Forces of magnitudes 3 and 4 units acting in the directions $\;$ $6 \hat{i} + 2 \hat{j} + 3 \hat{k}$ $\;$ and $\;$ $3 \hat{i} - 2 \hat{j} + 6 \hat{k}$ $\;$ respectively act on a particle which is displaced from $\left(2, 2, -1\right)$ to $\left(4, 3, 1\right)$. Find the work done by the forces.
Let $\overrightarrow{p} = 6 \hat{i} + 2 \hat{j} + 3 \hat{k}$; $\;$ $\overrightarrow{q} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$
$\left|\overrightarrow{p}\right| = \sqrt{\left(6\right)^2 + \left(2\right)^2 + \left(3\right)^2} = \sqrt{49} = 7$
$\left|\overrightarrow{q}\right| = \sqrt{\left(3\right)^2 + \left(-2\right)^2 + \left(6\right)^2} = \sqrt{49} = 7$
Let $F_1 =$ force of magnitude $3$ units in the direction of $\overrightarrow{p}$
Then, $\overrightarrow{F_1} = \dfrac{3}{7} \left(6 \hat{i} + 2 \hat{j} + 3 \hat{k}\right) = \dfrac{18}{7} \hat{i} + \dfrac{6}{7} \hat{j} + \dfrac{9}{7} \hat{k}$
Let $F_2 =$ force of magnitude $4$ units in the direction of $\overrightarrow{q}$
Then, $\overrightarrow{F_2} = \dfrac{4}{7} \left(3 \hat{i} - 2 \hat{j} + 6 \hat{k}\right) = \dfrac{12}{7} \hat{i} - \dfrac{8}{7} \hat{j} + \dfrac{24}{7} \hat{k}$
$\begin{aligned}
\therefore \; \text{Resultant force } = \overrightarrow{F} & = \overrightarrow{F_1} + \overrightarrow{F_2} \\\\
& = \left(\dfrac{18}{7} \hat{i} + \dfrac{6}{7} \hat{j} + \dfrac{9}{7} \hat{k}\right) + \left(\dfrac{12}{7} \hat{i} - \dfrac{8}{7} \hat{j} + \dfrac{24}{7} \hat{k}\right) \\\\
& = \dfrac{30}{7} \hat{i} - \dfrac{2}{7} \hat{j} + \dfrac{33}{7} \hat{k}
\end{aligned}$
Initial position of the particle $= \left(2, 2, -1\right)$
$\therefore$ $\;$ Initial position vector of the particle $= \overrightarrow{s_1} = 2 \hat{i} + 2 \hat{j} - \hat{k}$
Final position of the particle $= \left(4, 3, 1\right)$
$\therefore$ $\;$ Final position vector of the particle $= \overrightarrow{s_2} = 4 \hat{i} + 3 \hat{j} + \hat{k}$
$\begin{aligned}
\therefore \; \text{Displacement of the particle } = \overrightarrow{s} & = \overrightarrow{s_2} - \overrightarrow{s_1} \\\\
& = \left(4 \hat{i} + 3 \hat{j} + \hat{k}\right) - \left(2 \hat{i} + 2 \hat{j} - \hat{k}\right) \\\\
& = 2 \hat{i} + \hat{j} + 2 \hat{k}
\end{aligned}$
$\begin{aligned}
\text{Now, work done by the forces } & = \overrightarrow{F} \cdot \overrightarrow{s} \\\\
& = \left(\dfrac{30}{7} \hat{i} - \dfrac{2}{7} \hat{j} + \dfrac{33}{7} \hat{k}\right) \cdot \left(2 \hat{i} + \hat{j} + 2 \hat{k}\right) \\\\
& = \dfrac{60}{7} - \dfrac{2}{7} + \dfrac{66}{7} = \dfrac{124}{7} \; \text{ units}
\end{aligned}$