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Homework Help: Minimum distance between skew vectors

  1. Oct 31, 2012 #1

    Satellites A and B are traced at a particular moment by a space base located at the origin, with respect to which they move at constant velocities:
    A0 = (a1,0,a3); B0 = (0,b2,b3)
    V of satellite A = (Va,0,0); V of satellite B = (Vb,Vb,0)

    Is the minimum distance between the satellites equal to the distance between the two skew vectors A = (a1,0,a3) + (Va,0,0)t and B = (0,b2,b3) + (Vb,Vb,0)t? Are they indeed skew?

    If so, the distance is a3-b3, which does not seem reasonable to me (as over simpified and does not involve too many of the problem's parameters).

    Any advice, please?
  2. jcsd
  3. Nov 3, 2012 #2
    I drew a quick picture which helped me. The answer to the first question is I don't know. The answer to the second question I think is yes (they certainly aren't parallel).

    Does minimum here mean the straightest path or some point in time? Who knows. However, you're coming up with a constant number which can't be right for either definition.

    a3-b3 (maybe take the absolute value because a3-b3 could be negative) is the normal distance between the two lines the satellites travel on. Both satellites are traveling in planes parallel to the xy plane because both have a velocity of 0 in the z direction. However, this doesn't mean the satellites were ever that close apart because Va doesn't equal Vb. a3-b3 is the minimum distance between the paths the satellites travel on.

    Here is what I think is the best answer to this question which is the distance formula at a moment in time: [((a1+Va*t)-(0+Vb*t))^2+((0+0*t)-(b2+Vb*t))^2+((a3+0*t)-(b3+0*t))^2]^.5
    Take the derivative to find the point in time when they are the closet.

    Does that help some?
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