Find the work done by a force of 2 newtons (N) acting in the direction $\mathbf{i}+\mathbf{j}+\mathbf{k}$ in moving an object 1 m from (0, 0, 0) to (1, 0, 0).

Solution We need to express the force $\mathbf{F}$ in terms of its magnitude and direction. The unit vector $\mathbf{u}$ in the direction $\mathbf{v}=\mathbf{i}+\mathbf{j}+\mathbf{k}$ is $$\mathbf{u}=\dfrac{\mathbf{v}}{\left\Vert \mathbf{v}\right\Vert }=\dfrac{ \mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{3}}=\dfrac{1}{\sqrt{3}}\mathbf{i}+ \dfrac{1}{\sqrt{3}}\mathbf{j}+\dfrac{1}{\sqrt{3}}\mathbf{k}$$

Since the force vector $\mathbf{F}$ has magnitude $2,$ we have $$\mathbf{F}=2\left( \frac{1}{\sqrt{3}}\mathbf{i}+\dfrac{1}{\sqrt{3}}\mathbf{j} +\dfrac{1}{\sqrt{3}}\mathbf{k}\right) =\dfrac{2}{\sqrt{3}}( \mathbf{i}+ \mathbf{j}+\mathbf{k})$$

The line of motion of the object from $( 0,0,0)$ to $\left( 1,0,0\right)$ is $\skew5\overrightarrow{\it AB}=\mathbf{i}$. The work $W$ is therefore $$W=\mathbf{F}\,{\cdot}\, \skew5\overrightarrow{\it AB}=\frac{2}{\sqrt{3}}(\mathbf{i}+\mathbf{j }+\mathbf{k})\,{\cdot}\ \mathbf{i}=\frac{2}{\sqrt{3}}\hbox{ joules}$$