Newton's Second Law with unit vectors Q: Why does physics express formulas with a single unit vector when a linear combination of basis vectors is needed?
So, instead of: $\vec{v} = v_x \cdot \hat{x} + v_y \cdot \hat{y} + v_z \cdot \hat{z}$ , which we learn in linear algebra, physics uses: $\vec{v} = ||\vec{v}|| \cdot \hat{v}$
If I combine both ideas, I get: $||\vec{v}|| \cdot \hat{v} = \vec{v} = v_x \cdot \hat{x} + v_y \cdot \hat{y} + v_z \cdot \hat{z}$  , where $||\vec{v}||$ is the magnitude. But I run into a problem when I am dealing with a  linear combination and unit vector notation mixed together in the same equation because I have multiple unknowns 
For example:
$||F_g|| \cdot \hat{r} = m\vec{a}$
$||F_g|| \cdot \hat{r} = m(a_x \cdot \hat{x} + a_y \cdot \hat{y} + a_z \cdot \hat{z})$
Now, what must I do to get $\hat{r}$ equal to a linear combination so that I use dot product on both sides to get my scalar equations?
 A: It is equivalent to write:
$$\vec v=v_x\cdot \hat x+v_y\cdot \hat y+v_z\cdot \hat z$$
and
$$\vec v=v\cdot \hat v+0\cdot \hat w+0\cdot \hat \omega$$
(The latter simplifies to $\vec v=v\cdot \hat v$. And $v$ happens to equal $||\vec v||$.)
Both are describing the same vector but in different coordinate systems.


*

*One is the usual and well-known (Cartesian) ($\hat x$,$\hat y$,$\hat z$)-coordinate system and

*the other is a new ($\hat v$,$\hat w$,$\hat \omega$)-coordinate system, which we have invented just for this situation. It is invented to have it's first axis along the vector, thus the 2nd and 3rd coordinates are zero.


You can invent such new coordinate system on the fly anytime. It contains three basis vectors - enough to fully describe the entire 3D physical world - so it is perfectly legal and valid. You can think of the new coordinate system as a "tilted" version of the usual one; tilted to fit the vector. Regardless of the coordinate system, the vector is the same - we look at the same object but change "perspective" so to speak.
This simplifies the math (reduces the number of coordinate terms) and is therefore smart. Changing coordinates is an entire discipline in mathematics and very useful in physics.
Now to your specific example
You write Newton's 2nd law like this:
$$||\vec F||\cdot \hat r=m\vec a$$
but why stop there? Why not continue the unit-vector notation like this:
$$||\vec F||\cdot \hat r=m||\vec a||\cdot \hat r$$
After all, the force and the acceleration both point the same way. They can both be written from the same unit vector, which you have chosen to call $\hat r$. We could expand it to show the complete linear combination:
$$||\vec F||\cdot \hat r+0\cdot\hat p+0\cdot\hat q=m(||\vec a||\cdot \hat r+0\cdot\hat p+0\cdot\hat q)$$
where the ($\hat r$,$\hat p$,$\hat q$)-coordinates are those we have invented for this specific situation so the first axis aligns with the vector.
Both the $\vec F$ and the $\vec a$ vectors could just as well have been written in usual Cartesian coordinates:
$$F_x\cdot \hat x+F_y\cdot\hat y+F_z\cdot\hat z=m(a_x\cdot \hat x+a_y\cdot\hat y+a_z\cdot\hat z)$$
But now fewer terms will be zero. This is more complicated, nevertheless just as correct.

Now, what must I do to get $\hat{r}$ equal to a linear combination so that I use dot product on both sides to get my scalar equations?

It seems in this last sentence of yours that you want the $\hat r$ to be written as a linear combination of $\hat x$, $\hat y$ and $\hat z$? If so, then you can essentially just divide with the $||\vec F||$ term on both sides, and then $\hat r$ is isolated:
$$\begin{align}
||\vec F||\hat r&=m(a_x\cdot \hat x+a_y\cdot\hat y+a_z\cdot\hat z)\quad\Leftrightarrow\\
~&~\\
\hat r&=\frac m{||\vec F||}(a_x\cdot \hat x+a_y\cdot\hat y+a_z\cdot\hat z)\\
&=\frac{ma_x}{||\vec F||}\cdot \hat x+\frac{ma_y}{||\vec F||}\cdot\hat y+\frac{ma_z}{||\vec F||}\cdot\hat z
\end{align}$$
Voila. Was this what you meant or did I misunderstand the question?
