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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 $(cos(t)$

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    $\begingroup$ This only makes sense if you make a vector space out of the base field $\endgroup$ – Brevan Ellefsen Feb 22 at 0:04
  • $\begingroup$ This question lacks context. $\endgroup$ – Carsten S Feb 22 at 10:37
  • $\begingroup$ Just edited it, thank you for helping $\endgroup$ – ItIsLastThursday Feb 22 at 11:47
  • $\begingroup$ I do not see how your edited question relates to your original question. $\endgroup$ – Servaes Feb 22 at 12:00
  • $\begingroup$ Because I have the dot product of $cos(t)$ with $(cos(t)+n)$ $\endgroup$ – ItIsLastThursday Feb 22 at 12:04
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This product has no meaning since the sum in the parentheses is undefined; you can not add a vector and a scalar.

$c$ must be a vector, then the sum is defined and the given product is a dot product.

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  • $\begingroup$ So is there any answer I can get? $\endgroup$ – ItIsLastThursday Feb 21 at 20:31
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    $\begingroup$ No, if $c$ is a scalar then there is no answer. $\endgroup$ – Maria Mazur Feb 21 at 20:32
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    $\begingroup$ You can add a vector and a scalar in a more advanced context. This is like when all of my grade school teachers said you couldn't take the square root of a negative number, and I was ahead so I knew they were lying. $\endgroup$ – Matt Samuel Feb 22 at 0:50
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    $\begingroup$ I know nothing about that. Would you care to write it down, as a new answer? @MattSamuel $\endgroup$ – Maria Mazur Feb 22 at 7:23
  • $\begingroup$ Also, how could I fix the problem to actually give me a sensible answer? $\endgroup$ – ItIsLastThursday Feb 22 at 12:42
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For an expression to be defined, all of its constituting parts need to be defined as well.

The sum of a vector and a constant $\vec b + c$ is not defined, and hence the expression $\vec a \cdot (\vec b + c)$ is not defined.

This applies to any scalar, not just constants, because $c$ can depend on another variable.


Although it probably makes sense to you that $$\vec a \cdot (\vec b + c) = \vec a \cdot \vec b + c \cdot \vec a$$ neither side of the equation is defined, which means that you cannot use distribution the way it was used above.

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    $\begingroup$ The RHS expression isn't defined either. $\endgroup$ – Hammerite Feb 21 at 22:53
  • $\begingroup$ @Hammerite You are right. What a silly mistake. Thank you! $\endgroup$ – Haris Gusic Feb 21 at 22:54
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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 $\vec a \cdot (\vec b + \vec c) = \vec a \cdot \vec b + \vec a \cdot \vec c$.

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  • $\begingroup$ So would this be my dot product: $(-sin(t)\hat{i} + (cos(t)+n)\hat{j}) \cdot (-sin(t)\hat{i} + cos(t)\hat{j}) = sin^{2}(t) + cos^{2}(t) +ncos(t)$ $\endgroup$ – ItIsLastThursday Feb 22 at 16:08
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This is perfectly fine. Grassman/Clifford/Geometric algebra interpretations have a perfectly sane explanation for these objects. For any number of dimensions, expand the vector into its components, where $e_1$ and $e_2$ are mutually perpendicular unit vectors. Dot product with parallel is 1 (ie: cos). Dot product with perpendicular is 0 (ie: cos)

$$ (\vec a) \cdot (\vec b + c) = (a_1 e_1 + a_2 e_2) \cdot (b_1 e_1 + b_2 e_2 + c) \\ = a_1 e_1 \cdot (b_1 e_1 + b_2 e_2 + c) + a_2 e_2 \cdot (b_1 e_1 + b_2 e_2 + c) \\ = a_1 b_1 + c a_1 e_1 + a_2 b_2 + c a_2 e_2 \\ = (a_1 b_1 + a_2 b_2) + c (a_1 e_1 + a_2 e_2) \\ = \vec a \cdot \vec b + c \vec a $$

This is the same as previous answer, but worked out with an explanation of why. Notice that this is a scalar plus a vector answer. The question also had a scalar plus a vector. When you add and multiply these, you can end up with something that ends up being real, or imaginary, or something else with a straightforward geometric meaning. You can use this to find things like vector plus rotation in plane $e_1 e_2$ (ie: an imaginary plane we can call $I$):

$$ s = (a_1 b_1 + a_2 b_2) \\ (s + c (a_1 e_1 + a_2 e_2)) e_1 e_2 \\ s e_1 e_2 + ( c a_1 e_2 - c a_2 e_1 ) \\ s I + c( -a_2 e_1 + a_1 e_2 ) $$

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    $\begingroup$ There are various inner products in geometric algebra that differ in how you handle taking the inner product of a vector and a scalar. There are reasons why you might choose one over another, and arguably the most compelling option is one of the contraction inner products for which $c\cdot\vec{v}=c\vec{v}$ but $\vec{v}\cdot c=0$ or vice versa as it comes in both left and right versions (and this asymmetry is indicated by using the notation $c\rfloor\vec{v}$ instead). So the formula is a bit ambiguous as it depends on a choice of inner product. $\endgroup$ – Derek Elkins Feb 22 at 6:23
  • $\begingroup$ @DerekElkins I noticed something fishy about treating dot and wedge as fundamental operations, and therefore try to just explicitly expand vectors and rely on them being unit and mutually perpendicular. I saw one place that said that wedge of two scalars is 1, and became confused by what dot of two scalars must be (0?). So, I kind of just dispense with it all and multiply multivectors explicitly. $\endgroup$ – Rob Feb 22 at 7:09

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