Minimum distance between skew vectors

  • Thread starter peripatein
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  • #1
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Hi,

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?
 

Answers and Replies

  • #2
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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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