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Collision problem using vectors

  1. Dec 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Two particles are moving on trajectories given by r=a+ut and r=b+vt respectively where a, b, u and v are constant vectors and t is the time. Show that the particles will collide if v.(bxu)=v.(axu) Obtain an expression for the time of the collision in terms of a, b, u and v. (I think I've done the parts so far.) Hence, or otherwise, show that the collision will take place at position r=b+((a.(bxu))/(v.(bxu))v

    2. Relevant equations

    If three vectors a, b, c are linearly dependent, a.(bxc)=o

    3. The attempt at a solution

    a+ut=b+vt
    ut=(b-a)+vt
    ut, vt, and b-a are coplanar. Since t is a scalar multiple, u, v, and b-a are coplanar.
    v.((b-a)xu)=0
    v.(bxu-axu)=o
    v.(bxu)=v.(axu)

    ut-vt=b-a
    (u-v)t=b-a
    t=mod(b-a)/mod(u-v)

    I can't see how this expression for to is equivalent to ((a.(bxu))/(v.(bxu)), so I'm stuck.
     
  2. jcsd
  3. Dec 28, 2011 #2
    Any suggestions would be great. :)
     
  4. Dec 28, 2011 #3

    Dick

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    Ok, here's a sample of how to do it. I haven't worked through the permutations to show that the answer is what same as what the book gives but here's a start.

    (u-v)t=(b-a)

    cross both sides with a, so,

    (uxa-vxa)t=(bxa)

    Now we want to turn both sides into scalars so we can divide, let's dot with v.

    ((uxa).v-(vxa).v))t=(bxa).v

    (vxa).v=0 so,

    (uxa).v)t=(bxa).v, hence,

    t=((bxa).v)/((uxa).v). Now use cyclic permutations and the result from the first part to turn that into the books answer, if you can. You might. I'm not sure. I haven't chased it that far yet. Nor am I sure the book answer is even right. I haven't checked that either.
     
  5. Dec 28, 2011 #4

    ehild

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    Start to cross both sides with u instead:
    -(uxv)t=(uxb)-(uxa),

    then dot product with a :

    -a(uxv)t=a(uxb)

    as (uxa) is perpendicular to a. The triple products are scalars, so

    t=-a(uxb)/a(uxv)=a(bxu)/a(uxv).

    You need to change the denominator to v(bxu) by cyclical permutation of the vectors and using the condition v.(bxu)=v.(axu)

    ehild
     
  6. Dec 28, 2011 #5

    Dick

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    Yeah, I figured there was some permutation of those things that would work. But I was kind of hoping Lucy Yeats would figure it out instead.
     
  7. Dec 29, 2011 #6

    SammyS

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    Instead of dotting with v, dot with u.

    I.E. look at:
    [itex]((\vec{u}\times\vec{a})\cdot\vec{u}-(\vec{x}\times\vec{a})\cdot\vec{u})t=(\vec{b} \times\vec{a})\cdot\vec{u}[/itex]​
    Of course, [itex](\vec{u}\times\vec{a})\cdot\vec{u}=0[/itex]

    Added in Edit:

    I must have been asleep (as in not paying attention) when ehild made this same suggestion!
    Sorry about that !
     
    Last edited: Dec 29, 2011
  8. Dec 29, 2011 #7

    ehild

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    Sorry Dick. But you directed the OP to a more difficult way than necessary, and I did not say more than you.

    ehild
     
    Last edited: Dec 29, 2011
  9. Dec 29, 2011 #8

    Dick

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    Just trying to explain why I was being too lazy to work out the exact solution.
     
  10. Dec 29, 2011 #9

    SammyS

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    BTW:

    The condition v.(bxu)=v.(axu), can be obtained in a similar manner. Start with
    [itex]\vec{a}+\vec{u}t=\vec{b}+\vec{v}t[/itex]​
    Cross that with vector [itex]\vec{u}[/itex] then dot that with vector [itex]\vec{v}\,.[/itex]
     
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