# 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.

• 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. – dpmcmlxxvi Jul 18 '15 at 4:32
• @dpmcmlxxvi; a point bisects a line – JMP Jul 18 '15 at 4:43
• @EmilianoSorbello: I'm not sure I quite understand what you mean. Can you further clarify (if possible)? – Grace Jul 18 '15 at 4:57
• i was answering a deleted comment so, mine probably makes no sense at all now – Dleep Jul 18 '15 at 4:58


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

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?

• I believe I understand the majority. Why would we divide the vector d by its own length? – Grace Jul 18 '15 at 18:17
• 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? – Rory Daulton Jul 18 '15 at 19:31
• Hi Mr. Daulton, yes, it makes sense now. If the length of d is $1$, would the vector d = $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}?$ – Grace Jul 18 '15 at 20:39
• @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. – Rory Daulton Jul 18 '15 at 21:08