Distance involving 3D lines and vectors.

Question

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.

Answer 1

$ \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 $$ \|\v - \a \| = t $$

Answer 2

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?