# Line Integrals of Vector Fields, Homework Conundrum

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 $$\mathbf{r}_2 \cdot (\mathbf{a} \times \mathbf{r}_1)$$.

• I don't see a triple product of coplanar vectors. How about writing $r=r_1+t(r_2-r_1)$ and actually doing the integration? – Lord Shark the Unknown Dec 10 '18 at 6:54
• @LordSharktheUnknown So I suppose I've learned that $\mathbf{r}$ and $d\mathbf{r}$ are not necessarily parallel. Thank you for the comment. I have solved the problem. – Davis Rash Dec 10 '18 at 7:44
• @LordSharktheUnknown Oh my God, I have just now realized how stupid I am. Of course $\mathbf{r}$ and $d\mathbf{r}$ are not parallel. Shame. – Davis Rash Dec 10 '18 at 7:47

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}