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Vector problem - collision

  1. Dec 27, 2012 #1
    1. The problem statement, all variables and given/known data

    (a) Show that if three vectors a, b and c are linearly dependent then
    a[itex]\bullet[/itex](b x c) = 0


    (b)Two particles are on the trajectories: r = a + ut and r = b + vt. Show that the particles will collide if v[itex]\bullet[/itex](b x u) = v[itex]\bullet[/itex](a x u).


    (c) Express the time for the collision in terms of a, b, u and v.

    (d) Hence or otherwise, show that the collision takes place at position

    r = b + v [a [itex]\bullet[/itex] (b x u)/v [itex]\bullet[/itex] (b x u)]

    (e) What must the time of collision be if a, b, u and v are coplanar?

    2. Relevant equations



    3. The attempt at a solution

    (a) Shown.

    (b) shown.

    (c) shown.

    (d) shown.

    (e) this meant a [itex]\bullet[/itex] (b x u) = v [itex]\bullet[/itex] (b x u) = 0

    Then what i have is t = 0/0 which doesn't make sense..

    Not sure what difference does co-planar make to the question, as parts (a) to (d) never worked under the assumption that the vectors are co-planar. Does it impose some sort of restriction?
     
  2. jcsd
  3. Dec 27, 2012 #2

    mfb

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    I don't see the significance of those two equations.
    The time of collision stays the same if you add an arbitrary vector to both a and b (you just shift both trajectories), but you can remove the 0/0-result with that.

    The time of collison stays the same if you replace u and v by 0 and v-u (which gives all the relative velocity to one trajectory), but you always get 0/0 then.

    I would expect that there is some equation which avoids this case, but it has no "physical" significance, it is just a problem of the formula.
     
  4. Dec 27, 2012 #3

    haruspex

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    You did not post your answer to (c). I assume it involves triple products, producing infinity if the condition for collision is not met?
    I'm puzzled about (b). If the four vectors are coplanar, the equality condition will be met, but that does not guarantee they collide, does it?
    I think the point of question (e) is to find another way of calculating the time that does not break down when coplanarity makes the triple products vanish. mfb seems to be suggesting you can add an arbitrary vector to a and b to fix it; maybe right, but I'd be surprised. I need to see your answer for (c).
    Meanwhile, in the coplanar case, there is a very simple way of finding the collision time if it can be assumed that collision occurs, and a correspondingly simple test for collision.
     
  5. Dec 27, 2012 #4

    mfb

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    Actually, I am surprised to see a and b on their own anywhere. I would expect that only their difference, a-b, appears. Coplanarity of (a-b), u, v is then required (but not sufficient) for a collision, and the time of collision has to satisfy t(v-u)=(a-b). This equation is problematic for v=u and a=b only, and that case corresponds to identical trajectories.
     
  6. Dec 28, 2012 #5
    Yes that is what i got for the expression of time as well (for the last part). Part (c) can be found by simply taking the 'time' component to answer of part (d).

    So what must the time be in part (e) if they are co-planar?
     
  7. Dec 28, 2012 #6

    haruspex

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    If you assume collision occurs, you can just take the dot product of each side of that with u+v, allowing division by a scalar to isolate t. But it may be that u-v and a-b are not collinear, in which case there will be no collision.
     
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