Answer
We verify that the work performed is equal to ${\bf{F}}\cdot\overrightarrow {PQ} $.
Work Step by Step
Let the coordinates of $P$ and $Q$ be $P = \left( {{x_1},{y_1},{z_1}} \right)$ and $Q = \left( {{x_2},{y_2},{z_2}} \right)$, respectively.
The work performed by the field ${\bf{F}}$ is
$W = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$
Since ${\bf{F}}$ is constant, we can write the work as
$W = {\bf{F}}\cdot\left( {\mathop \smallint \limits_P^Q {\rm{d}}{\bf{r}}} \right) = {\bf{F}}\cdot\left( {{\bf{r}}|_P^Q} \right) = {\bf{F}}\cdot\left( {{{\bf{r}}_Q} - {{\bf{r}}_P}} \right)$,
where ${{\bf{r}}_Q}$ and ${{\bf{r}}_P}$ are the position vectors of $Q$ and $P$, respectively.
Since ${{\bf{r}}_Q} - {{\bf{r}}_P} = \overrightarrow {PQ} $, therefore, $W = {\bf{F}}\cdot\overrightarrow {PQ} $.
Thus, the work performed by a constant force field ${\bf{F}}$ over any path $C$ from $P$ to $Q$ is equal to ${\bf{F}}\cdot\overrightarrow {PQ} $.
