How to determine if two points lie on a vector, given a unit vector If you have two points, $A$ and $B$, at $(1,1,1)$ and $(1,1,7)$ respectively, and a unit vector $C (0,0,1)$. What's a way to find if unit vector C, will cross B if C extends forever. (Unit vector C has a starting point at A)
Basically, imagine point A is a person, point B is an object. C is a unit vector. Is person A looking directly at object B given A's unit vector?
Since  $A\cdot B = ||A||\;||B||\cos\theta$
could I say, if the dot product of A and B is equal to $||A|| \; ||B||$, then it would mean $\cos θ = 1$, therefore $θ = 0$ and the line crosses B.
 A: This is the line going through point $A$ having direction $e_z$:
$$
x(\alpha) 
= A + \alpha e_z 
= (1,1,1) + \alpha (0,0,1) = (1, 1, 1 + \alpha)
$$
where $\alpha \in \mathbb{R}$.

We want point $B$ to be on the line as well, thus there should be a real value $\alpha$ such that $x(\alpha)$ is the vector from the origin $(0,0,0)$ to point $B$: 
\begin{align}
x(\alpha) &= B \iff \\
(1,1,1+\alpha) &= (1,1,7) \iff \\
1+\alpha &= 7 \iff \\
\alpha = 7 - 1 &= 6
\end{align}
So we have 
$$
x(\alpha = 6) = (1,1,1+6) = (1,1,7) = B
$$.
For general values of the points $A=(a_i)$, $B=(b_i)$ and the direction vector $d = (d_i)$ one has
$$
x = A + \alpha d = B
$$
which in three dimensions gives three equations $(i \in \{1,2,3\})$:
$$
\alpha d_i = b_i - a_i
$$
If $d_i = 0$ then $a_i = b_i$ must hold, otherwise there is no solution.
For $d_i \ne 0$ then $\alpha = (b_i - a_i) / d_i$ and we must get the same $\alpha$ for all such indices $i$ where $d_i \ne 0$.
A: 
If you have two points, A and B, at (1,1,1) and (1,1,7) respectively,
  and a unit vector C (0,0,1). What's a way to find if unit vector C,
  will cross B if C extends forever. (Unit vector C has a starting point
  at A)

First make a line J with A+λC = (1,1,1) + λ(0,0,1)
Now if they intersect, then B will intersect with J.
To check that, let's rewrite the line in a easier form: J=(1,1,1+λ)
Now if they intersect, there will be a value of λ for which J=B(1,1,7). Can you figure out the value?
And your dot product method doesn't work because A and B are coordinates, not vectors. 'Dot product-ing' them would assume the line comes from the origin, when it should actually be in the direction of unit vector C.
(It is 6)
A: A simple way to look at this problem is to consider the
vector that goes from $A$ all the way to $B$ and stops at $B$.
We compute it this way:
$$v = B - A = (1,1,7) - (1,1,1) = (0,0,6).$$
The question to ask now is, is this a positive multiple of $C$?
If so, then starting at $A$, if we move according to the vector
$C$ (or its extension) we will eventually pass through $B$.
Now it should not be hard to see that $(0,0,6)$ is a positive
multiple of $(0,0,1)$, so the answer to the main question is yes.
