## Precalculus (6th Edition) Blitzer

The work done by a force F in moving an object A to B is $\mathbf{F}\cdot \overrightarrow{AB}$.
Consider an object that moves from point A to B, but the force acting on the object is at an angle $\theta$. The work done $W$ is defined as the product of the magnitude of the force F in the direction of displacement and the magnitude of the displacement. The component of force $\mathbf{F}$ in the direction of displacement is, $\mathbf{F}=\left\| \mathbf{F} \right\|\cos \theta$ The magnitude of the displacement vector is, $\text{Magnitude of }\overrightarrow{AB}=\left\| \overrightarrow{AB} \right\|$ According to the definition of work-done. The work-done is, \begin{align} & \text{W}=\left( \left\| \overrightarrow{AB} \right\| \right)\left( \left\| \mathbf{F} \right\|\cos \theta \right) \\ & =\left\| \overrightarrow{AB} \right\|\left\| \mathbf{F} \right\|\cos \theta \\ & =\mathbf{F}\centerdot \overrightarrow{AB} \end{align} Hence, the work done by a force F in moving an object A to B is $\mathbf{F}\cdot \overrightarrow{AB}$.