Distance involving 3D lines and vectors. In this problem, a = \begin{pmatrix} 5 \\ -3 \\ -4 \end{pmatrix} and b = \begin{pmatrix} -11 \\ 1 \\ 28 \end{pmatrix}
Vectors p and d exist such that the line containing a and b can be expressed in form
v = p + d$t$.
Additionally, for a specific value of d, it is the case that for all points v lying on the same side of a that b lies on, the distance between v and a is $t$. What is the value of d?
Another problem I have no clue on as to how to solve, let alone begin. What should my "first step" be? Clarification on the last paragraph of the problem and hints are appreciated.
 A: $
\renewcommand{\v}{\mathbf{v}}
\newcommand{\vo}{\mathbf{v_0}}
\renewcommand{\p}{\mathbf{p}}
\renewcommand{\d}{\mathbf{d}}
\renewcommand{\a}{\mathbf{a}}
\renewcommand{\b}{\mathbf{b}}
$
First, WLOG assume $\p = \a$, so that $\v = \p + \d t = \a + \d t$. 
Then  the distance between $\v $ and $\a$ is 
$$
\|\v - \a\| = \|\p + \d t - \a\|  = \|\a + \d t - \a\| = \|\d \| t
$$
Since we want to make sure that $\|\v - \a\| =  t$, we need to choose $\d$ such that 1) $\p$ is collinear with $\b-\a$, and 2) $\|\p\| = 1$. 
The most obvious choice is $\d = \dfrac{\b - \a}{\|\b - \a\|}$. 
Then we have 
$$
\|\v - \a\| = \|\a + \d t - \a\|  = \| \d \| t = 
\left\| \dfrac{\b - \a}{\|\b - \a\|}\right\| t = 
\dfrac{\left\| \b - \a\right\|}{\|\b - \a\|} t = t
$$
Thus, for a vector 
$$\v = \p + \d t = \a + \dfrac{\b - \a}{\|\b - \a\|} t$$
we have 
$$
\|\v - \a \| = t
$$
A: That sentence should read something like this:

Additionally, for a specific value of d, it is the case that for all points v lying on the ray from a to b, the distance between v and a is $t$.

If you want something more long-winded and avoiding the concept ray:

Additionally, for a specific value of d, it is the case that for all points v lying on the line containing a and b where a is not between b and v, the distance between v and a is $t$.

This can be done first by finding letting p be point a and d be b-a. Find the coordinates of d. Then divide that vector by its own length, giving you a new d that has length $1$. That will then be the d that you want.
Is that clear?
