I am a student and I have a conflict with a given answer in the textbook. The question is the following:
Evaluate the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ for the given vector field $\mathbf{F}$ and the specified curve $C$.
$\mathbf{F} = \mathbf{a} \times \mathbf{r}$, where $\mathbf{a}$ is a constant vector, $\mathbf{r} = \langle x, y, z \rangle$, and $C$ is a straight line segment from $\mathbf{r}_1$ to $\mathbf{r}_2$.
Here is my solution:
$$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (\mathbf{a} \times \mathbf{r}) \cdot d\mathbf{r} = 0$$ because the triple product of coplanar vectors vanishes.
However, the solution given is $\mathbf{r}_2 \cdot (\mathbf{a} \times \mathbf{r}_1)$.
With the help of the comment by Lord Shark the Unknown, I let \begin{align} \mathbf{r} & = \mathbf{r}_1 + t(\mathbf{r}_2 - \mathbf{r}_1) \\ \Longrightarrow \quad d\mathbf{r} & = (\mathbf{r}_2 - \mathbf{r}_1)\,dt. \end{align}
We now have \begin{align} \mathbf{a} \times \mathbf{r} & = \mathbf{a} \times \mathbf{r}_1 + t(\mathbf{r}_2 - \mathbf{r}_1) \\ & = \mathbf{a} \times \mathbf{r}_1 + t\mathbf{a} \times (\mathbf{r}_2 - \mathbf{r}_1). \end{align}
And finally \begin{align} \int_C \mathbf{F} \cdot d\mathbf{r} & = \int_0^1 (\mathbf{a} \times \mathbf{r}_1 + t\mathbf{a} \times (\mathbf{r}_2 - \mathbf{r}_1)) \cdot (\mathbf{r}_2 - \mathbf{r}_1)\,dt \\ & = \int_0^1 (\mathbf{a} \times \mathbf{r}_1) \cdot (\mathbf{r}_2 - \mathbf{r}_1)\,dt \\ & = \int_0^1 (\mathbf{a} \times \mathbf{r}_1) \cdot \mathbf{r}_2\,dt = (\mathbf{a} \times \mathbf{r}_1) \cdot \mathbf{r}_2. \end{align}