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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.

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  • $\begingroup$ How are you defining points "on the same side of a"? Consider drawing a figure because it's hard to understand your description of the problem. $\endgroup$ – dpmcmlxxvi Jul 18 '15 at 4:32
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    $\begingroup$ @dpmcmlxxvi; a point bisects a line $\endgroup$ – JMP Jul 18 '15 at 4:43
  • $\begingroup$ @EmilianoSorbello: I'm not sure I quite understand what you mean. Can you further clarify (if possible)? $\endgroup$ – Grace Jul 18 '15 at 4:57
  • $\begingroup$ i was answering a deleted comment so, mine probably makes no sense at all now $\endgroup$ – Dleep Jul 18 '15 at 4:58
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$ \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 $$ \|\v - \a \| = t $$

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  • $\begingroup$ $\|(x,y,z)\|=\sqrt{(x^2+y^2+z^2)}$? $\endgroup$ – JMP Jul 18 '15 at 5:48
  • $\begingroup$ @JonMarkPerry Yes, so $\|b-a\| = \sqrt{(-11 -5)^2 + (1 + 3)^2 + (28 + 4)^2} = \sqrt{1296} = 36$ $\endgroup$ – Vlad Jul 18 '15 at 5:53
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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?

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  • $\begingroup$ I believe I understand the majority. Why would we divide the vector d by its own length? $\endgroup$ – Grace Jul 18 '15 at 18:17
  • $\begingroup$ Without that you do not get "the distance between v and a is $t$." That happens only if the length of d is $1$, and you make that happen by dividing d by its length. @Vlad's answer shows that in detail. I emphasized giving detail on "clarification on the last paragraph of the problem and hints" while Vlad gave details on why those hints are valid. Both answers are good: I answered the question you actually asked. Is the detail on d now clear? $\endgroup$ – Rory Daulton Jul 18 '15 at 19:31
  • $\begingroup$ Hi Mr. Daulton, yes, it makes sense now. If the length of d is $1$, would the vector d = $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}?$ $\endgroup$ – Grace Jul 18 '15 at 20:39
  • $\begingroup$ @Grace: No. That vector has length $1$ but so do many other vectors. In your example, the first d=b-a is $(-16,4,32)$, which has length $\sqrt{(-16)^2+4^2+32^2}=36$, so the final d is $\left(-\frac 49,\frac 19,\frac 89\right)$. That final d has length $1$ so it works. $\endgroup$ – Rory Daulton Jul 18 '15 at 21:08

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