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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

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    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

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    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.
     
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