TITLE: Dot product with a constant
QUESTION [6 upvotes]: What is $\vec{a}\cdot (\vec{b}+c)$, where $c$ is a constant? 

Am I able to just take out the constant, or do I multiply the constant by the vector as a scalar?
EDIT (for context): I was finding the work done on a particle moving on the curve $C$: $x=cos(t)$, $y=sin(t)$, from $2\pi \leq t \leq 0$, through the vector field $\vec{F}(x,y)= -y\hat{i} +x\hat{j}$. I found that the work done is $-2\pi$. I wanted to figure out how that work changes when that same circle is moved horizontally by $n$ units, so that $x=cos(t)+n$, where $n$ is a constant. Here’s my work which led me to this issue: 
$$\oint_{C} \vec{F} \cdot d\vec{r}$$
$$\oint_{2\pi}^{0} (-sin(t)\hat{i} +(cos(t) +n)\hat{j}) \cdot (-sin(t)\hat{i} +cos(t)\hat{j}) dt$$
But I didn’t know how to cross $(cos(t)+n)$ with $(cos(t)$

REPLY [2 votes]: Others have already addressed your question about adding a scalar to a vector (and then taking the dot product of that with another vector), but not the underlying issue that led you to attempt that ill-defined (unless you work with fancy geometric algebras) operation.
Simply put, you say that you want to translate the object's path by $n$ units along the horizontal axis.  Thus, the constant you need to add to the position of the object is not the scalar n, but the vector $n\hat j$ (where $\hat j$ is the horizontal unit vector, as it apparently is in your question).  Once you fix that, everything should work just fine.
In particular, the dot product distributes over the sum of vectors, so that $\vec a \cdot (\vec b + \vec c) = \vec a \cdot \vec b + \vec a \cdot \vec c$.