I'm leaving the question here but my comment answers most of what I was confused about.
The only thing I'm still curious about is why the dot product is restricted from 0 to pi.
I understand that the dot product formula is $a \cdot b = |a||b|\cos(\theta)$ and that to get the projection of a onto b, is to get the length of a along b like the shadow of it.
So $\cos(\theta) = |a_{\text{par}}| / |a| \implies |a| \cdot \cos(\theta) = |a_{\text{par}}|$, where $\cos(\theta)$ may be negative. I'm just wondering why sites list the restriction of θ∈[0,π] since it looks like it invalidates the absolute value. I'm guessing it's negative, despite being an absolute value, due to being the length of the shadow of vector a of which to go backwards in order to reach where b and a meet. Vector a is behind b.
I've noticed it also becomes possible for $|a_{\text{parallel}}| = (a \cdot b) / (|b|)$ to similarly end up as a negative value as well. $\langle 1,1,1 \rangle = a; \langle -1,1,-1 \rangle = b$ and is one example. Theta is $109.47$ degrees.
Is the absolute value being invalidated and what does it mean to get a negative vector projection?negative value as well. <1,1,1> = a <-1,1,-1> = b and is one example. Theta is 109.47 degrees.
Is the absolute value being invalidated and what does it mean to get a negative vector projection?
I suppose you mean that $\mathbf a_\text{parallel}$ is the projection of the vector $\mathbf a$ onto the vector $\mathbf b.$
A typical equation giving the projection of $\mathbf a$ onto $\mathbf b$ is this:
$$ \mathbf a_\text{parallel} = \frac{\mathbf a \cdot \mathbf b}{\lVert\mathbf b\rVert^2} \mathbf b. $$
Here's a way to build up this formula. Let's start with a unit vector in the same direction as $\mathbf b$: $$ \hat{\mathbf b} = \frac{1}{\lVert\mathbf b\rVert} \mathbf b. $$ Multiply this by any positive number $r,$ and you get a vector of length $r$ in the direction of $\mathbf b.$ The clue for dealing with negative numbers is that negating a vector makes it go in the opposite direction. So if $r$ is a positive number, $r\mathbf b$ is in the same direction as $\mathbf b,$ but $-r\mathbf b$ is in the opposite direction from $\mathbf b.$ That is, multiply by a negative number $-r,$ get a vector in the opposite direction.
Next, we use the dot product formula to derive the fact that $$ \lVert\mathbf a\rVert \cos(\theta) = \frac{\mathbf a \cdot \mathbf b}{\lVert\mathbf b\rVert}. $$
Finally, to get the projection of $\mathbf a$ onto $\mathbf b,$ we take the unit vector $\hat{\mathbf b}$ in the direction of $\mathbf b,$ and multiply by $\lVert\mathbf a\rVert \cos(\theta)$: \begin{align} \mathbf a_\text{parallel} &= \left(\lVert\mathbf a\rVert \cos(\theta)\right)\hat{\mathbf b} \\ &= \left(\frac{\mathbf a \cdot \mathbf b}{\lVert\mathbf b\rVert}\right) \left(\frac{1}{\lVert\mathbf b\rVert} \mathbf b\right) \\ &= \frac{\mathbf a \cdot \mathbf b}{\lVert\mathbf b\rVert^2} \mathbf b. \end{align}
When we do this, the fact that $\cos(\theta)$ can be negative is a useful feature of the formula, not an annoyance to be worked around. We get negative values of $\mathbf a \cdot \mathbf b$ when the angle between the vectors is obtuse and only when the angle is obtuse. The cases where the angle is obtuse are exactly the cases in which we want the projection vector to go in the opposite direction from $\mathbf b,$ which is exactly what multiplication of $\mathbf b$ by a negative scalar makes the projection vector do.