\\\& =15-10+35 \\\\& =40 \mathrm{~J}\end{aligned}$