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Say I have a $F=kQ_{1}Q_{2}/r^{2}$ and a direction vector $(x, y, z).$ How can I find the component forces $F_{x}$, $F_{y}$, and $F_{z}$?

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To find the components of any vector $\bf F$ using unit vectors, you can use the dot product between the vector and each unit vector.

So the x-component of $\bf F$ is $\bf F\cdot \hat i$ the y-component is $\bf F\cdot \hat j$ and the z-component is $\bf F\cdot \hat k$

If you have a "direction vector" $\bf u=(x,y,z)$ then its unit vector would be $$\bf \hat u=\frac{u}{\mid u\mid}$$ so that $$\bf F\cdot \hat u$$ is the projection of $\bf F$ in the direction of the unit vector $\bf \hat u$ or the component of $\bf F$ along $\bf u$.

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It's somewhat unclear from your question, but I interpreted $F$ to be just the magnitude of the force (a scalar), and you want to construct a force of that magnitude pointing along the given direction vector.

If the direction vector is a unit vector (a vector of length 1), then all you have to do is scale (resize) it. So it's just:

$$\vec F = F \cdot (x, y, z) = (Fx, Fy, Fz)$$

If the direction vector is not a unit vector, then you have to make it into one first:

$$\text{let }\space \vec d = (x, y, z)$$ Then it's magnitude squared is $$d^2= \vec d \vec d = x^2 + y^2 + z^2 \space\space\text{, and}\\ d = \sqrt{x^2 + y^2 + z^2} $$ so $$\vec F = F \cdot \frac{\vec d}{d} = (\frac{Fx}{d}, \frac{Fy}{d}, \frac{Fz}{d})$$

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  • $\begingroup$ Your equations seem to imply that you get a vector from a dot product. $\endgroup$
    – joseph h
    Sep 21 at 3:56
  • $\begingroup$ @josephh - no, the only dot product here is $\vec d \vec d$, producing a scalar. I'm using the OP's notation where $\vec F$ (with an arrow) is a vector, and $F$ is its magnitude. Perhaps the multiplication dot is confusing in the final row? Or my use of (x, y, z) shorthand? $\endgroup$ Sep 21 at 5:13
  • $\begingroup$ Yeah. I think the dot caught me off guard - then I read it again and looks OK. Cheers. $\endgroup$
    – joseph h
    Sep 21 at 5:27
  • $\begingroup$ Thank you, this helped a lot! $\endgroup$
    – harry
    Sep 22 at 6:14
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You need to know the direction of the force as well as its magnitude. The force’s component along the $x$ axis is then

$F_x = |\vec F| \cos \theta$

where $\theta$ is the angle between $\vec F$ and the $x$ axis etc.

If the force is radial i.e. $\vec F = |\vec F| \vec {\hat r}$ then its components at $(x,y,z)$ are

$\displaystyle F_x = |\vec F| \frac x r$

etc.

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