1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Distance from a 3 space line

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the distance between the line x=3t-1, y=2-t, z=t and the point (2,0,-5)


    2. Relevant equations



    3. The attempt at a solution
    So I'm assuming I need to use the formula D=[ax+by+cz+d]/root(a^2+b^2+c^2) and I'm guessing I need to use (3,-1,1), extracted from the parametric line equation as the parallel vector, and possibly also the point (-1,2,0), extracted from the same. But not sure how to proceed??
     
  2. jcsd
  3. Mar 13, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Minimize the distance from (x,y,z) to (3,-1,1). The distance D just a function of t. Use calculus. Hint: you'll find it's nicer to minimize D^2.
     
  4. Mar 13, 2010 #3
    There is a way to do this using vector projection, does anyone remember it?
     
  5. Mar 13, 2010 #4
    I think you are on the right track. What I would do, instead of minimizing the distance, is to take advantage of the geometry.

    Draw a vector [tex]\vec{p}[/tex] from the origin to the point. Draw another vector from the origin to the parameterized vector function: [tex]\vec{r}(t) = (3t-1, 2-t, t)[/tex].

    Now, let [tex]\vec{d}[/tex] be the vector from the point to the place on the line at time t. In other words, [tex]\vec{p} + \vec{d} = \vec{r}(t)[/tex].

    It might help to draw a picture. Once you do, you will notice that the minimum distance will occur when [tex]\vec{d}[/tex] is perpindicular to [tex]\vec{r}(t)[/tex] and hence also orthogonal to [tex]\vec{r}'(t)[/tex].

    So all you need to do is use what you know about the dot product and orthogonality to solve for when d and r'(t) are orthogonal, and you can find the distance from there.
     
  6. Mar 13, 2010 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Call the point on the line P, the point off the line Q, and a direction vector for the line D. Draw the perpendicular from Q to the line hitting the line at S. Let

    [tex]\vec V = PQ,\ \hat D = \frac {\vec D}{|\vec D|}\hbox{ and }\theta\hbox{ be the angle between }\vec V \hbox{ and }\hat D[/tex].

    From the right triangle PQS you have formed you can see the distance from the point to the line is:

    [tex]d = |\vec V| \sin\theta[/tex]

    But this is the same as the magnitude of the cross product:

    [tex] d = |\vec V \times \hat D|[/tex]

    That is the most direct way to get the distance because it is easy to write down

    [tex]\vec V \hbox{ and }\hat D[/tex]

    and take the cross product and its length.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook